How to Calculate Transistor Leakage?

[fuzzymuff]

Ok....I'm lost.  I've read R.G. Keen's article on sorting out transitors, http://www.geofex.com.  I made a breadboard rigging his transitor tester, and got this reading with one PNP AC128 transistor, (with switch on 2.2M) .42mv and (switch off 2.2m off) 1.09v.  

battery source is 9.42v.

[LucifersTrip]

this is the key sentence:

"Let's say the device really leaks 93uA, and has a gain of 110 - a prime specimen. What happens when we test? We 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."

once you chuck it in the socket, you'll get a voltage. say, .229V as above. to figure leakage:

.229 / 2472 = 9.3 x 10^-5

microamp = 10^-6 amp

[Arcane Analog]

I can try to offer a simple response.

First, if using the RG method, you must regulate your power supply to an even 9V. A simple pot works well for that as it can dump the excess juice from your adapter or battery and make it easy to hit the 9V that RG specifies - along with the 2.472K and 2.2M resistors.

Measure your transistor for leakage with the tester and the old multimeter:
0.345V

As a side note, a lot of people use uA to note/discuss leakage: uA = microamps - mA = milliamps - 1000 uA = 1 mA
0.345V divided by 0.002472 = ~139.6uA of leakage

Flip Switch for the 'total' gain reading:
1.666V

So take the 'total' gain reading and subtract the leakage:
1.666 - 0.345 = 1.321V

Take that number and multiply by 100 for your 'true' gain.
1.321 x 100 = 132

Hope that helps.

[fuzzymuff]

I can try to offer a simple response.

First, if using the RG method, you must regulate your power supply to an even 9V. A simple pot works well for that as it can dump the excess juice from your adapter or battery and make it easy to hit the 9V that RG specifies - along with the 2.472K and 2.2M resistors.

Measure your transistor for leakage with the tester and the old multimeter:
0.345V

As a side note, a lot of people use uA to note/discuss leakage: uA = microamps - mA = milliamps - 1000 uA = 1 mA
0.345V divided by 0.002472 = ~139.6uA of leakage

Flip Switch for the 'total' gain reading:
1.666V

So take the 'total' gain reading and subtract the leakage:
1.666 - 0.345 = 1.321V

Take that number and multiply by 100 for your 'true' gain.
1.321 x 100 = 132

Hope that helps.

THANKS!!!!  I'm such a noob on these things.......  I'm using a 5k trimmer for the 2.472k and found a 2.2m metal film resistor that was spot on, but for the 9v, what value pot would you recommend?

[Arcane Analog]

I already replied to your PMs but for posterity's sake a 5K or 10K  is fine. You just need to dump the extra and hit 9V.

[fuzzymuff]

Thanks!!  Now, I'll be making a rig to test a bunch of OC125s.   

[Arcane Analog]

I am sure the 'experts' use their vastly superior Atlas tester but I highly recommend making something like this if you aim to test any decent amount of germanium:



You can watch the transistor fluctuate and/or settle which gives you indications on the device's quality that an Atlas cannot. I own an Atlas and I rarely use it other than for rough sorting.

[fuzzymuff]

I am sure the 'experts' use their vastly superior Atlas tester but I highly recommend making something like this if you aim to test any decent amount of germanium:



You can watch the transistor fluctuate and/or settle which gives you indications on the device's quality that an Atlas cannot. I own an Atlas and I rarely use it other than for rough sorting.

Wow!!!  That's sweet!!!  Now, I'm getting some ideas.  NICE!  Very clever on the use of that speaker wire connection terminal. 

[Henry89789]

I apologize for bringing this back but I can't figure this out. I have a few questions about the math. Otherwise I think I got it.

I am confused by this:

As a side note, a lot of people use uA to note/discuss leakage: uA = microamps - mA = milliamps - 1000 uA = 1 mA
0.345V divided by 0.002472 = ~139.6uA of leakage

If I get a readings of    .70    and 1.73    do I just subtract .70 from 1.73  to get gain =  1.03 X 100=  103 gain?   Is this correct?

and leakage is 700 mA. Is this correct?  Or is it more complicated than this?  thanks. Please help folks I am wasting way too much time on this.

[Henry89789]

Wait I think I figured it out.    is it this:

first reading:   .70

press switch:  1.73;    1.73  minus   .70  =   103 gain

Leakage=     .70  divide by .002472  = 283.2uA  =  .2832mA

Gain= 103   and leakage is .283ma   Is this right?

[Arcane Analog]

If I get a readings of    .70    and 1.73    do I just subtract .70 from 1.73  to get gain =  1.03 X 100=  103 gain?   Is this correct?

Yes

and leakage is 700 mA. Is this correct?  Or is it more complicated than this?  thanks. Please help folks I am wasting way too much time on this.

No.

0.7 divided by 0.002472 = 283uA Leakage

As I mentioned in your other thread you need to make sure you have exactly 9V at the powersupply or this will not work properly.

[Arcane Analog]

Wait I think I figured it out.    is it this:

first reading:   .70

press switch:  1.73;    1.73  minus   .70  =   103 gain

Leakage=     .70  divide by .002472  = 283.2uA  =  .2832mA

Gain= 103   and leakage is .283ma   Is this right?

Yes

[Henry89789]

Thank you very much sir. I appreciate your patience and help.  I inserted a B10K pot between the red wire of the battery and the emitter of the transistor.  Then I make it exactly 9v before every test. Is that the right way to do that? I only have 2.4 k on the resistor though. The 2.2m is right on. As I read I can put a trim pot on the 2.4k to make it 2.472k.  I will do that later right now I just want to be sure I understand the complete procedure.   

BTW: How much would  a variation of a given amount on any of these variables affect the Hfe and leakage results?

[Arcane Analog]

You can calculate the difference with a little math. Really though the nature of germanium means getting close (2.4K) will work fine. You can always use two resistors if you do not have a trimpot.

[DiscoVlad]

To test the total gain, press the switch that connects the 2.2M resistor to the base. This causes a touch more than 4 microamps of base current to flow in the base. The transistor multiplies this by its internal gain, and the sum of the leakage (which doesn't change with base current) and the amplified base current. If the transistor has a gain of 100 and no leakage, the voltage across the 2.4K resistor is then (4uA)*(100)*(2472) =  0.9888V - which is almost exactly 1/ 100 of the actual gain. Pretty neat, huh?

Here's a little math which hopefully is helpful:

The current for the switch being opened (leakage) and closed ("apparent gain") is just I= V/R (9V / 2.4k)

The 4µA base current comes from:
V_supply / R_base (9V ÷ 2.2Meg resistor) - This should actually be a little less (0.1-0.3V) than 9V because of the base-emitter voltage drop but 9V is close enough as explained in RG's article.

Now, we don't know the current gain of the transistor (lets's call it well... Gain), but we do know the current through the 2.4k resistor from both leakage (V₁, Base Current = 0), and when the switch is closed (V₂, Base current ≈ 4µA). Both cases are V_collector ÷ R_collector. As per RG, the actual gain can be calculated using the difference of the two measured voltages.

Rearranging the equation in RG's quote gives a general form of:
Gain = (V₂ - V₁) ÷ (( V_supply / R_base) x R_collector)

This ends up as: gain =  measured Voltage / (base current x collector resistance) which helpfully passes the sanity check of all the units that make up Voltage, current, and resistance (Joules of energy, coulombs of charge and seconds of time) get cancelled out -> Gain has no units.

Pop all of the above into a spreadsheet, and it does all the heavy lifting for you:

R1 is the base resistor, R2 is the collector resistor.
Measured values in the yellow boxes.
Leakage Current is VR2 / R2 (multiplied by 1,000,000 to make the units right) and some conditional formatting to colour code whether the amount of leakage is good or bad.
Gain (hfe) calculation as displayed.

[Henry89789]

More questions and some possibly interesting observations:

I set up my tester and did an experiment. I monitored an AC 125 (VI) transistor and found that as the air conditioner cycled on and off during a three hour period the  gain readings  varied between 2.21 when warm and 1.77 when cool; and the leakage readings varied between 1.23 when warm and .82 when cool.  The gain itself was 95 when cool and 98 when warm. The leakage was .497 when warm and .332 when cool.
Observation: Gain increased a little but leakage increased substantially.
Conclusion: Heat increases gain and leakage?       Or are my observations and conclusions completely wrong?

Which readings would be the most accurate indicator of true gain and leakage ?   warm or cool?    Or an average of the two?

Arcane Analog:

I have a couple of questions about the tester you built. The knob on the left side obviously adjusts  the voltage in.  Is that a jack for a 9v power supply on the top side (in the pic) ? Then if the red and black leads coming out the top are to connect a DMM to monitor the gain and leakage readings, then where do you connect a DMM to measure the voltage in?    On my tester I have a DMM connected to the breadboard to constantly measure hfe/lkg and a second DMM to measure the voltage in at the pot.



Discovlad:

That is interesting how you did that on a spreadsheet. I took a look at my Excel program and found a number of functions under the Sum link. Could you explain a little more how to do that calculation on Excel?   Which function did you use? Thanks.

[DiscoVlad]

More questions and some possibly interesting observations:

I set up my tester and did an experiment. I monitored an AC 125 (VI) transistor and found that as the air conditioner cycled on and off during a three hour period the  gain readings  varied between 2.21 when warm and 1.77 when cool; and the leakage readings varied between 1.23 when warm and .82 when cool.  The gain itself was 95 when cool and 98 when warm. The leakage was .497 when warm and .332 when cool.
Observation: Gain increased a little but leakage increased substantially.
Conclusion: Heat increases gain and leakage?       Or are my observations and conclusions completely wrong?

Which readings would be the most accurate indicator of true gain and leakage ?   warm or cool?    Or an average of the two?

All of the readings are the most accurate  The electrical resistance of Germanium (and Silicon, and Carbon, and metals) changes with temperature. This affects the gain and leakage.
Silicon changes a lot less with temperature than Germanium does, which is partly why the semiconductor industry moved to Silicon.

Ideally, if you're testing a large number of devices you should test them all at a specific controlled temperature e.g. 25ºC.

Discovlad:

That is interesting how you did that on a spreadsheet. I took a look at my Excel program and found a number of functions under the Sum link. Could you explain a little more how to do that calculation on Excel?   Which function did you use? Thanks.

The only functions I used were basic arithmetic.

Ileak = (VR2 / R2) * 1,000,000

The spreadsheet function is then: =(B6/B4) * 1000000
Additionally, there's conditional formatting on the cell where <300 = green (good), 300 to 500 = orange (tolerable), >500 = Red (bad)

Hfe is gained (hah!) from rearranging RG's equation: "(4uA)*(100)*(2472) =  0.9888V" -> Base current * Hfe * Collector resistance = Measured Collector Voltage

Making Hfe the subject gives: Hfe = Measured voltage / (base current * collector resistance)

Measured Collector Voltage is The two collector voltages subtracted from each other (e.g. from my picture, 1.31 - 0.6 = 0.71 V)
Base Current is Vsupply / R1 ( 9 / 2.23 MegΩ)
Collector resistance is R2: 2396 Ω

Jam all this together and you get (1.31 - 0.6) / ((9V / 2.23MegΩ) * 2396 Ω) = 73 (the actual gain of the transistor)
Which on the spreadsheet is: =(B7-B6) / ((B2/B3) * B4)

I hope this clarifies things! 

[Arcane Analog]

Arcane Analog:

I have a couple of questions about the tester you built. The knob on the left side obviously adjusts  the voltage in.  Is that a jack for a 9v power supply on the top side (in the pic) ? Then if the red and black leads coming out the top are to connect a DMM to monitor the gain and leakage readings, then where do you connect a DMM to measure the voltage in?    On my tester I have a DMM connected to the breadboard to constantly measure hfe/lkg and a second DMM to measure the voltage in at the pot.

Look at your setup. Where is the 9V being applied to the transistor? Where is the ground?

[Henry89789]

Arcane:

The schematic for the tester shows that the 9v is applied to the transistor at the emitter. So I have the 9v going into the pot and then coming out of the pot to the emitter leg ofr the transistor on the breadboard. When I need to "dump the extra " voltage I measure the voltage at the out lug of the pot and at the connection of the 9v black wire, and the 2.472k and 2.2M resistors.  Isn't the ground at the point where the 9v black wire connects to the resistors?  It seems to work. I hope you are using the Socratic method and that we have a profound lesson coming. LOL.

[Arcane Analog]

Your question pertained to where I measure the voltage to achieve 9V. Based on the schematic and my box, where do you think I apply the DMM's probes to determine that?