|
|
|
12 Volt 30 Amp Power
Supply
|
|
Regulated Power
Supply |
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 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
maximum 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.
|
|
|
