Could someone explain the variables in this equation: (equation in attachment)

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

 

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.

 

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?

 

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?

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.

 

IQ1 is the transistor collector current. It is shown a littrle too low down on the schematic.

pilko

 

IQ1 is the transistor collector current. It is shown a littrle too low down on the schematic.

pilko

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)

 

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!

 

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?

 

Is your negative supply going into the regulator capable of a 3A load?

 

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.

 

Apologies, I mistyped.

It's MJE13005A. It's this:

http://www.alldatasheet.com/datasheet-pdf/pdf/172983/UTC/MJE13005-TA3-T.html

 

What is the value of your input voltage?

 

Around 30-33 volts.

 

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.

 

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.

 

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.

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