Electronics Voltage Regulators

How to drive high curret system with voltage regulator ?

Lets say we need 2A and the power regulator gives only 1A …

Easy Solution

Put two voltage regulators in parralel and thats all. It is not recommended due to the fact that they will have different load and will heat up differently, but if you connect a proper heat sink, then all will be fine.

Hard Solution

Use PNP Transistor (BD536) as explained here (THE DATASHEET): here page 17 figure 18.
The explanation is that you connect:

  1. the Emmiter to the Vin (e.q 12V)
  2. the Base to VinRegulator
  3. the Collector to Vout Regulator
  4. put a resistor between Emmiter and Base (recommended by the book is 3 Ohm)

For fun I tried to think on how iit actually works ….

  • So the whole point is that the PNP transistor opens up when the voltage on the Base is pulled low relative to the emitter (wikipedia).
  • Therefore when too much current is taken by the regulator, a voltage on the Resistor is developed, which opens up the transistor, which boosts the output current.
  • the equations as shown in THE DATASHEET, are easy, the only thing which is not clear is how to calculate the Iq1/Bq1. By thinking on the equation (R1=) it seems that the vaule is actually the current on the Base.
  • Q1 seems to be the Hfe of the transistor.

Lets try to apply math and find the values for the Resistor. Lets assume that we need 2A current with 5Voutput and the power source we have is 12V. The voltage regulator we will use is 7805.

  • To find Base current we apply the following equation : Total output current = Current on Regulator + Current Collector, which is 2Amp=1Amp (max?) + 25(Hfe from datasheet )*0.04A (calculated).

The resistor equals to : now I am a bit hesitant … because I do not know if my thoughts are right … here is my point :

In my case, the Base current is 0.04A … plus 1A on the voltage regulator, makes it 1.04A on the Resistor.
Which should be around 0.8V on the Vbe (taken from the transistor performance graph) … Therefore R=0.8V/1.04=0.77Ohm …

But, lets assume that we can put a higher resistor, for example 3Ohm … as suggested by THE DATASHEET, then it seems that all will happen is that the transistor will get into the saturation earlier and most of the current will fail on the Transistor and not on the Voltage Regulator … and since the transistor is capable of 6Amps, the output of the circut is 6 Amps !!! wow ;)

Do you think that this makes sence ? please drop me a line !

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