A Question regarding RG's germanium transistor testing setup

[trjones1]

First of all, thanks for this germanium transistor testing method, RG.

I tried as hard as I could but the closest I could get to 2.472k ohms for the resistor was 2.35k.  When I'm figuring the leakage I divided the voltage across that resistor by 2.35k to find the actual leakage current.  My question is where does the 2.472k value come from?  Is that a constant coming from somewhere or is that just the value you need so that you can get gain readings by multiplying the difference in voltages by 100?  How does using a different value of resistor affect the measured gain?  The main thing I'm wondering about is this equation from RG's article: "(4uA)*(100)*(2472) =  0.9888V - which is almost exactly 1/ 100 of the actual gain."  If I figure out the actual current flowing into the base depending on the actual voltage of the battery and the exact value of the 2.2M resistor, should I use this equation with 2350 in place of 2472 because of the value of the resistor I used, or do I need to use 2472?

Finally, if we can figure out the leakage of the transistor, and therefore know what “false” gain this would cause a dmm to read, can we just measure the transistor on a dmm and subtract the false gain that we have measured?

I tested 10 transistors last night and got hfe's from the high 30's to the low 50's and I'm wondering if these might be a little low because of my testing setup or because this batch just happened to be low gain.  I realize that even with any inaccuracies from the testing setup taken into account that these transistors won’t get close to the ff range, but still…

Thanks in advance for any light anyone could shed on this.

[cd]

AFAIK you can use Ohm's law (V=IR) to unravel everything.

With an exact 9V supply and 2.2Mohm resistor, that's 9/2.2M = 4.1uA current.

Say you have a transistor with a gain of 100, and put 4.1uA into the base, and you want the V to be a nice round number (1V) to make it easier for later calculations (just multiply by 100 for the true gain without leakage) - what resistance do you have to use? So:

R=V/I
R=1V/(4.1uA * 100) (transistor current gain)
therefore R = 2444.4 ohms

What's the closest standard value to that?  2400 ohms.

If you have a 2350 ohm resistor, an exact 9V supply, and exact 2.2M resistor, you'd be off 100x slightly (instead of 1V w/4.1uA and a 100hfe transistor with no leakage you'd get .9635V or 1/103.7 the true gain).  

So say your transistor, with the 2350 ohm resistor, leaks 100uA for a voltage reading of .235V.  Close the switch and it reads 1.5V.  Subtract .235V from 1.5V = 1.265V * 103.7 = 131hfe.  Close enough for rock n' roll?  Well say you mutliplied by 100 instead of 103.7 - you'd get a gain of 126 - off by 5.  Close enough!

Here's an example with my own setup.  My 9V power supply puts out 8.88V, my 2.2M resistor is 5% tolerance (really 2.283M) and my 2.4k 5% resistor is 2.403k.

8.88V/2.283M = 3.89uA
ideal 100hfe transistor with no leakage: 3.89uA*100*2403=.9346V, or 1/106.9th true gain
voltage reading w/3.85uA (no switch) = .265V
voltage reading w/3.85uA (switch) = .531V
true gain is .531-.265 = .266 x 106.9 = 28

Had I just assumed 4ua and gone with the voltage readings: .266 x 100 = 27 close enough!!

[RLBJR65]

Great explanation cd  

You can do what I did. Get a precision[/b] 10 turn 5k trim pot. that way you will be able to adjust it if you are using an unstable power supply like a 9v battery. Just measure the battery voltage do the math and adjust.