Adjustable Voltage Regulator Advice

[BrianG]

I am making a simple 1.25v to 12v adjustable voltage regulator capable of 3 amps. The input is 18v DC. I came across a circuit (below) but I have a question about the resistor between the base and emitter of the PNP pass transistor.

As I see it, the 317T will be supplying the power to the load until the current through the 33 ohm resistor exceeds 21mA (assuming a .7 Vbe on the PNP). At that point the voltage drop across the 33 ohm resistor will be enough to turn the transistor on and do the work. However, I happen to have a TIP42 PNP transistor with the following specs: Ic=6A, Ib=2A, hfe (when Ic=3A) is 15 minimum. Assuming the min current gain of 15, that would make the maximum current allowed from the pass transistor to be 318mA correct? (21mA x 15) This is not even close to 3A.

So, I was planning to use a 3 ohm resistor in place of the 33 ohm resistor. That value would require 233mA of current to flow through it before it develops enough of a voltage drop (.7v) to turn on the pass transistor. Then, 233mA times the current gain of 15 would allow for 3.5A from the pass transistor.

I would like to know if I am correct in my assumptions of how the PNP pass transistor is biased and being used. Also, are the adjustment resistors correct for the output voltages I am looking for?

It's been a while since I've worked with electronics and want to be sure I'm calculating correctly.

I've already taken into account the maximum heat dissipated across the 317T and the pass transistor and have heat sinked them accordingly, and the device tabs have been insulated from the heatsink. The 3 ohm resistor is a bit overkill at 5W, but it's what I have. The adjustment resistors are a 1/4W 470 ohm, and a 5k pot in parallel with 33k ohm for a 4k ohm pot.

**broken link removed**

Thanks in advance for your assistance, and sorry if this is a simple question...

 

[eblc1388]

So, I was planning to use a 3 ohm resistor in place of the 33 ohm resistor. That value would require 233mA of current to flow through it before it develops enough of a voltage drop (.7v) to turn on the pass transistor. Then, 233mA times the current gain of 15 would allow for 3.5A from the pass transistor.

This is where your calculation have gone wrong. The 233mA is not supposed to flow along the 33R resistor, it's path is from E of PNP to its base. You will have a forward biased diode junction when the voltage across the E-B is higher than 0.65V.

So what you want is 233mA flowing out of the PNP base to turn it ON and supplying bulk current to the load. You do that by using a relatively high value resistor(33R) such that LM317T is forced to get large part of its input current via the E-B junction of the PNP to turn PNP on to provide large portion of the load current.

However, sometimes designer would use a smaller value resistor if he wants the whole setup to have thermal dissipation limiting when the designed load current is reached.

For the resistor values, I would suggest to use a 560R(to replace 470R) and the 5K pot with no parallel resistor. Your original resistor values can't reach 12V(~11.89V max) when used on the LM317T.

 

[Roff]

Here is an annotated schematic. I think 0.7V is too low for Vbe@Ic=3A, so I assumed Vbe~1V, although the difference in current is not significant.
Keep in mind that you probably won't want to draw 3A with Vout=1.25V, because your transistor dissipation will be around 50 watts. This will require an almost infinite heat sink.

 

[BrianG]

OK, I understand, I think. Just one more question; The current through the 33 ohm resistor is easy enough to understand, but what determines the 200mA through the E to B of the transistor?

Yeah, I figured I wouldn't want to draw too much current at 1.25v because of the heat. The heatsink is pretty large (CPU heatsink), but definitely not infinite.

Thanks eblc and Ron...

 

[eblc1388]

OK, I understand, I think. Just one more question; The current through the 33 ohm resistor is easy enough to understand, but what determines the 200mA through the E to B of the transistor?

Beta of the pass transistor. Higher beta, lower base current for a certain collector load.

 

[BrianG]

Doh!

You know, for some reason I have trouble when looking at circuits with PNP transistors (rather than NPN), I don't know why. Now that you said that, it all makes sense; the base current is determined by the load current / beta.

Thanks. Boy, sometimes I feel dumb!