Easy Analog Circuit

High frequency model of transistor

2015-12-13T09:14:00.001-08:00

We discussed small signal analysis of BJT under the key word "Small signal analysis". When input signal to an amplifier is in the range of ten to hundred Kilo Hertz, a small signal-low frequency model of the transistor can be used for analysis. But as the frequency increases, internal capacitance of the transistor will strongly effect it's performance. A low frequency model cannot work well in this situation. To accommodate these performance changes of the transistor, a separate model is developed for high frequency operations. This high frequency model is given in Fig 1. A high frequency hybrid-pi model is also known as Giacoletto model.

Fig 1 High frequency model

Fig 2 Low frequency model

Low frequency model is given in Fig 2 for quick reference. It is given in the post 'Transistor - small signal model and small signal parameters'. Input resistance (r_π), Transconductance(g_m) and Output resistance (r_o) are well explained there. Plz go through it, just to refresh your brain cells.

High frequency model- explanation

In high frequency model a resistor and two capacitors are added in addition to the components in low frequency model.

1) Resistor rx is known as Base-spreading resistor

In Fig1, B' is a point internal to the transistor and is a part of base region. rx denotes the resistance of the silicon material between external base terminal B and internal base terminal B'. It's value is usually less than 100 ohms and has significant effect in high frequency response. It does not have any roll in low frequency analysis. rx << r_π .

2) Base - Emitter capacitance C_π

This capacitance occurs due to the combined effect of emitter junction diffusion capacitance Cde and emitter junction depletion capacitance Cje. (Diffusion and depletion capacitance are in your 'Solid State Devices' text book.) It's value is in between few pF to few tens of pF.

3) Collector - Base capacitance Cμ
It is the capacitance of the Collector- Base junction of the transistor. It is ranging from fraction of pF to a few pF.

Fig 3 Internal capacitance of BJT

####Note: In Fig 3, E-B junction is forward biased. Both diffusion and depletion(space charge) capacitance are associated with a forward biased junction. So Cde and Cje forms C_π. C-B junction is reverse biased and depletion capacitance appear at this junction and is represented as Cμ. ####

For Input resistance (r_π), Transconductance(g_m) and Output resistance (r_o) plz use this link.

Integrator & Differentiator-B Tech university Question & answer

2015-09-11T23:33:00.000-07:00

Q1. Explain the working of RC low pass filter, how it can be act as an integrator.

Answer key: circuit(2), working+fre response(4), act as integrator(5), output(1) (12 marks)

To know, how to prepare answer key, visit link.
Answer:
Consider a first order RC circuit. It can be act as a low pass filter and allows to pass only the low frequency signals.

Circuit diagram:

fig 1

where,
AC – AC source that provides the input(Vin)
Output – Output voltage
Resistors(R) and Capacitors(C) are connected in series with Vin.

Working of a low pass filter:

The reactance of capacitor is Xc =

where, f – frequency of input AC c - capacitance of the capacitor C

Equation reveals that Xc is inversely proportional to frequency f. So,

At low frequency input signal, Xc become very large. So Xc >> R. Therefore voltage drop across the capacitor- which is the output, is very large. In a similar way, at high frequency, Xc << R. So output is low.

Frequency response of first order Low pass filter

Summarising :
For low frequency AC / for DC -- Vc >> VR -- larger output voltage

For high frequency AC -- Vc << VR -- smaller output voltage

This circuit produces a high output voltage for low frequency inputs and it attenuate the output voltage for high frequency input signals. That why it has the name ‘low pass filter’.

Integrator

An integrator produces an output which is proportional to integral of its input.

Low pass RC circuit itself is acting as an integrator provided it obey one condition. Lets check that condition.
Equation for output voltage in fig1 is
                                       1/CS –reactance offered by C in terms of Laplace transform

Therefore,

Laplace transform s=jw and w=2πf (w-omega)

Low pass circuit act as Integrator:

In the above equation,if we are adopting a condition as wRC >> 1, then , in denominator 1+jwRC is nearly equal to jwRC. Then equation become



Re arranging,

                  Here 1/RC is taken as a constant

Taking inverse Laplace transform,



Above equation shows that Vout is proportional to integral of input voltage. So if we have a low pass RC circuit which obeys the condition wRC >> 1, it will act as integrator.
In order to achieve a good integration, the following conditions must be satisfied:

The time constant RC of the circuit should be very large as compared to the time period of the input signal.
The value of R should be 10 or more times larger than Xc.

Output of integrator for square wave input is a triangular wave.

Q2. Explain the working of RC high pass filter, how it can be act as an differentiator.

Answer key: circuit(2), working+fre response(4), act as integrator(5), output(1) (12 marks)

To know, how to prepare answer key, visit link.
Answer:
Consider a first order RC circuit. It can be act as a high pass filter and allows to pass only the high frequency signals.

Circuit diagram:

Input is an AC source, Output is taken across resistor. Contains one pair of R & C.

where,
AC – AC source that provides the input(Vin)
Output – Output voltage
Resistors(R) and Capacitors(C) are connected in series with Vin.

Working of high pass filter:

Voltage across R is taken as output voltage. Let’s see what happened to this voltage at low and high frequencies.

For high frequency input:

Xc is very low(Xc inversely proportional to frequency). So Vc is also low. So Vc << V_R,where V_R is the output voltage. In conclusion, for high frequency input, output is higher.

For low frequency input:
Xc is very high.Vc is also high. Therefore for lower frequencies V_R<< Vc. So output voltage V_R is much less for low frequency input.

Frequency response of High pass filter

Summarising :
for high frequency input signal ----- output voltage is larger
for low frequency input signal ----- output voltage is lesser
Then we can call it as a “high pass”.

High-pass RC circuit as Differentiator:

In differentiator output voltage, Vout is proportional to the rate of change of input.

When the high pass circuit obey some condition, it act as a differentiator.

Lets write the equation for output voltage.

Laplace transform s=jw (where w=2πf) and RC is the time constant.Substitute for 's'.

Rearranging,we get

Take the condition that RCw << 1, then

We know jw = S, Laplace transform. So therefore, Vout = RC S Vin .

The above equation is a Laplace equation and we take inverse Laplace, then equation become,

Equation shows that output voltage is proportional to derivative of input voltage. So the circuit represents a differentiator circuit. Output of the differentiator circuit for different inputs is given below.

Small Signal - Low frequency - Analysis of common collector amplifier-continue

2015-02-20T04:01:00.000-08:00

Analysis of amplifier circuit includes derivation of equation for 1)input resistance 2)output resistance 3)voltage gain and 4)current gain. Derivation and analysis of input resistance is given in - 'Analysis of Common Collector amplifier (Emitter Follower) -begins here' . Remaining parameters will be discussed here.

Before beginning, recollect the common collector amplifier circuit and it's small signal model. It was discussed in detail in the -'begins here' post. For your convenience once again giving the circuits.
Here is the amplifier circuit:

fig 1

This is the small signal equivalent of amplifier circuit:

fig 2

Next step is to find out second parameter, i.e output resistance.

b)Output resistance (R_o)

Output resistance of the circuit can be calculated by 'looking into the output side of the circuit' from the mark as given in fig 3. We can see that R_B, R_S , r_e, and r_o resistances are contributed for R_o. Among these R_B and R_S are in input side and their impact on output resistance is marked as R_ob, in fig 3.

fig 3

Looking towards R_B and R_S from arrow mark shows that they are parallel to each other( R_B ‖ R_S ). It's presence in base terminal affects the emitter side resistance and is given as R_ob= ( R_B ‖ R_S ) / (1+ β).

(To know how to write this equation, Please see 'Analysis of Common Collector amplifier (Emitter Follower) -begins here'.

Now we can complete the derivation of R_o. Look into the circuit through the arrow marked along with R_o. We can see that r_ois parallel with (r_e + R_ob ). ## r_e and R_ob are in series.

Then equation for R_o = r_oll (r_e + R_ob ) ------(1) substitute for R_ob.

R_o = r_oll { r_e + [( R_Bll R_S ) / (1+ β)] } -------(2) Equation for output resistance.

Note: Output resistance of Common collector amplifier(Emitter follower) is low.

Reason behind it: r_ois much larger than{r_e + [( R_Bll R_S ) / (1+ β)]}. So result of parallel combination of these two terms is less than the smaller quantity. So this result is nearly equal to the smaller quantity i.e, { r_e + [( R_Bll R_S ) / (1+ β)] }.

R_o=̃ r_e+ [( R_Bll R_S ) / (1+ β)] -------(3)
Next parameter we have to find is voltage gain.

c): Voltage gain (Av)

Voltage gain for an amplifier is output voltage/input voltage. For the circuit in fig 2,
Voltage gain, Av = Vo/Vs
We cannot directly write an equation for voltage gain using fig2. Therefore consider it's Thevenin's equivalent and thus make it simpler. Two steps are there in this simplification:
First, consider the input side only which includes Vs, Rs and R_B (fig 4a) and find it's Thevenin's equivalent. Then the circuit in fig 4a can be replaced with it's Thevenin's equivalent.

fig 4

In order to calculate R_TH,voltage source Vs is short circuited(fig 4b). R_TH = Rs ll R_B. As in fig4c, V_TH is the voltage across R_B. V_TH = R_BVs / (Rs + R_B).

The second step is to take all resistances in fig 2 to one side- either to input side or to output side. Here resistances in the output(emitter) side are take to base side, using the shortcut method described in 'Analysis of Common Collector amplifier (Emitter Follower) -begins here'. The effect of Emitter side resistances on base side is obtained by multiplying emitter side resistances with (1+ β).

Now apply the result of above two steps in fig 2.

fig 5

Here in fig 5, input side(as in fig 4a) is replaced with V_TH and R_TH. Also output resistances are transformed in a way so that now we can consider the circuit as a resistance network(no more worry about amplification factor between base and emitter terminals). Rearranging circuit in fig 5 give fig6.

fig 6

Now write equation for voltage gain. VoVTH=(1+β)(rollRL)(RsllRB)+(1+β)re+(1+β)(rollRL)--

-------(4)

Substitute for V_TH. V_TH = R_BVs / (Rs + R_B).
VoVs=RB(Rs+RB)(1+β)(rollRL)(RsllRB)+(1+β)re+(1+β)(rollRL) ------(5)

Observing equation(5) for voltage gain, we can see that the denominator has a greater value than denominator. Therefore the voltage gain is less than unity. But under certain conditions it becomes very close to unity. The conditions are 1)Rs << R_B and 2) (R s ll RB) << (1+ β) re + (1+ β) (ro ll R L ). Voltage gain is nearly equal to unity means - voltage at the output side(Emitter) follows very closely the voltage at the input side. Thus the name Emitter follower.

d): Current gain (A _is)

Short circuit current gain = ie/ib = (1+ β) . Current gain is high.

Voltage buffer

Because of high input resistance and low output resistance, Emitter follower act as a voltage buffer. It is used for connecting a high resistance source to a low resistance load.

This analysis is in the text book: Micro Electronics by Sedra & Smith. I am garnishing it so that you can Easily flip into.

Small Signal-Low frequency- Analysis of Common Collector amplifier (Emitter Follower) -begins here

2015-02-10T02:41:00.000-08:00

We are very familiar with common collector circuits- otherwise known as Emitter follower. Collector terminal is common to both input side and output side. During analysis, we are finding out the nature(characteristics) of the circuit by observing it's input and output resistances, voltage gain and current gain. Here discussing the analysis of CC amplifier using low frequency - small amplitude signals.

Before beginning, please go through the analysis of CE and CB amplifiers. Better have a look into CE analysis . In all these analysis, we use similar steps -only parameters will vary. Once you understand CE analysis, CB and CC analysis will become easy.

Let's begin with Circuit ....
1. Consider the CC amplifier circuit.

Here we are starting with a simple - common collector amplifier circuit. Any way there may be some unfamiliarity but will clear as we go on. Since collector to be 'common', it is tied into a power supply and now it's available to the entire circuit. Base is taken as input terminal - an input source Vs is attached to it through Rs and C1. Output is taken across R_L, connected to the emitter of the transistor. A constant current biasing is provided at the output terminal-it consists of a constant current source(I) and -V_{EE power supply.}

fig 1

So the first part is over - we got the circuit. Second part is to draw the equivalent circuit. Here it is recommended to use 'T-model'(this will easily lead us to the answer).

2. Obtain small signal equivalent model (Step by step explanation was given in 'Small signal model of a circuit' ).
Steps to obtain small signal model of the circuit in fig1 are as follows:
     Steps 1: D C voltage sources are replaced by ground
   2: D C current sources are open circuited
   3: Capacitors are shorted

fig 2

4,5: Rearrange the components to have a common ground and insert transistor model. Step by step explanation is in transistor model .

fig 3

3. Find out the parameters
In order to analyze behavior of a transistor we depend mainly on four parameters:
a): Input resistance (R_in)

In the input side there is a resistance R_S,but R_Sis taken as source resistance and usually not consider for input resistance calculation. Input resistance of the circuit can be calculated by 'looking into the input side of the circuit' from the mark as given in fig 3. We can see that R_B, r_e, R_L and r_o resistances are contributed for R_in. If rearrange the circuit in fig 3 as in fig 4, calculation of R_inbecome easy.

fig 4

Consider only that part of the circuit which is relevant in R_in calculation.

fig 5

Equivalent resistance offered by r_e, R_Land r_o is marked as R_iB.
In effect,    R_in= R_B ll R_iB    ---------(1)
Then we have to find an equation for R_iBand substitute it.
From figure 5, R_iB= V_B/ I_b . In terms of I_e,
                        R_iB= V_B(1+ β) / I_e .---------(2)     #recollect equation connecting I_b and I_e.
An equation can be used to substitute for V_B / I_e using resistances as,
I_e= V_B / [(r_oll R_L) + r_e] .-----------(3)
Substitute for V_B / I_e in equation (2).
                         R_iB= (1+ β) [(r_{o ll} R_L) + r_e] .-----------(4)

Note:{{ Here output side (Emitter side) resistances r_e, R_L and r_o have impact on input(base) side resistance R_in.
There is a Shortcut for finding this:
Impact of output side resistances on input resistance= (1+ β) * equivalent resistance in output side. It's clear from fig4 /fig5 that, equivalent resistance in output side = [(r_oll R_L) + r_e]. Now have a look on equ(4). Using short cut method, after equ(1), we can directly find equation for R_iBand thus will easily reach at equ(5).
The term (1+ β) arises due to the relation, I_b = I_e /(1+ β) . Conclusion :

"Reflection of emitter side resistance into base side = emitter side resistance * (1+ β) ". Similarly,

"Reflection of base side resistance into emitter side = base side resistance / (1+ β) ". }}

As per equ (1), input resistance R_in= R_B ll { (1+ β) [(r_oll R_L) + r_e] } ---------(5)

CC amplifier has a high input resistance. It happened because of the presence of (1+ β) in the term { (1+ β) [(r_oll R_L) + r_e] }. 'When two or more resistances are connected in parallel, the resultant resistance will always be less than the lowest among them' - as per this, R_in will be high only if R_Bis also high.

This analysis is in the text book: Micro Electronics by Sedra & Smith. I am garnishing it so that you can Easily flip into.

Remaining parameters will be in the next post as this become too lengthy.

b): Output resistance (R_o)
c): Voltage gain (Av)
d): Current gain (A _is)

Small Signal - Low frequency - Analysis of Common Base amplifier using hybrid model

2015-01-15T01:03:00.002-08:00

Small Signal - Low frequency - Analysis of Common Base amplifier.. Yes, the characteristics of a CB amplifier is analyzing using low frequency -small amplitude signal. Analysis of CE amplifier is in another post. Better to go through it before diving into this.

As discussed in previous posts, analysis requires three steps - 1)Taking the circuit which is to be analyzed 2)obtaining the small signal model and 3)finding out equations for parameters helps to get approximate values and it in turn will help to predict the characteristics of the circuit.

1. Consider the CB amplifier circuit.

The figure below shows a Common base amplifier. A signal ground is attached to the base terminal. Emitter is turn into input terminal by attaching an input source Vs through Rs and C1. Output is taken across R_L, connected to the collector of the transistor. DC voltage Vcc provides biasing at the output terminal and a constant current biasing is provided at the input terminal.

fig 1

So the first part is over, we got the circuit. Second part is to draw the equivalent circuit. As in earlier post we can take a hybrid π model. But here it is recommended to use 'T-model'.
2. Obtain small signal equivalent model (Please see 'Small signal model of a circuit' )
     Steps 1: D C voltage sources are replaced by ground
   2: D C current sources are open circuited
   3: Capacitors are shorted

fig 2

4,5: Rearrange the components to have a common ground and insert transistor model

fig 3

3. Find out the parameters
For analyzing behavior of a transistor we depends mainly on four parameters:
a): Input resistance (R_in)

Input resistance of the circuit can be calculated by 'looking into the input side of the circuit' from the mark as given in fig 3. This is because R_Sis taken as a source resistance and usually not consider for input resistance calculation.

From the above circuit,    R_in= r_e    ---------(1)
CB amplifier has a low input resistance because the value of r_e is usually a few ohms to few tens of ohms.
b): Output resistance (R_o)
Again to get the output resistance, look into the circuit from right side of arrow.
Then from the circuit, R_o= Rc    ----------(2)
c): Voltage gain (Av)
Voltage gain (Av) is defined as

Av = Vo / Vi               where,
                                Vo = output voltage    and
                                Vi = input voltage appear across the input terminal(here Base) of the amplifier.

Let's find out expression for Vi :

Vi = -i_er_{e -----------(3)}

Output voltage(Vo) is the voltage across the parallel combination of Rc and R_L.
                               Vo = -αi_e(Rc ‖ R_L)     ------------(4)
Now we can write an expression for Av.
                               Av = Vo / Vi

                               -----------(5)
For T-model, α/re = g_m_. Using this equ(5) can be written as

                               Av = g_m(Rc ‖ R_L)     ------------(6)
Here, Av has a positive sign.It means the output and input waveforms have the same polarity. So CB is a non inverting amplifier.

Another parameter related to voltage gain is overall voltage gain. The overall voltage gain for the circuit in fig 3 is given as Gv = Vo / Vs . We can easily get it as

Gv = Vo / Vs = (Vo / Vi ) x (Vi / Vs) .
                      Here Vo/Vi is in our hand. Now find an equ for Vi/Vs

        --------------(7)

Overall voltage gain is Gv is obtained by multiplying equ (5) and (6).

                  --------------(8)

d): Current gain (A _is)
Here short circuit current gain is considered. That is the current gain when R_L is shorted. From fig 9, when R_L is shorted, output current is

i_os= - α i_e .

The input current is marked in fig 3 as i_i = - i_e.
Short circuit current gain (A _is) = i_os / i_i .

A _is = α ------(9)

Current Buffer

Since α ≈ 1, we can conclude that, short circuit current gain of CB amplifier is nearly unity. Because of this property it can act as a current buffer. It means current at the output terminal is same as that at the input terminal.

This analysis is given in the text book: Micro Electronics by Sedra & Smith. It is presented here with detailed description.

Small Signal - Low frequency - Analysis of CE amplifier using hybrid π model

2014-12-23T23:49:00.001-08:00

Here we are stepping to a zone mostly liked by questioners for essay - Low frequency analysis of CE amplifier . Small signal analysis of common emitter amplifier can be done using hybrid π model. Here low frequency analysis is considered.

In order to do the analysis of certain circuit, first step is to draw the model of that circuit. This model will help us to easily find equations for certain parameters of the circuit. Usually, input and output resistances, voltage gain and current gain are these four parameters. The equations will give an idea about the approximate value of these parameters and this in turn will help us to comment about characteristics of the circuit.

Let's start...

1. Consider the CE amplifier circuit.

Circuit of a common emitter amplifier is given below. It consists of two bias dc voltages - Vcc and

V_E
E. An input voltage V_S is applied to the transistor through a resistor R_S. C₁,C₂are coupling capacitors and C_E is bypass capacitor.

fig 1

2. Obtain small signal equivalent model (Please see 'Small signal model of a circuit' )
     Steps 1: D C voltage sources are replaced by ground
   2: D C current sources are open circuited
   3: Capacitors are shorted

fig 2

4: Rearrange the components such as to have a common ground

fig 3

5: Simplify the circuit and insert transistor model

fig 4

3. Find out the parameters
For analyzing behavior of a transistor we depends mainly on four parameters:
a): Input resistance (R_in)

Input resistance of the circuit can be calculated by 'looking into the input side of the circuit' from the mark as given in fig 5. This is because R_Sis taken as a source resistance and usually not consider for input resistance calculation.

fig 5

Part of the circuit considering for R_in calculation is

fig 6

From the above circuit, R_in= R_b‖ r_π ---------(1)
Normally, we select R_b such that R_b >>r_π. Then R_inbecomes

                                           R_in≈ r_π     ----------- (2)
We can conclude that input resistance of CE amplifier is nearly equal to r_π and its value is typically few kilo ohms. So CE amplifier has a moderate value of input resistance.
Note: In fig 6, it is clear that    Vi = V_π. (Voltage across two parallel paths)
       b): Output resistance (R_o)
Again to get the output resistance, look into the circuit from right side of arrow.

fig 7

Then from the circuit, R_o= Rc ‖ r_O    ----------(3)
Typically, Rc << r_O . Therefore, R_o≈ Rc   ----------(4)
       c): Voltage gain (Av)
Voltage gain (Av) is defined as

Av = Vo / Vi               where,
                                Vo = output voltage    and
                                 Vi = input voltage appear across the input terminal(here Base) of the amplifier.

Let's find out expression for Vi :
In fig 6, we get R_in= R_b‖ r_π . Using this the input side of fig 5 can be redrawn as

fig 8

As in fig 8, Vi is the voltage across Rin. So,

                  Rin
Vi = Vs -----------            --------(5)
                Rin + Rs
Substitute for Rin using equ (2) and take Vi = V_π then

--------(6)

Keep this equ in our account and try to get an equ for Vo. Consider the output side of the circuit given in fig 4 or fig5. It can be redrawn as below.

fig 9

In fig 9,resistances r_o,Rc and R_L are parallel and -g_mV_π is the current flowing through this parallel combination. Using these facts we can write the equ for output voltage as

Vo = - g_mV_π ( r_o ‖ Rc ‖R_L ) ----------(7) but we have Vi = V_π

Therefore, Vo / Vi = - g_m ( r_o ‖ Rc ‖R_L ) which is the voltage gain Av = Vo / Vi .
So, Av = Vo / Vi = - g_m ( r_o ‖ Rc ‖R_L ) -------(8)

Another parameter related to this is overall voltage gain. The overall voltage gain for the circuit in fig 4 is given as Gv = Vo / Vs . We can easily get it as

Gv = Vo / Vs = (Vo / Vi ) x (Vi / Vs) . Here Vo/Vi is in our hand. We already have an equ for Vi/Vs (refer equ 5 & 6).Then

Gv = - g_m ( r_o ‖ Rc ‖R_L ) r_π

   -----------            --------(9)
                                                    r_π + Rs
Small signal parameter can be written as r_π = β / g_m   where β is the current gain.

Above equation becomes

   - β ( r_o ‖ Rc ‖R_L )
Gv =   ---------------------           --------(10)
                       r_π + Rs
As per equation, voltage gain for CE amplifier is high.
       d): Current gain (A _is)
For the CE amplifier circuit, short circuit current gain is considered. That is the current gain when is R_L shorted. From fig 9, when R_L is shorted, output current is

i_os= - g_mV_π .

We can write an equation for input current as i_i = Vi / R_in= V_π / R_in.
Short circuit current gain (A _is) = i_os / i_i .

A _is = - g_m R_in ------(11)

From the above equ, we can realize that CE amplifier has a large current gain.

This analysis is given in the text book: Micro Electronics by Sedra & Smith. It is presented here with fine descriptions of each step.

Small signal model of a circuit

2014-12-11T02:20:00.002-08:00

'Small signal analysis of CE, CB, CC configurations using hybrid π model' is a major topic in 'Micro electronics syllabus'. To do the analysis of a circuit, first we have to draw a model for that particular circuit(CE,CB or CC). How can we obtain a small signal model for a circuit?- It is explained below.

Steps to obtain a model from corresponding circuit :

Here we are interested in small signal analysis and so taking only small ac signals into consideration. If you need little more idea, please visit 'Transistor analysis - Small signal analysis and Large signal analysis'.

I will explain the steps by taking a CE amplifier circuit as example. (Once you get it for a particular circuit, then it will be Easy..Only thing to do is just follow the steps- one by one).Ok..Turn your head to first year class and catch - - - your favorite - - - RC Coupled Amplifier ( it is an example of CE circuit).

fig 1 R C Coupled amplifier

Please Don't mind red line around bias section. .

Step 1: D C voltage sources are replaced by ground(Because of small signal analysis, avoid D C sources)

fig 2

Step 2: D C current sources are open circuited
NO current sources here.

Step 3: Capacitors are shorted

fig 3

Step 4: Inductors are open circuited
NO Inductors here.

Step 5: Check for resistors which are parallel to short circuits (here R_E) and remove.

fig 4

Step 6: Rearrange the components such as to have a common ground

fig 5

Step 7: Simplify the circuit and insert transistor model

fig 6

Now fig 6 represents small signal model of fig 1.

Generally analysis of a given circuit is done by finding values(or expressions) of certain parameters such as:
1. Input Impedance
2. Output Impedance
3. Voltage gain    and
4. Current gain.

            These parameters give an idea about the behavior of the circuit.

This topic is given in the text book: Micro Electronics by Sedra & Smith. It is presented here with detailed description.

Transistor - small signal model and small signal parameters

2014-12-08T02:14:00.000-08:00

Analog Circuits /Electronics Circuits syllabus is packed with models.

( I can see twinkling in your eyes, but… plz ~~don’t~~ disappoint. Just scan through below lines.)

The model

To study the behavior of a transistor for different input conditions, we make use of different transistor models. These models are designed by taking various parameters of transistor into consideration. Hence having the same behavior as that of the corresponding transistor. To test the behavior of a transistor for a new input condition, instead of conducting a practical experiment, these inputs are applied on the corresponding transistor model and outputs are measured. Thus 'the model' helps to predict the nature of a transistor(or circuit) for various input or bias conditions without actually wiring it.
Here we are thinking about 'Small signal model of a transistor' only. Of course there is 'Large signal model' also.

Small signal model of a transistor

Small signal model is related to 'Small signal analysis'. Don't confused, it is here, of course in a chewable way...small signal analysis.
To conduct a small signal analysis, a small signal model of a transistor is used. The parameters used along with this small signal model are known as small signal parameters.

One of the widely using small signal model of a transistor is Hybrid - pi model.

There are two versions for this model as shown below.

fig 1

Fig 1 shows a voltage controlled current source version.

fig 2

Fig 2 is current controlled current source version.

Lets familiar with the parameters used in small signal model.

1. Transconductance(g_m)

As we know conductance is the ratio of current to voltage. For BJT, transconductance is find by taking ratio of output current to input voltage.

Transconductance g_m= i_c / v_be

One more equation for g_min terms of dc bias current Ic and V_T where V_T is the voltage equivalent of temperature.g_m= Ic / V_T

2. Input resistance (r_π)

Small signal input resistance between base and emitter is called as input resistance and denoted by r_π.
         r_π = v_be/i_b    It can also be expressed in terms of g_m as
       r_π = β / g_m   where β is the current gain.
We can find out more equations for r_π by substituting for g_m.
   r_π = βV_T / Ic   = V_T / I_B

3.Output resistance (r_o)

It is the output resistance r_o= V_CE/ Ic

T- model of transistor

Another small signal model of a transistor is T -model. Both pi and T models can be used alternatively. But choose the one which will make the circuit less complicate.

fig 3

T- model with output resistance is below.

fig 4

These are the different small signal models of npn transistor.

Transistor analysis - Small signal analysis and Large signal analysis

2014-10-03T22:53:00.000-07:00

We can divide analysis of diode and transistor into Small signal analysis and Large signal analysis.

Large signal analysis

While doing a large signal analysis, a large dc voltage is applied and find out the behavior of diode or transistor towards this signal(find an operating point of these components). The voltage applied are of the order of few volts to few tens of volts. Here we use current and voltage equations such as

to find bias points. This analysis is applicable to any value of current and voltage applied to the devices.

Small signal analysis

Small signal analysis approximate nonlinear behavior of electronic components with linear equations. This approximation is taken across the bias point.

While discussing about the operating point of a BJT amplifier, it is said that preferable position for the operating point is at the center of the load line. Suppose after fixing the operating point(by applying dc bias voltages only), we are applying the input-AC signal. Then the operating point will swing around the initial operating point according to the value of A C signal. During this if it enter into saturation or cut-off region the behavior of the amplifier become nonlinear. So while choosing the input A C signal for amplifiers care must be taken to select signals which are within the specified maximum value. Because of this reason input A C signals for amplifiers are usually small and this fact encourages small signal analysis.

Small signal model is used for active region of transistor and hence is used to analyze amplifier characteristics such as gain,input and output impedance etc.

Small signal parameters for BJT are :

Representation of Large signal and Small signal parameters:

Large signal signal parameters are represented by UPPERCASE LETTERS and UPPERCASE SUBSCRIPTS. eg: VBE

Small signal signal parameters are represented by lowercase letters and lowercase subscripts. eg: v_be

Total parameters (while combining small signal parameters with large signal parameters) are represented by lowercase letters and UPPERCASE SUBSCRIPTS. eg: v_BE

Small signal parameters and small signal analysis are on coming posts.

Stability in Collector-to-Base bias

2014-09-29T01:02:00.000-07:00

This is one of the circuit which is used to provide biasing for BJT. Voltage divider bias and fixed bias are other biasing circuits. In collector-to-base circuit, resistor R_B connects collector terminal of BJT to base terminal, the specialty of the circuit .. This resistor provides a negative feedback and it helps to reduce variations in the operating point.

Circuit diagram:


fig 1

R_Bstabilizes Ic against changes either due to variations in temperature or due to replacement of transistor by another one. How it happens?

Let us assume Ic is increasing.

What about V_CE?. It will decrease (Increase in Ic increases voltage drop across Rc. Since Vcc =Voltage across Rc +Voltage V_CE , any increase in voltage drop across Rc will reduce V_CE ).

What about I_B?. It also will reduce with V_{CE .}(Equation for V_CE is V_CE= I_BR_B+ V_BE. As per equation I_B and V_CE are in direct proportion.)

Reduction in I_B will in turn reduce Ic. Thus the increment in Ic get cancelled and the circuit obtain stabilization. Note that stabilization is obtained because of the reduction in I_B and it would not happen if R_B is not there. So credit for this negative feedback goes to R_B.

Stability factor for collector-to-base bias.

Take general equation for stability factor. It is given in the post 'Stability factors & Biasing circuits'.      --------(1)

In the above equation only variable is dI_B/dIc. Let us find an equation for it from the above circuit.

From the circuit,

Therefore,        differentiate with respect to Ic.(Vcc and V_BEare constants)

                      substitute this in equation for stability factor S.

               ---(2) This is the equation for stability factor for collector-to-base bias.

Please refer the equation for stability factor in case of fixed bias. It is . ------(3)
Compared to fixed bias, stability factor of collector-to-base bias is less. This ensures better stability for collector-to-base bias.

Fixed bias circuit - Stability factor

2014-09-25T08:59:00.000-07:00

Fixed bias circuit is the simplest biasing circuit for BJT. If an equation can be obtained for it's stability factor, we can easily comment about it's stability. Let's begin..

Circuit diagram for fixed bias

fig 1

The simplest biasing circuit but has least stability.

It consists of a dc voltage Vcc for biasing and base resistor R_B. Together they provide a 'fixed' base current(I_B). (For a given circuit, values of Vcc and R_B are constant, so they provide a fixed current).

In order to justify the above statement, let us write an equation for I_B.

_{For that, consider equation for Vcc, ------(1)}

Therefore, _{a constant ----(2)}

In an amplifier circuit, Ic and V_CE together define operating point for BJT. Equations for Ic and V_CE are as follows.

_-------(3)

_------(4)

Stability in fixed bias:

Take general equation for stability factor. It is given in the post 'Stability factors & Biasing circuits'. It is here for ~~your~~ our convenience, -----(5)
This equation contains two terms -β and dI_B/dIc. Among them β is constant and check dI_B/dIc . From equation(2), we know that, I_Bis constant and is independent of variations in Ic. So dI_B/dIc is almost zero. Now the equation become

_{. ------(6)}

This equation gives a very high value to stability factor(because β is very high), an undesirable condition. So it has least stability.

Voltage divider circuit - stability factor (Sβ)

2014-09-20T03:23:00.000-07:00

Myself Stability factor S_β. My address is .
To view my history visit Stability factors & Biasing circuits

Request you to visit my Seniors also- Voltage divider bias - Stability factor 'S'

and

Voltage divider circuit - stability factor (Sv)

Referring to Thevenin's equivalent of Voltage divider circuit(fig 1)- ( details are given in Voltage divider bias - Stability factor 'S' )- equation for Thevenins voltage is given below.

fig 1

-------(1) then,

substitute for I_B

re-arrange

differentiate this equation with respect to β

we have dIc/dβ = S_β

re-arrange

-------(2) Numerator of this equation can be substituted using equ(1).

From equation (1) we will get, Substitute in (2).

-------(3) divide numerator and denominator with R_TH+R_{E .}

-----------(4) Smelling some similarity with equation of 'S'. Verify equ(7).

Now we can follow conclusion for 'S' and also can substitute 'S' in the above equation (4).

Substitute for I_Bin terms of Ic.

Like 'S', S_βalso depends on values of R_TH and R_E.

For visiting other stability factors for voltage divider circuit - S and Sv .

Voltage divider circuit - stability factor (Sv)

2014-09-16T10:45:00.000-07:00

We are going to catch "Sv" for voltage divider circuit. "Sv" indicates a stability factor used to discuss stabilization of the circuit during variations in Ic with respect to VBE . Before starting, it's better to take a trip through Stability factor 'S' for voltage divider circuit . Also for getting more idea about stabilization and stability factors, plz visit Stability factors & Biasing circuits .

Stability factor (Sv)

Stability factor indicates the stability of the circuit which is under discussion. Stability means variations in operating point(mainly Ic). To find 'Which are the main components in the circuit that affects stability?' -we always depends upon the equation for the stability factor. Once we get it, we can easily find the components affecting stability and it will then lead us to 'different ways for increasing stability'.
Consider the Thevenin equivalent for Voltage divider circuit. A detailed description is given in the post Voltage divider bias - Stability factor 'S'. For convenience i am including Thevenin's equivalent of input side of voltage divider circuit(fig 1).

fig 1

On writing loop equation, we will get as,

re-arranging,

substitute for I_B

-----(1)

General equations for stability factors are given in Stability factors & Biasing circuits. Definition for the second stability factor 'Sv' is .
If equation (1) is is differentiated with respect to Ic then we can find an another equation for dIc/dV_BE.

   which is equal to Sv. Therefore,

   re-arrange this equation to include a factor R_TH/R_E.

    ------(2)

For good stabilization, stability factor Sv should be low. This will happen when (R_TH/R_E) ˂˂ 1+ β and β ˃˃ 1.

During these conditions equ (2) become, Sv =̃ -1/R_E

Sv mainly depends on R_E, when it is low .
Above equation revels that dominant factor in stabilizing Ic against variations in V_BEis the voltage drop across R_E.
Other stability factors for voltage divider circuit are S and S _β.

Voltage divider bias - Stability factor 'S'(Stabilization in voltage divider circuit)

2014-09-13T00:03:00.000-07:00

Various biasing circuits are designed for amplifiers(a refreshment is provided here). One which is familiar to you is Voltage divider bias, also known as Self bias. Here, we are discussing about various stability factors in self bias.

First, consider diagram for voltage divider bias :

fig 1

This is the commonly used biasing circuit as it provides good bias stability. It is connected in Common Emitter configuration. In order to connect as an amplifier, Emitter-base junction is to be
forward biased and Emitter-collector junction to be reverse biased. R1-R2 network provides the necessary base voltage. The emitter resistance R_E provides stabilization. Voltage drop across R_E ensures reverse bias across Emitter junction.

In order to understand how stabilization is obtained in potential divider circuit, lets do the analysis and derive equations for stability factors.

In amplifiers, stabilization means keeping the operating point fixed against variations in temperature and transistor parameters. Both V_CE and Ic together decides the operating point. Equation for V_CE is V_CE = Vcc - Ic(Rc + R_E ). This does not contains β term and so V_CE is independent of variations in β. So we need to bother only about variations in operating point due to changes in Ic

Stabilization in voltage divider circuit

Take only input side of fig 1 into consideration as shown below(inside green line) and find Thevenin's equivalent. This will reduce complexity of analysis.

fig 2

Redraw the input side, we will get

fig 3

Consider Thevenin equivalent model of input side( left side of Blue line).First find out R_TH and then E_TH.

fig 4

In order to calculate R_TH , replace voltage source(Vcc) by a short circuit in fig 3. Then look into the circuit as shown in figure 4. We can see that R1 is parallel to R2. R_TH =R1 ll R2.

Then E_TH is the voltage across R2 in fig 3.

fig 5

In fig 3, replace the section for which we are taking Thevenin's (left side of blue line) with Thevenin equivalent resistance R_TH and voltage E_TH obtained.

fig 6

With this simplified circuit, lets do some analysis about its stability. For that find an equation for Ic.

For that, first write an equation for I_B.

           Apply KVL,     E_TH - I_BR_TH - V_BE- I_ER_E= 0 ----(1) re-writing,

       -----------(2)

We have a general equation between I_Band I_E

_{-----------(3)}Using equ. (2) and (3) we can write an equ. for I_E .

    -------(4) ( Because I_E and Ic are nearly equal)

Stability Factors

Next we are going to find out equations of Stability factors for voltage divider circuit. In this post, i am including the first stability factor 'S' only. Sv and S _βare described in other posts.

Just recollect the definition: Stability factor (S) is defined as the rate of change of collector current(Ic) with respect to the reverse saturation current(Ico), keeping β and VBE constant.

at constant β and VBE .

Consider fig 6 for simplicity. Take euq (1) and modified as follows.

E_TH = I_BR_TH + V_BE+ (I_B+ I_c)R_E ----(5)

Consider equ for stability factor S, from previous post 'Stability factors & Biasing circuits'.

i) Stability factor (S)

Here we have a term dIB/dIc. We will get an equ for substituting for this if we differentiate equ(5) with respect to Ic(take V_BE and E_THas a constants).

From (5),

re-arranging we will get,

--------(6) Put this in equ for 'S'.

-------(7) take denominator and re-arrange as

------- multiply denominator and numerator by (R_TH/ R_E) +1

------------ (8)

Value of 'S' depends on two factors of the equation, (R_TH/ R_E) and β('1' is very less compared to them).
Minimum value of 'S' is '1' for small values of (R_TH/ R_E).
Maximum value of 'S' is 1+β for large (R_TH/ R_E).

## When (R_TH/R_E)˂˂1. [1+ β][(R_TH/R_E)+1] → 1+ β and 1+(R_TH/R_E)+β → 1+ β
When (R_TH/R_E)˃˃ 1 . [1+ β][(R_TH/R_E)+1] → [1+ β](R_TH/R_E) and 1+(R_TH/R_E)+β → (R_TH/R_E) ##

In the previous post 'Stability factors & Biasing circuits' it is given that for proper stability of an amplifier, 'S' should be small. For voltage divider circuit, we obtained that 'S ' is small for small values of (R_TH/ R_E). So R_TH should be small and R_E should be large. We can conclude that value of 'S' is determined by R₁,R₂ and R_E resistors.

While designing a voltage divider circuit, for good stability against variations in Ic with respect to Ico (see definition of 'S'), we should select small values R₁ and R₂ and high value R_E .

In this post,the stability factor 'S' for the voltage divider circuit is discussed. To get an idea about other stability factors for voltage divider circuit - visit Sv and S _β.

Stability factors & Biasing circuits

2014-09-07T03:27:00.000-07:00

Stability factors seems to be heavy???. Make a trip along with fundamentals of BJT- plz click here - BJT - first glimpse

Stability factors & Biasing circuits:

An idea about biasing and stabilization will help in the study of Stability factors -plz check the following paragraph.

Things to be ensured before wiring an amplifier:

Before setting up an amplifier circuit, operating point should be fixed at an appropriate position on the load line - preferably at the center of the load line. This can be done with the help of a biasing circuit. It is also to be ensured that the operating point does not shift due to changes in temperature or due to variations in transistor parameters(see explanation below) - in other words operating point should be stable.

The maintenance of the operating point stable is known as stabilization.

## We have, Ic = β I_B + (1+β)Ico . Since Ico and β are vary with temperature, Ic also varies with temperature - it will change operating point. Variations in transistor parameters such as β and VBE (These variations occurs when one transistor in a circuit is replaced with another of the same type) also will affect position of the operating point. ##

Next we discuss about Stability factor.

Stability factor (S) is defined as the rate of change of collector current(Ic) with respect to the reverse saturation current(Ico), keeping β and VBE constant.

at constant β and VBE . ------(1)

What is desirable for S - a large value or a small value? Smaller S indicates higher stability. ( S high means variations in Ic is large with change in Ico - it will vary operating point).

Now we are going to derive an equation for S. For that lets take this equation below,

----(1)

Differentiate this equation with respect to 'Ic', consider β as a constant(see definition of S).

-------(2)

From this, write equation for Stability factor, S

----(3)

One more

A second Stability factor, Sv can be defined by considering variations in Ic with respect to VBE . Here Ico and β are considered as constants.
at constant Ico and β ------(4)

One more ..

A third Stability factor, S_β is defined for expressing variations in Ic with respect to β. It is the rate of change of Ic with respect to β. at constant Ico and VBE ------(5)

BJT - first glimpse (Biasing in RC coupled amplifier and Q point)

2014-09-03T22:25:00.000-07:00

A vital part of an electronic circuit is a transistor . One which is familiar to you is Bipolar Junction Transistor(BJT). Just take a tour through the necessary points. Here include the basic equations related to transistor. Also explain the concept of biasing and Q-point with the help of an amplifier circuit.

fig 1

Equations we need:

Ic = β I_B

Ic = αI_E

α = β /1+β

I_E = Ic + I_B

_{Recollect functions of collector, emitter and base and related currents.Ok, then move to amplifier section.}

_{The below circuit is very familiar in your first year.}

_{RC coupled amplifier(CE amplifier)}

fig 2 BJT amplifier

_{This
is an amplifier circuit. If we give a small input(sine wave) as Vs, we
will get an amplified output in the output side Vo. To understand this
circuit, it can be divided into 3 sections:}
_{1) biasing section(inside red dots)}
It consists of resistors R1,R2,Rc,RE, dc voltage Vcc and ground.This is the first section to be wired while setting up an amplifier for lab experiments. Value of the dc source Vcc and value of the resistors in this section decide the position of operating point and region of operation of the transistor. While the transistor is working as an amplifier, it should be in the active region to ensure proper amplification. Otherwise some portions of the output sine wave will be cut off.

2) input section

It consists of input signal Vs, resistor in the source side Rs and capacitor C1. This section provides the input signal which to be amplified.

3) output side

It consists of capacitor C2 and sometimes a resistor Ro(not in figure). Amplified output can be obtained from here.

Circuit also consists of a bypass capacitor CE.

Why a biasing section is included in amplifier circuit:

As i already said, a biasing section should we wired before input and output sections in an amplifier. This biasing circuit ensures initial dc values of IB, Ic and VcE. After ensuring this initial value, if we apply the input sine wave, then new variations in the values of IB, Ic and VcE are around its initial dc values.

IB = Base current, Ic = Collector current and VcE = Voltage across Collector and Emitter

From fig 2, VcE = Vcc - Ic(Rc+RE) ---(1)

Minimum value of VcE is zero and if we apply this to (1),

Ic =Vcc/(Rc+RE) -of course this is the maximum value that Ic can be obtained and represented as Ic(sat).

Maximum value of VcE is Vcc and this happens when Ic = 0 in (1). Therefore minimum value of Ic = 0.

With these values of Ic and Vcc, if a graph is drawn , it is known as D C load line for an amplifier.

fig 3

Load line represents every possible values of (Ic , VcE ) combinations. It means if we measure values of Ic and VcE of an amplifier at various instants, these points (Ic, VcE ) will be on the load line (dotted line in above figure).

After familiarized with load line, we will go to Q-point(operating point).

Consider the biasing section in fig 2.(At this point we are taking that no input sine wave is applied - only Vcc is given). If Ic and VcE are measured at this point, they have specific dc values. These values of Ic and VcE can be represented as a point (Ic , VcE ) and it will corresponds to a particular point on dc load line. This point is known as Q-point. (Note that we are dealing with a zero input condition only bias voltage Vcc is applied).

Need for biasing:

In fig 3, Ic(sat) represents saturation current for the circuit. At this point transistor in the amplifier circuit is in saturation and VcE is zero (A transistor is said to be in saturation when collector(output) current is maximum and at this condition there is a short circuit from Emitter to Collector to provide maximum current. Therefore voltage across Emitter and Collector(VcE) is zero.).

When VcE = Vcc, Ic is zero(from graph) and an open circuit from Emitter to Collector. Therefore transistor is in cut off.

Taking these two conditions in fig 3,

i) if (Ic , VcE ) point on the load line is near to Y-axis, there is a chance for the transistor to be in saturation.

ii) if (Ic , VcE ) point on the load line is near to X-axis, there is a chance for the transistor to be in cut off.

During the above conditions, transistor cannot properly amplify the input sine wave. Some parts of the amplified output may get lost. To avoid this, (Ic , VcE ) point must be at the center of the load line (to reduce chances of saturation and cut off).

For proper amplification, it is desirable to have Q-point centered on the load line. This is ensured by a proper biasing circuit (as in fig 2).

After proper biasing, if we connect the input section and apply an input signal, then Ic and VcE both will vary. This variation is around their Q-point values which is already set by biasing circuit at the center of the load line.

Clamper

2014-08-31T10:37:00.000-07:00

The word Clamp means 'to hold things together or fasten two things together'. In electronics it means 'hold voltage level of a wave at a particular value which is different from its normal.

Clamping circuit is used to shifts a signal by a dc voltage. How it is possible to shift a sine wave?

Fig a

This dc level shifting is explained in Fig a. Dc level of sine wave (leftmost figure) is at zero volt. Sine waves having dc level at +2v and dc level at -2v are followed. A clamping circuit can be used for this purpose.

Clamper circuits -section I

A simple clamping circuit is given below in fig 1. The circuit consists of a capacitor and a diode. Across the diode output voltage is taking. Here the capacitor act as a battery(Capacitor has a charge storing capability and thus can be act as a cell. At the first instant of switching on the circuit, capacitor is charged to maximum value of input voltage(Vm). It will maintain this voltage in it until the input is switched off). In fig 1, positive side of the diode is at the top side and so leftmost plate of the capacitor is positive(current direction through the diode is from top to bottom and that in capacitor is from left to right - so left plate is taken as positive. Since right plate is negative, donation to output voltage by the capacitor is negative).

fig 1

To get an idea about how the output is obtained, keep these two things in mind:
1.During positive half cycle of input voltage, diode is forward biased and no input voltage is dropped across diode(so no donation to output voltage by diode). But our capacitor battery is there, and its voltage(Vc) is equal to maximum value of input sine wave(Vm). Then have a look into voltages in the circuit.

Positive half cycle of the input sine wave is shown with Vm=10v, its polarity is also indicated. Capacitor is charged to 10v with polarity as shown. Since diode is forward biased it is represented with a small forward resistance. Output voltage is taking across points 'a' and 'd'(Vad). Vad = Voltage across 'abcd' = -Vm+Vin = -10v+ Vin. (Vin is varying from 0 to +10 then to 0). So -10v +Vin will be as follows:

This is how the output in fig 1 is obtained during positive half of input.

2. During negative half cycle of input, diode is in reverse bias and it will behave as an open circuit.

Now have a look into polarities of input voltage and capacitor. Since capacitor charges only during the positive half cycle of input sine wave, polarity is same. But because of negative half cycle of input, polarity of Vin is reversed. Output voltage is taken across points 'a ' and 'd' (Vad). Vad = Voltage acrosss 'abcd' = -Vm -Vin = -10v -Vin.

This explanation is only for keeping in your mind, don't pour it into your papers. Only one thing you need to do is --- assign a polarity for the capacitor according to forward bias condition of diode. If -ve polarity is nearer to diode, keep in mind that output will be in the -ve side of Y axis. Keeping circuit of fig 1 in mind, it is easy to find outputs of fig 2 and fig3. Lets see..


fig 2

Compare outputs of fig 1 and fig 2. If output of fig 1 is slightly shift towards top we will get output of fig 2. This shift is because of the presence of positive cell voltage Vr.

fig 3

Similarly if we shift the output of fig 1 towards bottom, output of fig 3 will be obtained. This is due to negative battery voltage Vr attached to the diode.

Clamper circuits - Section II

It is easy to diegest the remaining circuits if we relate them to that in section I. Just reverse the direction of diode in fig 1, then look below. Yaa, next circuit.

fig 4

With the reversal of diode, don't forget to reverse the polarity of capacitor - here current direction through the capacitor is taken from right plate to left plate. A tip to remember - if the capacitor plate nearer to the diode is positive then output wave will be in the positive side of y axis.

Follow the same route of transition from fig 1 to fig 2, for fig 4 to fig 5..

fig 5

When a positive bias is introduced into fig 4, we get fig 5 and waveform shifts up.

Then again find an old route, that is from fig 1 to fig 3 and follow it for fig 4 to fig 6.

fig 6

With an extra negative bias in fig 6 compared to fig 4, waveform shifts down.

Easy???

Combinational Clipper Circuit

2014-08-26T23:12:00.000-07:00

This is a continuation of 'Clipping Circuits'. In a simple clipper circuit either +ve or -ve half cycle of sine wave can be altered at a time. Using a combination clipper circuit, both half cycles can be shaped using a single circuit.

fig a

Circuit in fig a is a combination of fig 2 and fig 5 in 'Clipping circuits'. Its easy to understand the output waveform by considering +ve half cycle of fig 2 output and -ve half cycle of fig 5 output.

fig b

Circuit in fig b is a combination of fig 2 and fig 6. Here also the output can be easily find out by combining output of fig 2 and fig 6. (In fig 6, bottom side of +ve half cycle of sine wave is removed. In fig 2, topside of +ve half cycle of sine wave is removed. Apply some logic and combine two - in a +ve half cycle of sine wave top and bottom side are removed - that is given in above fig b.Please forget about the -ve side of sine wave - there is no alternation for -ve side in fig 2 and nothing is there in the -ve side of fig 6.)

fig c

Circuit in fig c is obtained from fig 3 and fig 5. By removing both sides of -ve half cycle of sine wave, output of fig c is obtained.

Clipping Circuits

2014-08-22T11:20:00.000-07:00

Clipping circuits are used as wave shaping circuits. Usually it is used to shape sine waves by cutting off some of its parts, but it can be used for other waves also.The idea will be clear if we familiar with the simplest member of clipper family.

An example of a clipper circuit is Half wave rectifier. Circuit, Output voltage ('Vo' vs 't' -obtain using CRO) and Transfer characteristics (' Vo' vs 'Vin' -obtain using CRO) are shown.

fig 1

Here i take two circuits -one in top and other in the bottom position, both give same output, hence same transfer chara. Though you know very well about the working, am just briefing it - for my convenience. Circuit in the top consists of a resistor(R) and a diode(D). Output takes across D ie. CRO probes are connect across two ends of D.

For positive half of input sine wave, D is in forward bias - current is flowing without any resistance, so voltage zero(diode ON). //A voltage is developed only across a resistance, if resistance of a component is zero then voltage across it is also zero. But this an ideal situation only - not practical. Every component, even 'diode in forward bias' have a small resistance (ON resistance), therefore a small voltage across it ie. 0.7v //

For negative half cycle of sine wave, D is in reverse bias. No current flow through D. Diode act as 'Open circuit'. So whatever the voltage applied in the input side, is completely appeared across D - in other words, output follows sine wave's(input) negative half cycle. //When there is no current flow through a component, resistance of it is taken as infinity(resistance=voltage/current). Comparing to this infinite resistance, all other resistances in the concerned circuit are negligible(even though it has any value -K ohm/M ohm/..). We know V=RI, since resistance is infinite, voltage is very high(not infinite). Therefore whatever the voltage given as input to the circuit, that full voltage is dropped across this infinite resistance. ie. voltage across an 'open circuit' = input voltage --all other components in the concerned circuit have negligible voltage.//

We can easily draw transfer chara using input and output waves. Below given explanation will help you for it.

Transfer characteristics : It means input vs output.When both input and output are sine waves, transfer chara is

fig a

When Vin increases +vely (in transfer chara, from (0,0) towards right), Vout also increases +vely(in transfer chara, from(0,0) towards top). Result is a straight line with 45 degree slope. Similarly for -ve half cycle also.

Now let's see how transfer characteristics is obtained in fig 1. Take fig a, as reference. Consider transfer chara of fig 1. For +ve half of Vin, Vout is constant at 0.7v - so transfer characteristics is a straight line at 0.7v(output is constant at 0.7v irrespective of change in input). It is indicated both in output waveform and transfer chara. For -ve half of Vin, fig 1 is same as fig a.

Hope the above explanation is satisfactory. Now move to next clipper circuit.

fig 2

Explanation for obtaining such an output is as follows: We are adding a +ve bias (+ve voltage) in the basic circuit(fig1) using a battery and it shifts the cut off voltage of diode from 0.7v to (Vr + 0.7v). Vr is the voltage of battery(we can use 5v or 7v or any other, but should be less than Vin). You can easily understand its transfer chara by comparing it with fig a.

Next clipper is obtained by adding a -ve bias to circuit in fig 1. It shifts the cut off voltage to (-Vr + 0.7v).

fig 3

It is easy to remember the output of fig 2 and fig 3 by relating it to output of fig 1.

(i)adding Vr to output of fig1 (iii)output of fig2

(i)adding -Vr to output of fig1 (iii)output of fig3

No need to study theoretical explanation of how to obtain the output for a particular clipper circuit. Such essay questions are not yet asked for theory exams. But short answer questions may be asked and it will be like 'given a transfer chara. draw the corresponding circuit'. To answer such questions, first you draw the output wave form using given transfer chara and then the circuit.

Next look into next set of clipper circuits. Here circuit in fig 4 is taken as the basic circuit and is obtained by reversing the diode in fig 1.

fig 4

In fig 4 cut off is at (-0.7v) and is an example of a half wave rectifier. Now see what happens if a negative bias is added to circuit in fig4.

fig 5

A - ve bias of (-Vr) is added to fig 4 to obtain circuit in fig 5. Then cut off voltage changes from -0.7 to (-0.7-Vr). It is adopted in output waveform and transfer characteristics. Similarly if a +ve bias of Vr is added to circuit in fig4, we will get circuit in fig 6. Cut off voltage is shifted to (-0.7+Vr). //For example, battery voltage Vr =5v. Then cut off voltage is -0.7+5 = +4.3//

fig 6

These are the simple clipping circuits. Once you familiarized them, move to combination clippers

RC Differentiator

2014-08-16T00:34:00.000-07:00

High-pass RC circuit as Differentiator:

Adding some ‘salt and pepper’ over high pass RC circuit will give differentiator.
In differentiator output voltage, Vout is proportional to the rate of change of input.

Circuit diagram:

As in the case of integrator, when the high pass circuit obey some condition, it act as a differentiator. Shall we see how it is?

Lets write the equation for output voltage.

//output is taken across R and R+1/CS is the total resistance of the circuit. So applying voltage divider rule, Vout = Vin{ R/(Total resistance) } //

Laplace transform s=jw (w=2πf) and RC is the time constant.

Rearranging,

Take the condition that RCw << 1, then

// When RCw << 1, then 1/RCjw become large, so 1+1/RCjw is approximately equal to 1/RCjw. Now the equation become
//

We know jw = S, Laplace transform. So

therefore, Vout = RC S Vin

The above equation is a Laplace equation and we take inverse Laplace, then equation become,

Equation shows that output voltage is proportional to derivative of input voltage. So the circuit represents a differentiator circuit.
OK!! then go through output of the differentiator circuit for different inputs.

//In lab hours, select different R and C and check output.Also check output just by changing input frequency(by adjusting front nob of signal generator) //

If we make a chicken fry, what is next?. Check its taste, whether salt, chilly etc are ok. Likewise if an output is obtained some people check its different properties and we people study it. So two properties of output wave form.

Rise time (tr): The time taken for the output voltage to rise from 10% to 90% of final value.
//if peak is 100 v, then how much time the wave will take to rise from 10 v to 90v.//

Delay time (td): The time taken for the output voltage to rise from 0 to 10% of its final value.

Low pass RC - integrator and high pass RC - differentiator are very important in lab exams. It will ask in combination with Rectifiers. For theory exams it is not a usual question. Something bigger is going to happen...

Come back with some interesting circuits..

RC Integrator

2014-08-11T01:09:00.000-07:00

How Low pass RC filter work as an Integrator ?- It is explained here.

Low pass RC circuit is explained in detail here - Low pass RC filter , plz visit. Then move to next course - but have same ingredients -that's Integrators . Oops ...Integration. Is this our AC1 or Maths??? ....

Arranging the table for serving...

In electronics also there are Integrators and differentiators because of the following reasons. Many circuits need signals other than sine wave, like square wave, cosine wave, triangular wave, pulses etc. But the basic signal available as input to any device is our Alternating Current, which is sine wave. Square wave can be produced easily either from sine wave or from a dc voltage with the help of some simple circuits. By integrating and differentiating these two waves(square and sine), it is possible to produce other waves also. In order to realize integration and differentiation, we are using integrator and differentiator circuits.

Integrator

An integrator produces an output which is proportional to integral of its input.

Circuit diagram:

AC – AC source that provides the input - Vin
Output – Output voltage
Resistors(R) and Capacitors(C) are connected in series with Vin.

Low pass RC circuit itself is acting as an integrator provided it obey one condition. Lets check that condition.

Equation for output voltage
                                       1/CS –reactance offered by C in terms of Laplace transform



Laplace transform s=jw and w=2πf (w-omega)


if we are adopting a condition as wRC >> 1, then , in denominator 1+jwRC is nearly equal to jwRC. Then equation become



Re arranging,

                  Here 1/RC is taken as a constant

Taking inverse Laplace transform,



//Those who are not familiar with Laplace-- while converting integral into Laplace 1/s is adding with the equation. During inverse Laplace, 1/s indicates an integration.//

Above equation shows that Vout is proportional to integral of input voltage. So if we have a low pass RC circuit which obeys the condition wRC >> 1, it will act as integrator.

//The product RC is known as time constant(T). With the value of RC there will be changes in output waveform.//

In order to achieve a good integration, the following conditions must be satisfied:

The time constant RC of the circuit should be very large as compared to the time period of the input signal.
The value of R should be 10 or more times larger than Xc.

Output for different input waves:

1. For Step input:

2. For square wave input

Three output waveforms are shown. These are with increase in values of time constant ‘T’.

Applications:
Use to perform integration in certain electronic circuits.
Use to convert square wave into triangular wave.

High pass RC circuit

2014-08-05T04:06:00.003-07:00

We can hear so many sounds in nature. These sounds may have different frequencies. How we can separate a particular type of sound(waves) from others?..We can do it by using a filter circuit. A filter circuit helps us to 'filter out' that waves having particular frequencies. Many circuits are used as filters. Here, explain the circuit diagram and working of a high pass RC circuit(high pass filter). Frequency response is also given. This is only a stepping stone to differentiator.

Passive High pass filter

A first order high pass RC circuit contains one pair of Resistor(R) and Capacitor(C). It act as a simple high pass filter - it allows high frequency (input)signals to appear at the output side while attenuating low frequency signals.

Circuit diagram:

Input is an AC source, Output is taken across resistor.Contains one pair of R & C.

Working:

Voltage across R is taken as output voltage. Let’s see what happened to this voltage at low and high frequencies.For refreshments of basic concept see Refreshments .

For high frequency input:

Xc is very low(Xc inversely proportional to frequency). So Vc is also low. For very high frequencies, magnitude of Vc is very less and so we can say that Vc << V_R,where V_R is the output voltage. In conclusion, for high frequency input, output is higher.

For low frequency input:
Xc is very high.Vc is also high. Therefore for lower frequencies V_R<< Vc. So output voltage V_R is much less for low frequency input.

If you are comfortable with 'working', directly go to 'Summary'.
Feel something odd.. just look below.

When working with filter circuit , we are keeping the amplitude of input AC constant(here 6v) and varying its frequency.
First case: we give a high frequency input(fig a), say 10MHz. For this suppose Xc = 10. Then the 6v input is divided as Vc : V_R = Xc : R = 1:2. V_R = 4v. Then Vc= ?? (2v).Vc <V_R .
Second case: increase input frequency as 50MHz. Suppose in this case Xc =5. But R doesn’t depend on frequency, so fixed value R=20 . Then the 6v input is divided as Vc : V_R = Xc : R = 5:20 = 1:4 . V_R = 4.8v. Then Vc = 1.2v.
This example shows that as frequency increases, Vc is also increases. So at higher frequencies it is taken as Vc>>V_R.

Then check for a low frequency input(fig b), say 10KHz. Suppose Xc = 40. Find V_R. Then you will get, why V_R << Vc for low frequencies?.

Summarising :
for high frequency input signal ----- output voltage is larger
for low frequency input signal ----- output voltage is lesser

Then we can call it as a “high pass”.

Equation for output voltage

Vout = V_R = where Vin indicates amplitude of input signal

Frequency response of first order High pass filter:

A frequency response gives you an idea about 'the amplitude of wave at different frequencies'. So in X-axis, we plot 'frequency' and in Y-axis, Gain(Gain= amplitude of output wave/amplitude of input wave).

Frequency response of High pass filter

Peeping into AC1 class room and low pass filter

2014-07-30T03:56:00.000-07:00

Analog Circuit I is included in B Tech Syllabus under M G University for Electronics & Communication Engineering(EC 010305) and Applied Electronics & Instrumentation Engineering(A010 305).

Hope this blog will show an easy way for Analog Circuit I for u S3ans.

Its better now we think about module 1 only.

Module I includes study of RC Circuits,Rectifiers,Voltage regulators and clipping & clamping circuits. RC circuits and rectifiers are familiar in your first year(OR a nightmare, no problem i will help you to give a twist.Together we can lit a torch into that darkness ).

You have to ~~study~~ chew the following things -yaa,"The Syllabus"
('The' is for showing respect - not for comparing it's size with that of 'The Sun' and 'The Earth").

Section I

1. RC circuit:
   Response of low pass and high pass RC circuits to –
       sine wave, step, pulse and square wave.
        Properties of such output waves - Tilt and Rise time
Differentiator, Integrator
2. Rectifiers:
Analysis of Rectifiers – Half wave, Full wave, Bridge
Analysis of Filters – L, C, LC, π
3. Clipping and Clamping circuit
4. Voltage Regulators :
5. Zener voltage regulator
Transistor series (with feedback)
Short circuit and foldback protection
Shunt voltage regulators
6. Small signal diode model for low and high frequencies

RC Circuit

NEED some Refreshments

1. Resistor (R)- If there is a resistor in a circuit, there is a voltage (voltage drop) across it. That voltage drop is in proportion to the value of resistance. Certainly it is in direct proportion because V = RI.

2. Capacitor(C) – Easily allows AC (high frequency) but opposes DC (low frequency). In other words, Capacitor offers low resistance to AC and high resistance to DC. This resistance is represented as Xc.

Just for pushing the above fact into brain, watch this:

3. Voltage division

     V =    = 6.67v

      The voltage divided across 20Ω and 10Ω in 2:1.
       Voltage across 10Ω = 10-6.67 = 3.33v

First note: A simple RC circuit can be act as a passive filter ( don’t afraid, here passive means ‘no transistors’ and filter means it filtered out some portion from input and fed the output side with the rest). It may be either a low pass filter or a high pass filter depending upon the positions of R and C relative to each other.

Low pass RC Circuit:

We all are equipped with a low pass filter in our body. Guess which one is that organ...Our ears- rejects all the frequencies above 20K Hz. SHHH....accurately it is a band pass (20Hz to 20 KHz). Great scientists always adopt some assumptions, then why we can’t???. We also can take as 20Hz is near to zero. So 0Hz to 20 KHz and low pass.

Let’s Start !!!

Consider a first order RC circuit. \\first order = one pair of R and C \\

Circuit diagram:

Working: How it act as a low pass filter?

The reactance of capacitor is Xc = \\ reactance = resistance of a capacitor \\

f – frequency of input AC (In India frequency of AC is ?? 50Hz) c - capacitance of the capacitor C (eg: 10 µF, 50 µF etc)

Equation reveals that Xc is inversely proportional to frequency f. Now just think that we have options to change the frequency of input AC. \\ Its very easy with an equipment called Signal generator which is in your college lab \\

At first we give a low frequency input signal, look into equ, what happens to Xc? Xc become very large. \\ Just imagine the case when ‘f’ is nearly zero, Xc is as large as infinity. Then you compare Xc and R.R has a fixed and finite value. So Xc >> R\\ .

Then consider voltages across Xc and R. Which one is greater? Yaa, that across Xc. Therefore voltage drop across the capacitor is very large compared to voltage drop across the resistor. In a similar way, at high frequency, Xc << R. So voltage drop across R is greater than that across Xc.

Summarising : for low frequency AC or for DC Vc >> VR larger output voltage
for high frequency AC Vc << VR smaller output voltage

\\ say, we are changing the frequency of input AC from 10Hz to 1,00,000Hz (but keeping its amplitude fixed eg:10v). For low frequency range, output voltage is nearly 10v(may vary depends on filter type). For higher frequency range, output will be much less than 10v.//

From the above statements it is clear that this circuit produces a high voltage for low frequency inputs and it attenuate the output voltage for high frequency input signals. That why it has the name ‘low pass filter’.

Equation for output voltage

Vout = Vc = Vin indicates amplitude of input signal

Frequency response of first order Low pass filter:

Analog Circuits I Syllabus(M G University - B Tech)

2014-07-25T02:53:00.000-07:00

Module 1
RC circuit:

          Response of low pass and high pass RC circuits to –
   sine wave, step, pulse and square wave.
                                                   Properties of such output waves - Tilt and Rise time

Differentiator, Integrator

Rectifiers:
        Analysis of Rectifiers – Half wave, Full wave, Bridge
        Analysis of Filters – L, C, LC, π
Clipping and Clamping circuit
Voltage Regulators :
       Zener voltage regulator
       Transistor series (with feedback)
       Short circuit and foldback protection
       Shunt voltage regulators
       Small signal diode model for low and high frequencies

Module 2

DC analysis of BJTs –
      BJT as amplifier
      Small signal equivalent circuits (Low frequency π and h Models only). Transistor Biasing circuits
      Stability factors
      Thermal runaway.

Small signal analysis of CE, CB, CC configurations using approximate hybrid π model (gain, input and output impedance)

Module 3

MOSFET-

I-V relation, load lines

small signal parameters, small signal equivalent circuits

body effect

Biasing of MOSFETs amplifiers

Analysis of single stage discrete MOSFET amplifiers –
Expressions for   a. Small signal voltage gain and current gain
b. Input and output impedance of
        1)Common Source amplifier with source bypass capacitor
                  2)Common Source amplifier without source bypass capacitor
                  3)Source follower
                  4)Common Gate amplifier

Module 4

Highfrequency equivalent circuits of 1)BJTs 2) MOSFETs,

Miller effect, short circuit current gain, s-domain analysis, amplifier transfer function.
Analysis of high frequency response of 1)CE, CB, CC amplifiers 2) CS, CG, CD amplifiers.

Module 5

Power amplifiers: Class A, B, AB and C circuits – efficiency and distortion.

Biasing of class AB circuits. Transformer less power amplifiers.

Feedback amplifiers – Properties of negative feedback.
The four basic feedback topologies-
1) Series-shunt 2) series-series 3) shunt-shunt 4) shunt-series.
Analysis and design of discrete circuits in each feedback topology
Loop gain, input and output impedance. Stability of feedback circuits
Voltage amplifiers, Current amplifiers, Trans conductance amplifiers and Trans resistance amplifiers

References:

1) Sedra and Smith: MicroelectronicCircuits, 4/e, Oxford University Press 1998.

2) B. Razavi , “Fundamentalsof Microelectronics”, Wiley

3) Donald A Neamen. : ElectronicCircuit Analysis and Design, 3/e, TMH.

4) Millman and Halkias: IntegratedElectronics, TMH, 2004.

5) Spencer & Ghausi: Introductionto Electronic Circuit Design, PearsonEducation, 2003.

6) Roger T. Howe, Charles G. Sodini: Microelectronics: An IntegratedApproach, Pearson

Education,1997.

7) R E Boylstead and L Nashelsky: Electronic Devices and Circuit Theory, 9/e, Pearson

Education

Easy Analog Circuit

High frequency model of transistor

High frequency model- explanation

Integrator & Differentiator-B Tech university Question & answer

Q1. Explain the working of RC low pass filter, how it can be act as an integrator.

Circuit diagram:

Working of a low pass filter:

Integrator

Low pass circuit act as Integrator:

Q2. Explain the working of RC high pass filter, how it can be act as an differentiator.

Circuit diagram:

Working of high pass filter:

High-pass RC circuit as Differentiator:

Small Signal - Low frequency - Analysis of common collector amplifier-continue

b)Output resistance (Ro )

c): Voltage gain (Av)

d): Current gain (A is)

Voltage buffer

Small Signal-Low frequency- Analysis of Common Collector amplifier (Emitter Follower) -begins here

Small Signal - Low frequency - Analysis of Common Base amplifier using hybrid model

Current Buffer

Small Signal - Low frequency - Analysis of CE amplifier using hybrid π model

Small signal model of a circuit

Steps to obtain a model from corresponding circuit :

Transistor - small signal model and small signal parameters

Analog Circuits /Electronics Circuits syllabus is packed with models.

The model

Small signal model of a transistor

1. Transconductance(gm)

As we know conductance is the ratio of current to voltage. For BJT, transconductance is find by taking ratio of output current to input voltage.

Transconductance gm = ic / vbe

2. Input resistance (rπ )

3.Output resistance (ro)

T- model of transistor

Transistor analysis - Small signal analysis and Large signal analysis

Large signal analysis

Small signal analysis

Representation of Large signal and Small signal parameters:

Stability in Collector-to-Base bias

Circuit diagram:

Stability factor for collector-to-base bias.

Fixed bias circuit - Stability factor

Circuit diagram for fixed bias

Stability in fixed bias:

Voltage divider circuit - stability factor (Sβ)

Voltage divider circuit - stability factor (Sv)

Stability factor (Sv)

Voltage divider bias - Stability factor 'S'(Stabilization in voltage divider circuit)

Stabilization in voltage divider circuit

Stability Factors

Stability factors & Biasing circuits

Stability factors & Biasing circuits:

Things to be ensured before wiring an amplifier:

One more

One more ..

BJT - first glimpse (Biasing in RC coupled amplifier and Q point)

RC coupled amplifier(CE amplifier)

Why a biasing section is included in amplifier circuit:

Need for biasing:

Clamper

Clamper circuits -section I

Clamper circuits - Section II

Combinational Clipper Circuit

Clipping Circuits

fig 1

RC Differentiator

High-pass RC circuit as Differentiator:

Circuit diagram:

RC Integrator

Integrator

High pass RC circuit

Passive High pass filter

Circuit diagram:

Working:

Frequency response of first order High pass filter:

Peeping into AC1 class room and low pass filter

RC Circuit

NEED some Refreshments

Low pass RC Circuit:

Let’s Start !!!

b)Output resistance (R_o)

d): Current gain (A _is)

1. Transconductance(g_m)

Transconductance g_m= i_c / v_be

2. Input resistance (r_π)

3.Output resistance (r_o)

_{RC coupled amplifier(CE amplifier)}