Testing Gain on NPN Ge transistors

[soggybag]

I bought a lot of a AC128's. These are Ge NPN types. I want to test them for gain using the method on Geofex. I'm guessing I can swap the Emitter and Collector for NPN type transistors and test as described, or is there something I'm missing?

[greaser_au]

AC128?  aren't they PNP?  

but to answer your question: Because of the way a  transistor is made: effectively (for PNP) emitter-PN-base-NP-collector, if you swap the emitter & collector, you don't swap the polarity  - the DUT will still behave as a PNP transistor, albeit with different characteristics because of the differences in the way the collector and emitter are doped.

The way to test NPN transistors with this is explained somewhere on the same page at geofex. Basically you need to change the circuit polarity completely.
* reverse the DMM leads
* reverse the  battery
* use an NPN transistor

david

EDIT: it's explained in the last paragraph!

[tonyharker]

AC128 are PNP,  AC176 are NPN!

[soggybag]

Thanks for the plies, my error the transistors in question are AC176.

[soggybag]

A question and some observations on testing gain using the Geofex method.

I notice that the leakage tends to start high and and drop over time. The same goes for gain. It can take as much as 30 secodns for the numbers to settle. Should I wait for the numbers to settle is the more accurate value?

[reveal]

Yes, let them settle.  Germaniums are very sensitive to heat, even the heat from your hands.  After you get the reading on one and it settles, touch it again and see how the readings change.  The heat sensitivity is mentioned in the GEO article.

[soggybag]

Thanks for the reply. I want to make sure I got the math correct from the Geo article.

Measuring the leakage I take the voltage measured across the 2472 ohm resistor and divide by 2.472 to get milliamperes. Then switch on the base and measure voltage again to get gain. Gain should be the voltage across the same resistor less the leakage calculated above.

[rockhorst]

AC127 would be NPN too. You can let the numbers settle in, but you'll quickly find that the gain measurement doesn't change a lot over time and the leakage measurement will drop a little as it cools down, but that effect is only little. So say you measure 1.2V as gain and start off with 0.3V for leakage, the real gain would seem 90. A minute later it might be 1.2V - 0.25V = 0.95. Guesstimate it after you've gotten some experience with the measurements and allow for a 5-10% error. Will save a lot of time.

[soggybag]

Thanks for the reply, sounds like you just subtract the first voltage (leakage) reading from the second (gain). Then find the amount of leak by dividing the first reading by 2472.

For example, if I measured 1.9v leakage and 2.5v gain, the adjusted gain would be 60, with 768 micro amps leakage?

[ECistheBest]

Gain or hfe for transistors is given as: Ic/Ib. divide the collector current by the base current. what i'd do is have a 2M2 resistor on the base, and a 470R resistor on the collector, both going to 9v, and the emitter going to ground.

collector current is [9v - Vc]/2M2

base current is [9v - Vb]/470R

divide Ic by Ib, and u get hfe.



"gain" in this case is current gain. there is supposed to be current into the base. that's what BJTs do. current ratio between the collector current and base current is the "gain" or hfe of a BJT.