Quick question on testing transistors for FF

[Joe Hart]

I read the GEO "Technology of Fuzz Face" article and I am now testing germanium transistors, but I'm a little unclear on a few things. The article uses some terms that I'm not too familiar with and I searched Google with no luck. Mainly converting mV to V to uA and all that.

So, if I test a transistor and get a leakage reading of 43mV and a total gain reading of 105mV, does that mean that the leakage is about 174uA? And then the total gain would be 105mV - 43mV = 62? So this transistor would be a little on the low side, but still usable?

Thank you!!

-Joe Hart

[brett]

You've almost got it.  The mV of leakage is just that - mV.  It's measured across a 2.5k (approx) resistor, so to get the results in uA (millionths of an Amp), you need to divide the volts by 2,500 and multiply by 1 million.  Anything under 100uA is great for a PNP Ge transistor.  Over about 300 uA and you'd be throwing it away.  (So what's the equivalent critical voltage, I don't hear you ask?  It's about 300*2500/1000000 = 0.75V)  

As you have already worked out, gain is equal to the difference between the leakage and total measurement (leakage+gain).  Although the hfe of a transistor is usually measured as the ratio of two currents, RG has organised the tester so that the readout in V is equal to 1/100th of that ratio.  So if you've got 0.2V of leakage and 1.2V total, the true gain is (1.2-0.2)x 100 =100.  

In summary, with a good transistor you'll see around 0.1 to 0.6V of leakage (ie <200uA leakage) and about 1V more than that (ie 1.1 to 1.6V) of leakage+gain voltage (ie gain of about 100) .  

Get back if you have more questions.

Have fun!