7812 IC 12 Volt 30 Amp PSU Circuit

Using a distinct 7812 IC voltage regulator and assorted outboard canyon transistors, this ability accumulation can bear achievement amount currents of up to 30 amps.

Notes:

The ascribe agent is acceptable to be the best big-ticket allotment of the absolute project. As an alternative, a brace of 12 Volt car batteries could be used. The ascribe voltage to the regulator charge be at atomic several volts college than the achievement voltage (12V) so that the regulator can advance its output. If a agent is used, again the rectifier diodes charge be able of casual a actual aerial aiguille advanced current, about 100amps or more. The 7812 IC will alone canyon 1 amp or beneath of the achievement current, the butt actuality supplied by the outboard canyon transistors. As the ambit is advised to handle endless of up to 30 amps, again six TIP2955 are active in alongside to accommodated this demand. The amusement in anniversary ability transistor is one sixth of the absolute load, but able calefaction biconcave is still required. Best amount accepted will accomplish best dissipation, so a actual ample calefaction bore is required. In because a calefaction sink, it may be a acceptable abstraction to attending for either a fan or baptize cooled calefaction sink. In the accident that the ability transistors should fail, again the regulator would accept to accumulation abounding amount accepted and would abort with adverse results. A 1 amp agglutinate in the regulators achievement prevents a safeguard. The 400mohm amount is for analysis purposes alone and should not be included in the final circuit.

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 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 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.

author:
e-mail:
web site: http://www.mitedu.freeserve.co.uk/