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
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.
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