# Pass Transistors

Discussion in 'The Projects Forum' started by PCBoy, Jan 23, 2012.

1. ### PCBoy Thread Starter New Member

Dec 14, 2011
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Could someone explain the variables in this equation: (equation in attachment)

I'm solving for R1, what is Ireg and what is Iq1?

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2. ### PaulEE Member

Dec 23, 2011
423
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Ireg is the current entering the input pin on the regulator, and IQ is the current coming out of the output.

IQ is not the same as IO because the difference between the two is DC current from the transistor.

This series-pass transistor configuration allows the regulator to source more output current to a lower impedance load.

3. ### PCBoy Thread Starter New Member

Dec 14, 2011
26
0
Say I were to want 2A to 3A output, does that mean Iq = 2A/3A?

How do I determine Ireg? Where would I get that?

4. ### PaulEE Member

Dec 23, 2011
423
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Suppose the regulator can source 1 amp by itself. This implies that your series-pass transistor will have to source the rest. This usually means that, at approximately 1 amp, you want the transistor to "turn on". This is usually done via a resistor at the base pin that drops 0.7V when the current reaches 1 amp. In this image, the transistor is "on" at approximately 0.7v/3 ohm = 230 mA or so. Note also that the voltage at the base is at a lower potential than the emitter because of the PNP transistor used.

IQ is not 2A/3A. IQ should be a current not exceeding the regulator's max. output current. IQ, then, in our example here, is about 1 amp. IQ + Itransistor = IO...which yields 1A/2A for the Itransistor for your 2A/3A output.

5. ### pilko Active Member

Dec 8, 2008
213
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IQ1 is the transistor collector current. It is shown a littrle too low down on the schematic.

pilko

6. ### PaulEE Member

Dec 23, 2011
423
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OH! Missed that. I thought they were illustrating something else the way it was written.

In that case, everything I said in my last post is true. Basically, the regulator can only source about 230 mA. At that point, the transistor kicks on and picks up the slack. The total output current is equal to the regulator output plus transistor output.

Good to go?

P.S. - I'm not sure what transistor that is, but I'd use a pretty hefty one with a heat sink if you're planning to pull 3A out of it. Possibly a TIP42, etc.

EDIT: Definitely use a different transistor. The one in that diagram does 2A rated collector current. (for the 3A plan, anyway)

7. ### PCBoy Thread Starter New Member

Dec 14, 2011
26
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I'm planning on giving an negative regulator a current boost.

I have a 7915 regulator and I hooked it up with an NPN power transistor. It's set the same way as the illustration only this time it's for a negative regulation.

My NPN transistor is MJE10035A, says on the datasheet it's a 4A transistor.

Currently I'm randomly testing out low resistors by trial and error on my setup.

I'm able to light a 12V 25W bulb with my 7915 regulator hooked with my transistor with a resistor of 2.2ohms. Problem is, I get a voltage drop of 2V. I would prefer if it the drop was 0.5 max. Any thoughts what the problem may be?

Thanks for the help so far!

8. ### PCBoy Thread Starter New Member

Dec 14, 2011
26
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On a side note, I'm also getting -14.6V instead of -15. Could bad soldering be a reason?

Could also this low voltage be a reason for poor current boosting?

9. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
Is your negative supply going into the regulator capable of a 3A load?

10. ### Ron H AAC Fanatic!

Apr 14, 2005
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A Google search for MJE10035A comes up with nothing. Did you mistype the part number of your NPN?

IQ1 is the transistor collector current. It is not the current coming out of the output.

Dec 14, 2011
26
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12. ### Ron H AAC Fanatic!

Apr 14, 2005
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What is the value of your input voltage?

13. ### PCBoy Thread Starter New Member

Dec 14, 2011
26
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Around 30-33 volts.

14. ### Ron H AAC Fanatic!

Apr 14, 2005
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If you want 2 or 3 amps, your transistor will be dissipating around 30 to 50 watts. Removing the heat will not be a simple task.

15. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
Have you looked into switching power supply options?

For the current, you'll probably be looking at an external MOSFET switch, and the inductor won't be very cheap, but it would get your supplies stable with the current required without 100 Watts of heat to dissipate from the power supply alone.

16. ### PaulEE Member

Dec 23, 2011
423
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Alternatively, start with a lower input voltage such that the difference between the output voltage and input voltage is minimal; this will prevent the transistor from dissipating prohibitive amounts of power