Transmogrifox's germanium testing equation...

[brad]

EDIT: I'M A DOOFUS  :shock:

Scroll down  :oops:

[davebungo]

You have forgotten to take into account the 1K in the collector.
edited Oops , I fired too quickly.

[davebungo]

((8.29-1.12)/1000)/((8.32-4.08)/1000000) = 1691 - seems rather high.  Perhaps you need to measure across the 1Meg base resistor and increase the other base resistance (to decrease base current) by a smaller factor (say 100K).  You also need to ensure that the transistor is biased in the correct place i.e. collector roughly middle of the supply and only adjust the base resistance to cause a medium deviation in collector voltage say 1 to 2V i.e. make sure the tranny isn't going into cut-off or saturation.

[brad]

Um...this is really REALLY embassing  :oops:

I think I had the...TRANSISTOR BACKWARDS *NOOB ALERT*

I redid the whole thing with RG's proper instructions and found that:

My 2K4 resistor = 2390 ohms
My 2M2 resistor = 2.14M
My battery = 8.96v

8.96V / 2.14M = 4.18uA
ideal 100hfe transistor with no leakage: 4.18uA*100*2390= 0.999v or, roughly 1/100th of true gain.
voltage reading w/3.85uA (no switch) = 0.13v
voltage reading w/3.85uA (switch) = 0.22v
true gain is .22-.13 = .09 x 100 = 90

hfe = 90  

Now the only thing I'm not certain about is working out the leakage.  Is it:
V/R = I
0.13v/2390 = 0.000054
I= 54uA ?

[brad]

btw, thanks for the help even though the original info I posted was incorrect, davebungo...  

[cd]

Actually if your voltage difference is .09V, the gain is 9, not 90.  Leakage is the V with no switch (no current into base).  If that's .13V, it's leaking 54ua.  Not leaking much, but the gain is too low to be useful.  With RG's method you want a difference of at least .75V of a volt.

[brad]

A gain of...9?  :?   I hope the rest aren't that crappy.  The transistors I'm testing are Phillips AC125's with a black spot on the emitter.  Would transistors with such low gain be of any use at all (except as diodes?)

btw, thanks for pointing out that 0.09 x 100 = 9 and not 90, cd!  Doh again.

The spots on germanium transistors ARE the emitter...right?  :twisted:

[davebungo]

I'm a little confused now...can we review things a little.  Just to confirm, you have a 2M2 resistor (Rb) from +9V to the base, and a 2K4 resistor (Rc) from +9V to the collector, and the emitter is connected to ground, right?

You need to adjust the base current to achieve something near 4.5V on the collector - this is not critical, you just need to ensure a good operating point before attempting to measure hfe.  To adjust the base current, you could adjust the base resistor value until you get a reasonable collector voltage.

Measure either side of Rb and Rc and take the difference for each resistor.  You now have VRb1 and VRc1 from which you can work out the current in each resistor at the chosen operating point Ib1 and Ic1.  

Then adjust the base resistor value by an amount which causes a reasonable deviation in the collector voltage, say 1 or 2V - amount not that critical.  Measure again VRb2 and VRc2, and calculate Ib2 and Ic2

You now have 2 sets of readings.  

Calculate hfe as (Ic2-Ic1)/(Ib2-Ib1) ensuring that you use the same units for current in each case.

You should now have a reasonable measurement of hfe.

Incidentally, may I ask why hfe is critical?  Are you matching devices?  Does your circuit provide enough negative feedback in the transistor stage to lessen the effects of individual tranny hfe?

[cd]

I'm a little confused now...can we review things a little.  Just to confirm, you have a 2M2 resistor (Rb) from +9V to the base, and a 2K4 resistor (Rc) from +9V to the collector, and the emitter is connected to ground, right?

If you're talking about RG's testing method, no.  With a PNP germanium transistor, E to +9V, 2M2 from B to ground, 2k4 from C to ground.  Measure the voltage drop across the 2k4 resistor with the 2M2 resistor connected, then disconnected.  Take the difference and multiply it by 100 for the transistor's true gain without leakage.

As for transistor dots - I've seen them both ways, the dot marks C or the dot marks E.  Stick it in the Hfe checker in your multimeter to verify which lead is which.

[brad]

Thanks for the info.  Germanium transistors are kinda new to me  :oops:

For some back-story into this whole morass:  I basically bought 10x AC125 PNPs and 10x AC127 NPNs and want to seperate the useful ones with usable gains from the leaky ones.

...and it turns out that the dot is the collector for both types of transistors.  Which means my calculations are out of wack for the millionth time in a row.

hfe:
voltage reading w/3.85uA (no switch) = 1.51v
voltage reading w/3.85uA (switch) = 2.67v
true gain is 2.67 - 1.51 = 1.16 x 100 = 116
hfe = 116

Leakage:
V/R = I
1.15/2390 = 0.0006317991631799164
= 631uA

So my hfe is good, but the transistor is too leaky.

Btw, thanks for replying to my thread even though it's a shambles.  This has been a learning experience.