Slightly noobish question re: matching ge transistors

[Cardboard Tube Samurai]

So I've made a GE transistor tester circuit from here: http://www.geofex.com/Article_Folders/fuzzface/fffram.htm with the hope of matching up some of the 60 GE transistors I bought off ebay. Problem is, no matter how many times I read the part of the article that explains how to test them, I end up stupidly confused. I have left it and come back to it with a fresh mind on several occasions now but still can't make head nor tail of it. Can someone please try and explain it in layman's terms? Y'know, really dumb it down for me.

[niggez]

Hi. I just built the circuit a few days ago to go through my transistor stash.
Basically you put in a transistor (watch out for the correct pins/sockets), then read the voltage on your dmm. Then you push the button, and read the voltage on your dmm again. The first measurement is leakage gain, the second minus the first is your actual gain.
Correct me if I'm wrong! But it should be ok 

[R.G.]

So I've made a GE transistor tester circuit from here: http://www.geofex.com/Article_Folders/fuzzface/fffram.htm with the hope of matching up some of the 60 GE transistors I bought off ebay. Problem is, no matter how many times I read the part of the article that explains how to test them, I end up stupidly confused. I have left it and come back to it with a fresh mind on several occasions now but still can't make head nor tail of it. Can someone please try and explain it in layman's terms? Y'know, really dumb it down for me.

Let me try. The gain in question is the current gain of the transistor. If you put in X current into the base, you will see Y current in the collector, so the gain is Y/X. So if you put in a known current, like 100 micro-amperes, into the base, and measure 10 milliamperes in the collector, the gain is (10ma / 100uA)= (10ma/ 0.1ma) = 100. Simple, right?

If only it were that easy. Germanium leaks some current through even with zero base current. So if you leave the base open, there will be some collector current. It's common for germanium transistors to leak 50uA up to a few milliamperes with the base open. Fortunately, the leakage is kind of constant at a given temperature, so we can do this:
1. measure the collector current with no base current; that tells us the leakage current.
2. measure the collector current with a known, fixed base current into the base; that tells us multiplied base current plus leakage.
3. Subtract out the leakage; that gives us the change in collector current that was only due to the base current.
4. Now we can do the division trick; divide the total collector current minus leakage by the base current, and get a real gain answer.

That's what's really going on. The first test leaves the base open. You are measuring the voltage across a resistor, and we can always tell the current by I = voltage/resistance. This first reading is the LEAKAGE CURRENT. There is no base current, it's open. So the first reading is the voltage which leakage current causes. The higher, the worse.

When you press the button, the base resistor causes about 4 micro amperes of base current to flow. This is multiplied in the transistor by its current gain, and that much current adds to the leakage current. We want to know what the DIFFERENCE in the first reading (the leakage) and the second reading (the leakage plus the gain current) is.

Since I knew the base current would always be 4microamps, I eliminated some of the math for you, and scaled the test resistor so that the voltage across the test resistor tells you the gain.

Here's an example. Pretend I have a transistor with 102uA of leakage and a gain of 82. I put it into the test socket, making sure to get the leads in the right holes.
First step: read the meter: it leaks 102uA, and that causes a voltage of 102uA*2472 ohms = 0.252V to appear as read by my meter. I write that down.
Second step: press the button and read the meter again. That causes 4uA times the gain of 82 or 328uA to flow in the resistor, but it adds to the 102uA of leakage we can't get rid of. So the meter now reads 1.063V.
Now for calculating the gain. The good-stuff gain is the difference between the second reading and the first one. That's 1.063v-0.252V = 0.81098V.

And now for the scaling: because I chose that funny 2472 ohm resistor, the voltage you get when you subtract is 1/100 of the real gain. So the real gain is 100*0.81098 = 81.

Oops - we're one off. But that's an error of about 1.2%, and the transistor will drift more than that.

So here's the drill again:
1. Put a transistor into the socket, preferably with a gloved hand or pliers to avoid warming it.
2. Read the meter. Write that reading down.
3. Press the button. Read the meter again. Write that down too.
4. Subtract the first number from the second. (You'll probably come out with numbers between 0.2 and 2.0.) Now multiply by 100 by moving the decimal point two places to the right. So a result of 0.2 becomes a gain of 20; a result of 0.94 becomes a gain of 94, and a result of 2.0 becomes a gain of 200.

All done. Oh, if the first voltage is over 0.74 VOLTS that's a leakage of 300uA; transistors over that much deserve suspicion as being just too leaky.

Did that help, and which part did I obscure?

[Cardboard Tube Samurai]

Well, I guess that just about covers it then. Thank you very much, it was very thorough. I guess the most important part that made the most sense was  this bit:

So here's the drill again:
1. Put a transistor into the socket, preferably with a gloved hand or pliers to avoid warming it.
2. Read the meter. Write that reading down.
3. Press the button. Read the meter again. Write that down too.
4. Subtract the first number from the second. (You'll probably come out with numbers between 0.2 and 2.0.) Now multiply by 100 by moving the decimal point two places to the right. So a result of 0.2 becomes a gain of 20; a result of 0.94 becomes a gain of 94, and a result of 2.0 becomes a gain of 200.

[Cardboard Tube Samurai]

So after testing 6 transistors I have a matched pair, though they are very low gain. Can anyone shed some light on the "ideal" ranges of matched transistors? Should they be exact or as close as possible to?

[joegagan]

i remember reading at geofex in the fuzz face article:

RG Keen wrote:
quote


The right *real* gains are from 70 to about 130. Within that range, people report the best sounds. Some people prefer equal gains, others prefer having a lower gain of 70-100 for the first transistor and 90 to 130 for the second.

unquote

in my experience with fuzz faces i found that i prefer a gain of around 125 for Q2 and the first one doesn't matter too much.
a 2k emitter resistor ( instead of a 1k fuzzpot) and 47 uf cap to ground is cool as well ( collector resistor may need tweaking to get good bias on Q2)

my personal FF preference is it is better to get heavy fuzz , then throttle it back at the front with a 250k pregain control

i actually prefer a 2n3906 si as Q1 anyway, tightens up the bass while still giving a fat heavy fuzz. still sounds like a ge fuzz too.

[Cardboard Tube Samurai]

i remember reading at geofex in the fuzz face article:

RG Keen wrote:
quote


The right *real* gains are from 70 to about 130. Within that range, people report the best sounds. Some people prefer equal gains, others prefer having a lower gain of 70-100 for the first transistor and 90 to 130 for the second.

unquote

in my experience with fuzz faces i found that i prefer a gain of around 125 for Q2 and the first one doesn't matter too much.
a 2k emitter resistor ( instead of a 1k fuzzpot) and 47 uf cap to ground is cool as well ( collector resistor may need tweaking to get good bias on Q2)

my personal FF preference is it is better to get heavy fuzz , then throttle it back at the front with a 250k pregain control

i actually prefer a 2n3906 si as Q1 anyway, tightens up the bass while still giving a fat heavy fuzz. still sounds like a ge fuzz too.

Muchly obliged! Thankyou