12V 20Amp Power Supply by IC 7912 + 2N3055
Here is 12 volt power supply 20A , USE LM7912 and 2N3055 .
It easy circuit , Low cost. see detail more in image.
This power supply is the result of the use of components at hand (junk box). The core of the project is a number of 2N3055 transistors and a transformer with a 16 volt 18 amp secondary winding.
As it is constructed using the parts I had available at the time it may not be the most cost effective design if all parts are purchased new.
Circuit
Rather than use a more conventional arrangement with a positive supply it was easier to use a negative supply and make use of a 7912 negative regulator as the voltage control. Then by using available NPN transistors to boost output, an output up to 20 amps can be obtained at 12 volts.
Similar circuits : Regulator 12V 10A by IC 723+2N3055 , Power Supply 13.8V 30-40A with LM723 and BUZ24 or IRF150 , 12 Volt Low Voltage Droput Precision Regulator with LM324 , 12V Switching Car PSU by UC3843 + 74LS02 , 12V 5A power supply Regulator with LM2678-12
Read More Source:http://radarclub.tripod.com/Homebrew.htm
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Categories: High current power supply, Power supply Tags: 12 volt power supply, 24v to 12v converter circuit, High current power supply, LM7912
12 Volt Low Voltage Droput Precision Regulator with LM324
Most regulator circuits require at least 3 to 5 volts between input and output to operate reliably. This regulator features a P-Channel MOS-FET pass element that will allow for a very low drop out voltage. In my particular application reliable operation was noted down to nearly 0.1 volt between input and output at nearly 0.8 amperes of load current. Useful applications include automotive and battery applications were regulation is desired and available input-output differential is less than 0.5 V. By using the LM324 Op-Amp and the LM336 Precision Reference Diode, both single supply operation and precision output regulation were easily achieved.
I originally designed this for a 12V camera. The camera was a surplus property item and contained only the camera and a small adapter box. The 1″ x 2″ 1″ adapter box contained a connector from which video, audio and power adapter connectors were attached. My intent was to use a 12V 1 amp adapter from Radio Shack to power the camera. Unfortunately the adapter provided an unregulated 15V under no load and slightly more than 12.3V under full load. This necessitated the use of a very low drop-out regulator circuit. By using a MOS-FET I could minimize the drop-out voltage and have the added benefit of very low power dissipation. This design allowed for minimal heatsinking and could be packaged very tightly in nearly any small enclosure. In my final design the entire regulator circuit fit easily in the 1″ x 2″ x 1″ adapter case.
Read More Source :http://www.drbob.net/project/powersupply/linear/12v1alvdfxd/default.html
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Categories: Power supply Tags: 12 volt power supply, lm324 regulator circuit, voltage dropout regulator
High Current Power Regulator 12Volt 30Amp by 7812+2N3055
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:
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 ICwill 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:
similar circuits : Regulator 5V,6V,9V,12V 1A By IC 7805,7806,7809,7812 , Regulator 12V 10A by IC 723+2N3055 , 12 Volt Low Voltage Droput Precision Regulator with LM324 , 12V Switching Car PSU by UC3843 + 74LS02 , 12V 5A power supply Regulator with LM2678-12 , power supply Switching Regulator 12V 3A by LM2576-12
Power Supply 13.8V 30-40A with LM723 and BUZ24 or IRF150
Calculations:
This circuit is a fine example of Kirchoff’s current and voltage laws. To summarise, 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 theTIP2955. 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 Kirchoff’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.
source: http://www.mitedu.freeserve.co.uk/
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