Testing Transistors?

Started by bluefireamps, December 25, 2009, 01:40:40 PM

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bluefireamps

I wanted to start testing germanium transistors for gain and leakage so I fabbed up a fixture based on RG's FF test rig. I'm wondering if it's "normal" for the leakage test numbers to start out high and slowly go down after about ten minutes. When I put a transistor into the socket my meter first reads about 700 or 800 millivolts and slowly starts to drop to around 300 millivolts, depending, of course, on the transistor.
In the text of RG's description of the test rig it alludes to being able to directly calculate the gain by subtracting the leakage gain from the total number when you connect the 2.2M resistor to the base. It doesn't exactly come right out and say that. I assume that is correct?
Thanks for the help and Merry Christmas.

Dave

jdub

Germanium trannys are very temperature-sensitive, and the readings on your meter will fall as they cool down from handling.  I've found that picking them up by the leads w/ pliers reduces the temp change- you can also wear gloves when handling them.  R.G. Keen recommended fanning Ge transistors with a piece of paper or something while they're in the socket for testing- makes them reach a stable reading somewhat quicker. Seems to work.
A boy has never wept nor dashed a thousand kim

bluefireamps

jdub,
Thanks! I kind of figured that, but I wasn't 100% sure. Any idea about directly calculating the gain?
Dave

jdub

QuoteWe chuck the thing in the socket, and read (93uA)*(2472) = .229V. Then we press the switch, and read 1.330V. To get the real gain, we subtract 0.229V from 1.330V and get 1.101V. The true gain is just 100 times the reading. 

My understanding is that this procedure allows one to take voltage readings on the meter which are scaled to current gain.  For example, you put a tranny in the socket, allow it to stabilize, and get a reading of say .250 volts (250mV).  This is the false leakage gain. Then, connect the 2M2 and take the reading again.  This time you might get, say, .880 volts (880mV).  Subtract the first number from the second and you get .620 which, when multiplied by 100, gives you a gain of 62.

I hope this understanding is correct, 'cuz that's how I've been doing it!
A boy has never wept nor dashed a thousand kim

bluefireamps

jdub,
That's what I got from reading RG's explaination. He doesn't exactly come right out and say that so I was unsure. Thanks a bunch!
Dave