Increasing Regulator Current

Circuit : Andy Collinson
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An outboard pass transistor used to increase the current output of a voltage regulator IC. Notes
Although the 78xx series of voltage regulators are available with different current outputs, you can boost the available current output with this circuit. A power transistor is used to supply extra current to the load the regulator, maintaining a constant voltage. This transistor is known as an outboard bypass transistor.

Currents below 600mA will flow through the regulator. Above 600mA the input current flowing through the 1 ohm resistor develops a voltage. As this voltage increases above 0.6V (600mA through 1 ohm) then the TIP2955 power transistor starts to conduct, supplying the extra current to the load. The 10 ohm resistor limits excessive base current. The power transistor requires an adequate heat sink as it is likely to get very hot. Suppose you use a 12v regulator, 7812. The minimum input voltage should always be a few volts higher than the regulator output voltage to allow for voltage drops.

The 1 ohm resistor needs to be rated 3 Watts for load currents up to 3 amp and rated 7 Watts for load currents of 5 amps. As the HFE of a power transistor falls with increased collector current it is not recommended to draw more than 5 amps with this circuit. The 10 ohm base resistor drops less power and a 0.5 Watt resistor can be used at all output currents.

Power Dissipation in Bypass Transistor
Assume a supply of 20 volts and that the load will draw 5amps. The power dissipation in the transistor will be:

Vce * Ic

Vce = Vcc - Vreg

Pdiss = (20-12) * 5 = 40 Watts

It may keep you warm in the Winter, but you will need a large heatsink with good thermal dissipation. If however the input voltage was 15V then the dissipation would be reduced to just 15 Watts. If you want to increase the output current with a negative regulator, such as the 79xx series, then the circuit is similar, but an NPN type power transistor is used instead.

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