power supply

A simple 5 Volt regulated PSU featuring overvoltage protection.

2011-04-23T06:07:00.000-07:00

Description:
A simple 5 Volt regulated PSU featuring overvoltage protection.

Notes
The 5 volt regulated power supply for TTL and 74LS series integrated circuits, has to be very precise and tolerant of voltage transients. These IC's are easily damaged by short voltage spikes. A fuse will blow when its current rating is exceeded, but requires several hundred milliseconds to respond. This circuit will react in a few microseconds, triggered when the output voltage exceeds the limit of the zener diode.

This circuit uses the crowbar method, where a thyristor is employed and short circuits the supply, causing the fuse to blow. This will take place in a few microseconds or less, and so offers much greater protection than an ordinary fuse. If the output voltage exceed 5.6Volt, then the zener diode will conduct, switching on the thyristor (all in a few microseconds), the output voltage is therefore reduced to 0 volts and sensitive logic IC's will be saved. The fuse will still take a few hundred milliseconds to blow but this is not important now because the supply to the circuit is already at zero volts and no damage can be done. The dc input to the regulator needs to be a few volts higher than the regulator voltage. In the case of a 5v regulator, I would recommend a transformer with secondary voltage of 8-10volts ac.

By choosing a different regulator and zener diode, you can build an over voltag trip at any value. I have a simulated transient graph of this over voltage protection circuit in the Design section

Circuit Modification
I have a circuit modification from one of my viewers in Sweden, Ulf Kylenfall. The modified circuit is shown below and the electrolytic capacitor is moved to a position directly after the rectifier bridge and before the fuse.

With the original circuit, the thyristor had to [rapidly] discharge the capacitor in its effort to protect the downstream circuitry. This current surge may otherwise destroy the thyristor depending on what quality that has been used for the capacitor.

With the modified circuit, shown above, the fuse needs to blow in such a short time that the integrated energy dissipation over the thyristor is within its limits. Moving the capacitor to a position before the fuse ensures this. (Provided that a sufficiently fast-blow fuse has been used).

Thank you Ulf, for highlighting this design problem, and your solution to improve the circuit, kind regards Andy Collinson.
this circuit by : Andy Collinson.

Unregulated Power Supply

2011-04-23T06:05:00.000-07:00

Description:

A basic full wave rectified power supply is shown below. The transformer is chosen according to the desired load. For example, if the load requires 12V at 1amp current, then a 12V, 1 amp rated transformer would do. However, when designing power supplies or most electronic circuits, you should always plan for a worst case scenario. With this in mind, for a load current of 1 amp a wise choice would be a transformer with a secondary current rating of 1.5 amp or even 2 amps. Allowing for a load of 50% higher than the needed value is a good rule of thumb. The primary winding is always matched to the value of the local electricity supply. Notes
An approximate formula for determining the amount of ripple on an unregulated supply is: Vrip = Iload * 0.007 / C where I load is the DC current measured through the load in amps and C is the value of the capacitor in uF.The diagram below shows an example with a load current of 0.1 amp and a smoothing capacitor value of 1000uF.
The calculated value of ripple is (0.1 * 0.007) / 1000e-6 = 0.7 volts or 700mV. The value of peak-peak ripple measured from the graph is 628mV. Therefor, the equation is a good rule of thumb guide for choosing the correct value for a smoothing capacitor in a power supply.
this circuit by: Andy Collinson

Variable DC Power Supply Voltage range: 0.7 - 24V

2011-04-23T05:42:00.000-07:00

Circuit diagram:

Parts:

P1____________500R   Linear Potentiometer
P2_____________10K   Log. Potentiometer

R1,R2___________2K2  1/2W Resistors
R3____________330R   1/4W Resistor
R4____________150R   1/4W Resistor
R5______________1R     5W Resistor

C1___________3300µF   35V Electrolytic Capacitor (see Notes)
C2______________1µF   63V Polyester Capacitor

D1,D2________1N5402 200V 3A Diodes
D3_____________5mm. Red LED

Q1____________BC182  50V 100mA NPN Transistor
Q2____________BD139  80V 1.5A  NPN Transistor
Q3____________BC212  50V 100mA PNP Transistor
Q4 __________2N3055  60V 15A   NPN Transistor

T1_____________220V Primary, 36V Center-tapped Secondary
               50VA Mains transformer (see Notes)

PL1____________Male Mains plug

SW1____________SPST Mains switch

Device purpose:

A Variable DC Power Supply is one of the most useful tools on the electronics hobbyist's workbench. This circuit is not an absolute novelty, but it is simple, reliable, "rugged" and short-proof, featuring variable voltage up to 24V and variable current limiting up to 2A. Well suited to supply the circuits shown in this website. You can adapt it to your own requirements as explained in the notes below.

Notes:

P1 sets the maximum output current you want to be delivered by the power supply at a given output voltage.
P2 sets the output voltage and must be a logarithmic taper type, in order to obtain a more linear scale voltage indication.
You can choose the Transformer on the grounds of maximum voltage and current output needed. Best choices are: 36, 40 or 48V center-tapped and 50, 75, 80 or 100VA.
Capacitor C1 can be 2200 to 6800µF, 35 to 50V.
Q4 must be mounted on a good heatsink in order to withstand sustained output short-circuit. In some cases the rear panel of the metal box in which you will enclose the circuit can do the job.
The 2N3055 transistor (Q4) can be replaced with the slightly less powerful TIP3055 type.

12 Volt 30 Amp Power Supply

2011-03-05T05:36:00.000-08:00

Description

Using a single 7812 IC voltage regulator and multiple outboard pass transistors, this power supply can deliver output load currents of up to 30 amps. The design is shown below:

Notes

The input transformer is likely to be the most expensive part of the entire project. As an alternative, a couple of 12 Volt car batteries could be used. The input voltage to the regulator must be at least several volts higher than the output voltage (12V) so that the regulator can maintain its output. If a transformer is used, then the rectifier diodes must be capable of passing a very high peak forward current, typically 100amps or more. The 7812 IC will only pass 1 amp or less of the output current, the remainder being supplied by the outboard pass transistors. As the circuit is designed to handle loads of up to 30 amps, then six TIP2955 are wired in parallel to meet this demand. The dissipation in each power transistor is one sixth of the total load, but adequate heat sinking is still required. Maximum load current will generate maximum dissipation, so a very large heat sink is required. In considering a heat sink, it may be a good idea to look for either a fan or water cooled heat sink. In the event that the power transistors should fail, then the regulator would have to supply full load current and would fail with catastrophic results. A 1 amp fuse in the regulators output prevents a safeguard. The 400mohm load is for test purposes only and should not be included in the final circuit. A simulated performance is shown below:

Calculations

This circuit is a fine example of Kirchhoff's current and voltage laws. To summarize, the sum of the currents entering a junction, must equal the current leaving the junction, and the voltages around a loop must equal zero. For example, in the diagram above, the input voltage is 24 volts. 4 volts is dropped across R7 and 20 volts across the regulator input, 24 -4 -20 =0. At the output :- the total load current is 30 amps, the regulator supplies 0.866 A and the 6 transistors 4.855 Amp each , 30 = 6 * 4.855 + 0.866. Each power transistor contributes around 4.86 A to the load. The base current is about 138 mA per transistor. A DC current gain of 35 at a collector current of 6 amp is required. This is well within the limits of the TIP2955. Resistors R1 to R6 are included for stability and prevent current swamping as the manufacturing tolerances of dc current gain will be different for each transistor. Resistor R7 is 100 ohms and develops 4 Volts with maximun load. Power dissipation is hence (4^2)/200 or about 160 mW. I recommend using a 0.5 Watt resistor for R7. The input current to the regulator is fed via the emitter resistor and base emitter junctions of the power transistors. Once again using Kirchhoff's current laws, the 871 mA regulator input current is derived from the base chain and the 40.3 mA flowing through the 100 Ohm resistor. 871.18 = 40.3 + 830. 88. The current from the regulator itself cannot be greater than the input current. As can be seen the regulator only draws about 5 mA and should run cold.

Initial Testing and Faulting

For the initial test, do not connect a load. First use a voltmeter across the output terminals, you should measure 12 Volts, or very close to it. Then connect a 100 ohm, 3 Watt resistor or other small load. The reading on the voltmeter should not change. If you do not see 12 Volt, power off and check all connections.

I have heard from one reader whose supply was 35 Volt, not the regulated 12 Volts. This was caused by a short circuited power transistor. Should a short in any of the output transistors, occur, all 6 need to be un-soldered. Check with a multimeter set to resistance and measure between collector and emitter terminals. Power transistors usually fail short circuit so should be easy to find the faulty one.

12V SUPPLY UNIT

2011-03-05T05:30:00.001-08:00

The ±12 V supply required by the OP amps is provided by the 12V supply circuit arrangement. The circuit diagram of ±12V power supply unit is shown in figure. Initially 230 V AC supply is reduced to (15V-0-15V) with the help of a step down transformer having a capacity of 1A and the center tap of the transformer is grounded. This low voltage is rectified with the help off bridge rectifier. Since the input voltage to the regulator IC should be more than its output voltage, transformer secondary voltage is 15V-0-15V.The ripples are minimized with the help of capacitor filter to get a smooth DC supply. The rating of the chosen capacitor filter is 1000µF.
The regulated DC output voltage is obtained by using regulator ICs. For regulated +12V DC supply, IC 7812 is used and for regulated -12V DC supply, IC 7912 is used. In the case of IC 7812 the unregulated DC voltage is applied to Pin 1, and the output is taken at Pin 3 and Pin 2 is grounded. In the case of 7912 ,the unregulated DC voltage is applied to Pin 2, the output is taken at Pin 3 and Pin 1 is grounded. The pair of capacitors of 10µF is connected at the output as shown in figure to eliminate the voltage oscillations at the output due to the large voltage oscillations at the input of the regulator.

12V switching power supply by LT3844

2011-03-05T05:26:00.000-08:00

This 12V switching power supply electronic circuit is based on LT3844 DC DC controller circuit that can be used for medium power devices . The LT3844 IC offers a wide 4V to 60V input range (7.5V minimum start-up voltage)and can implement step-down, step-up, inverting and SEPIC topologies.
This power supply circuit will provide a 12 volts output voltage with a maximum current of 4 amperes . This 12V switching power supply electronic circuit requires an input voltage between 15 and 60 volts . The maximum output power that can be supplied by this circuit is around 50 watts .
D1,D2 can be an 1N4148 type and D3 can be an PDS5100H type .
The T1 transistor used for this project must be an Si7850DP type .
If you want to see more details about how you can to design a power supply circuit using the LT3844 IC ,please consult the manufacturer datasheet .

12V DC 500mA using NCP4629 low drop out regulator

2011-03-05T05:24:00.000-08:00

This 12V DC 500mA electronic project is based on the NCP4629 low drop out regulator manufactured by ON Semiconductor . This NCP4629 is a CMOS 12V DC 500mA low drop out regulator which features a high input voltage range while maintaining a low quiescent current. Several protection features like current limiting and
thermal shutdown are fully integrated to create a versatile and robust device.
The NCP4629 regulator is available in three output voltages configurations that will provide an output voltage range between 3 and 12 volts .
A high maximum input voltage (36 V) and wide temperature range (−40°C to 105°C) makes the NCP4629 an ideal choice for high power industrial applications.
A 470 nF ceramic input decoupling capacitor should be connected as close as possible to the input and ground pin of the NCP4629.
A 10 uF ceramic output decoupling capacitor is sufficient to achieve stable operation of the IC.
The enable pin CE may be used for turning the regulator on and off. The IC is switched on when a high level voltage is applied to the CE pin.
The IC includes internal thermal shutdown circuit that stops operation of regulator, if junction temperature is higher than 160°C. After that, when junction temperature decreases below 135°C, the operation of voltage regulator would restart.

5Vdc 2A circuit with LM317

2011-03-05T05:08:00.000-08:00

The electronic circuit for DC voltage 5-12 Volts or -12Volts regulator isolated from double LM317 IC-12V. Power supply for 4V to 6V, current 0.3A.

Details of this circuits.

A power supply of low voltage 5V. function of regulation by LM317 When +5VDC want but the total of the input voltage is 5.1V only slightly.

Heavy possible that we see in this fuel system low-voltage devices to regulation by the LM317.

The circuitry by low job control have low voltage drop in the circuit a lot. The important point for this new type of circuit use IC LM317 very popular. Then make sure that no heavy equipment safely. If the circle of thinking, the use of LEDs ensures that a voltage regulated then mainly in the direction of the work of the racetrack. Normally, this route, the input voltage of 5.1V to 12V. Q1 TIP42 also improve current tallly carry up to 2A. Completion of this course is really useful, a friend of the joy of the Lord Mail.

Applications
1.Flyback controller
2. Multiple-output regulator
3. Simply promoting regulator
4. Forward Converter

5V Dc converter using ADP2303 ADP2302

2011-03-05T05:03:00.000-08:00

A very simple and high efficient DC-DC 5 volts power supply circuit can be constructed using ADP2302 or ADP2303 nonsynchronous step down regulator , designed by Analog Devices .
Both of this circuits support a wide input voltage range from 3 volts up to 20 volts .
The difference between this two fixed frequency , current mode control, step down regulators is that ADP2302 support a maximum load current of 2A and the ADP2303 regulator support a maximum load current of 3A .

This schematic circuit is a step down dc converter that will provide a 5 volts output voltage at a maximum 2A load , from a 12 volts input voltage .
The output voltage of the ADP2302 , ADP2303 can be set down to 0.8 V for the adjustable version, while the fixed output version is available in preset output voltage options of 5.0 V, 3.3 V, and 2.5 V.
This circuit use a 700 kHz operating frequency that allows small inductor and ceramic capacitors to be used, providing a compact solution.
For this DC DC converter circuit we will use the ADP2303ARDZ-5.0 which is the ADP2303 fixed voltage 5 volts version .

The recommended vales for the Shotky diodes that can be used for this project are : Vishay SSB43L or SSA33L , ON Semiconductor MBRS330T3 , Diodes Inc. B330B .
If you need to choose another part for the coil , please read the datasheet to see the recommended parts .

5V DC power supply

2011-03-05T04:13:00.000-08:00

A regulated 5 volt DC supply is essential for powering micro-controller and TTL based circuits. The output of most wall-warts and adapters is to rippled and impure for use in digital circuits. Lets build an inexpensive power supply using some discrete components and a fixed voltage regulator IC.

You will need:

a step down transformer [12V]
four silicon diodes [1N4007]
a resistor [47Ω]
capacitors [2 x 220µF, 0.3µF and 0.1µF]
three terminal voltage regulator IC [LM7805]
a small general propose PCB, some wires, and a suitable output port (I use a
audio connector).

The circuit of the supply is as given below:

The circuit consists of three main blocks, the rectifier, filter and regulator. The rectifier is used to transform the mains AC voltage to a suitable DC voltage. The output of the rectifier is however an impure DC signal so we use a filter to clean the signal and finally a regulator to deliver precisely 5 volts, irrespective of the load connected to the output.

Rectifier

It consists of a transformer and a diode bridge. The diodes are standard silicon 1N4007 diodes. We have chosen a 12V transformer because the regulator IC needs at-least 7.5V of input voltage to function properly.

Filter

We use a pi-filter here. The two 220µF capacitors and resistor form the filter. Pi-filters are great for light load applications like digital circuits. Be sure to check the polarity of the capacitors before connecting them. The resistor shown above is one rated to dissipate up-to 5W of power across it. You may use smaller 3W resistors, but availability may be an issue. Don't use the tiny 0.25 W or 0.5W ones though.

Regulator

The voltage is regulated by three terminal voltage regulator IC – LM7805. This regulator provides stable 5V DC output against large fluctuations in input voltage and load. It also has internal protection circuits which 'brownout' the device when overloaded. To decide the pin-outs, hold the regulator with its face towards you and legs pointing upward, the pin to the right is the input, middle pin is ground and left most pin is output.

Questions and suggestions are welcome, please use the comment form below.

1.2 - 15V/3A adjustable DC power supply

2011-03-05T03:59:00.001-08:00

Specifications


This kit provides a variable output power supply ranging from 1.2 to 15 V @ 3 A. It uses Low Dropout Positive Regulator LM1084 in TO220 package for delivering variable output voltage. Input - 18 VAC/DC Output - variable output from 1.2 ~ 15 V @ 3 A Regulated low ripple DC voltage Heatsink for regulator IC On board bridge rectifier to convert AC to DC LED indication at input of IC Zener trimmed band gap reference, current limiting and thermal shutdown (provided by IC feature) Power Battery Terminal (PBT) for easy input and output connection Onboard PCB mounted Potentiometer (POT) for varying the output voltage Filter capacitors for low ripple DC output Four mounting holes of 3.2 mm each PCB dimensions 46 mm x 58 mm

Schematic

Parts List

PCB

Adjustable power supply using LM317

2011-03-05T03:53:00.001-08:00

I +-------+ O
    Vin (+) o-----+---| LM317 |---+--------------+-----o Vout (+)
                  |   +-------+   |              |
                  |       | A     /              |
                  |       |       \ R1 = 240     |
                  |       |       /              |           ___
                 _|_ C1   |       |            +_|_ C2      |_0_|  LM317
                 ___ .01  +-------+             ___ 1 uF    |   |  1 - Adjust
                  |  uF   |                    - |          |___|  2 - Output
                  |       \                      |           |||   3 - Input
                  |       / R2                   |           123
                  |       \                      |
                  |       |                      |
    Vin(-) o------+-------+----------------------+-----o Vout (-)

For the LM317:

R2 = (192 x Vout) - 240, where R2 in ohms, Vout is in volts and must be at between 1.2 V and 35 V.
Vin should be at least 2.5V greater than Vout. Select a wall adapter with a voltage at least 2.5 V greater than your regulated output at full load
Maximum output current is 1 A. Use proper heatsink for LM317 if it has to dissipate more than 1W.
The tab of the LM317 is connected to the center pin.

Equation for calculating the output voltage when R1 and R2 are known:

Vout = 1.25V * (1 + R2/R1) + Iadj * R2

Where Iadj is typically in the range of 50 microamperes.

5V/9V/12V DC fixed power supply

2011-03-05T03:52:00.000-08:00

Circuit Diagram (Schematic Diagram)-1

L1 = Step down transformer with i/p of 230 AC 50 Hz and output of (XX ) - 0- (XX)) volts(rms).
XX = Required DC output voltage.

Here is the table for different voltages

Output voltage (DC Volts)	Transformer rating (rms Volts)
5	230: 5-0-5
9	230:9-0-9
12	230:12-0-12
15	230:15-0-15

The current rating has to be more than 1 Amp.
D1, D2 = Diodes 1N4003
D3 = Diode 1N4003/ 1N4001 (optional)
C1 = 1000 Micro Farad aluminum electrolytic capacitor (For loads less than 100mA you can sustitute with 220 microfards capacitor), Voltage rating = 2.5 times of Output Voltage.
C2 = 10 Micro Farad aluminum electrolytic capacitor
IC1 = 7805 for + 5V DC output
       =7809 for +9V DC output
       =7812 for +12V DC output
       =7815 for +15V DC output

5V/9V/12V fixed power supply (Negative) at 1 Ampere current rating

Circuit Diagram (Schematic Diagram)-2

L1 = Step down transformer with i/p of 230 AC 50 Hz and output of (XX ) - 0- (XX)) volts(rms).
XX = Required DC output voltages.

Output voltage (DC Volts)	Transformer rating (rms Volts)
5	230: 5-0-5
9	230:9-0-9
12	230:12-0-12
15	230:15-0-15

The current rating has to be more than 1 Amp.

D1, D2 = Diodes 1N4003
D3 = Diode 1N4003/ 1N4001 (optional)
C1 = 1000 Micro Farad aluminum electrolytic capacitor(For loads less than 100mA you can sustitute with 220 microfards capacitor). Voltage rating = 2.5 times of Output Voltage.
C2 = 10 Micro Farad aluminum electrolytic capacitor
IC1 = 7905 for -5V DC output
       =7909 for -9V DC output
       =7912 for -12V DC output
       =7915 for -15V DC output

Additional note: It's safer to put one heat sink to 78XXX and 79XX IC for safeguarding the IC from overheating
In case you are using both the power supplies the ground connection of both positive and negative power supplies can be shorted.

5V DC power supply

2011-03-05T03:50:00.000-08:00

Summary of circuit features

Brief description of operation: Gives out well regulated +5V output, output current capability of 100 mA
Circuit protection: Built-in overheating protection shuts down output when regulator IC gets too hot
Circuit complexity: Very simple and easy to build
Circuit performance: Very stable +5V output voltage, reliable operation
Availability of components: Easy to get, uses only very common basic components
Design testing: Based on datasheet example circuit, I have used this circuit succesfully as part of many electronics projects
Applications: Part of electronics devices, small laboratory power supply
Power supply voltage: Unreglated DC 8-18V power supply
Power supply current: Needed output current + 5 mA
Component costs: Few dollars for the electronics components + the input transformer cost

Circuit description

This circuit is a small +5V power supply, which is useful when experimenting with digital electronics. Small inexpensive wall tranformers with variable output voltage are available from any electronics shop and supermarket. Those transformers are easily available, but usually their voltage regulation is very poor, which makes then not very usable for digital circuit experimenter unless a better regulation can be achieved in some way. The following circuit is the answer to the problem.
This circuit can give +5V output at about 150 mA current, but it can be increased to 1 A when good cooling is added to 7805 regulator chip. The circuit has over overload and therminal protection.

The capacitors must have enough high voltage rating to safely handle the input voltage feed to circuit. The circuit is very easy to build for example into a piece of veroboard.

Component list

7805 regulator IC
100 uF electrolytic capacitor, at least 25V voltage rating
10 uF electrolytic capacitor, at least 6V voltage rating
100 nF ceramic or polyester capacitor

Modification ideas

More output current

If you need more than 150 mA of output current, you can update the output current up to 1A doing the following modifications:

Change the transformer from where you take the power to the circuit to a model which can give as much current as you need from output
Put a heatsink to the 7805 regulator (so big that it does not overheat because of the extra losses in the regulator)

Other output voltages

If you need other voltages than +5V, you can modify the circuit by replacing the 7805 chips with another regulator with different output voltage from regulator 78xx chip family. The last numbers in the the chip code tells the output voltage. Remember that the input voltage muts be at least 3V greater than regulator output voltage ot otherwise the regulator does not work well.