How to Calculate Transistor Leakage?

[RobertJay]

None of this makes any sense and doesn't answer my rather simple question of.... Do I have the formula figured out from what I read in the earlier thread.

[RobertJay]

The hfe you are getting on a regular meter is not true, due to ................the leakage, not being figured in. For a silicon transistor, yes it would be a good reading. Go to RG'S website geoeffects and see how to build a simple germanium tester. That being said, even testing silicon transistors on various meters, you will get different readings, due to each brand of meter using a different testing current.

YEs, I DID use the keen method, I built that tester and got my numbers accordingly. But the keen instructions don't actually tell you how to calculate the leakage. He does many calcualtions and uses numbers but doesn't say where they come from. WTH is 100E-6? HE talks about calculating total gain and gain loss and he mentions leakage in uA.... But how to calculate the leakage itself? I had to read that from some other guy asking the same question. I want to make sure I understood THAT guy's formula and if we're all right about this...

[RobertJay]

Hi, Let's try this again. And if correct, we should post for the world as it's some kinda secret. Make the Keen tester. Read the voltage with the base, then without, subtract the 2 voltages and divide by .002472 and that's our leakage in uA. Yes?
That's the formula I thought I read up in the thread, but now see it looks like that user was told to take the voltage reading without the base and divide THAT by .002472. But how is that leakage? Doesn't leakage ratings need the difference in voltage to be calculated? IDK.

[antonis]

YEs, I DID use the keen method, I built that tester and got my numbers accordingly. But the keen instructions don't actually tell you how to calculate the leakage. He does many calcualtions and uses numbers but doesn't say where they come from. WTH is 100E-6? HE talks about calculating total gain and gain loss and he mentions leakage in uA....

You should know (or ask for) that E means X10 and -6 is the exponent to which 10 is raised (e.g 100E-6 equals 100 x 10^-6 or 100 x 10^-6) and equals to the number before it multilpied by 0.000001 (μ or micro).
So, for current, 100E-6 means 100μA..

P.S.
I should dare to berate R.G. for his slip to type capital "E" instead of smal "e", the later almost exclusivelly used as power of ten..

[RobertJay]

YEs, I DID use the keen method, I built that tester and got my numbers accordingly. But the keen instructions don't actually tell you how to calculate the leakage. He does many calcualtions and uses numbers but doesn't say where they come from. WTH is 100E-6? HE talks about calculating total gain and gain loss and he mentions leakage in uA....

You should know (or ask for) that E means X10 and -6 is the exponent to which 10 is raised (e.g 100E-6 equals 100 x 10^-6 or 100 x 10^-6) and equals to the number before it multilpied by 0.000001 (μ or micro).
So, for current, 100E-6 means 100μA..

P.S.
I should dare to berate R.G. for his slip to type capital "E" instead of smal "e", the later almost exclusivelly used as power of ten..

But I don't suppose you can help me with the formula to get the uA leakage rating?.... What number am I dividing by .002472. The voltage of the collector without the base? Or the difference in the 2 voltages with and without base?...

[RobertJay]

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.

Here it is... I'm confused here. His first reading waS .70V. THAT'S What we divide by .002472? or do we subtract the 2 voltages and divide THAT answer by .002472?
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.

[RobertJay]

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.

Here! This one! Here it is! Leakage is calculated by taking the lower voltage reading of the collector without base and dividing it by .00247. I don't know why I thought it was the difference of the 2 voltages that gets divided.

[RobertJay]

The hFE is easy enough to test using a cheap tester, but the uA leakage is another factor. Tranny 1&2 have to be less than 100uA and #3 between 100uA-300uA. Ok, so here's what I figured out... I took my measurements with and without the Base, subtracted the 2 and divided that by .00247. So,  .33v-.06v=.27v/.002472=109uA... Yes?

Looks OK to me.  Also the values are reasonable.

Off hand,  RG's site (geofex) spells it out.

Or Not? I think the leakage is calculated by the lower voltage reading of the collector without base divided by.002472. I don't know why I thought it was the difference of the 2 readings that got divided....

[antonis]

I think the leakage is calculated by the lower voltage reading of the collector without base divided by.002472.

No..

https://www.diystompboxes.com/smfforum/index.php?topic=45481.0

[RobertJay]

I think the leakage is calculated by the lower voltage reading of the collector without base divided by.002472.

No..

https://www.diystompboxes.com/smfforum/index.php?topic=45481.0

Then What? Please. That's wrong? But someone said it was right. I keep getting links to forums where it's not said in plain english. The link you sent me has this...
2. What is the leakage value
    Need calculator, vc/2K472, so I used 1K 1% resistor series with 2K trimmer set to 1K472
    V1K * leakage (ex 100uA) would read 0.1V read as 0.1mA, a lot easier

VC/2k472. Isn't that what I thought? vc/.002472?

[r080]

VC/2k472. Isn't that what I thought? vc/.002472?

It seems to get the answer you want, but I might be misunderstanding what Antonis is trying to point out.

Assuming RC = 2472 Ohms and VC= voltage across RC, and you are measuring the VC voltage with no voltage on the base:

VC (volts) / 2472 (ohms) = collector leakage current in amps

VC (volts) / 0.002472 (megaohms) = collector leakage current in microamps

[RobertJay]

Another inquiry on the DIYStompboxes FB page gave the same calculation. VC with no base voltage divided by 2472 then multiplied by 1000000. or just divide the vc by .002472
These are what the voltages are and what I see the calculation is, but my uA leakages are too low then for this Ge transistor. A transistor that should have a min leakage of 100uA are calculating to be 24uA.
.06 / .002472 = 24uA. Either this is the wrong way to measure and calculate, or the transistors are not as advertised.
I Just Don't Know!!!

[antonis]

You're free to not trust your power supply, your trasistors, your items values or your DMM but NOT Ohm's Law..

If you measure a voltage drop of 60mV across a 2k472 resistor then the current through this resistor indeed is 24μA..!!

[andy-h-h]

Given that you know the method, why not just cross check your results using an online calculator?  This one is good. https://stompboxelectronics.com/resources/germanium-transistor-gain-hfe-tool/

I made a basic one on my blog for my own use, this one is better as you can add your own variables such as voltage, resistance etc. 

There will definitely be occasions when germanium transistors will give you unexpected results, which is why it's a good idea to measure them / test on a breadboard or in sockets before you reach for the soldering iron.

[Rob Strand]

It seems to get the answer you want, but I might be misunderstanding what Antonis is trying to point out.

Assuming RC = 2472 Ohms and VC= voltage across RC, and you are measuring the VC voltage with no voltage on the base:

VC (volts) / 2472 (ohms) = collector leakage current in amps

VC (volts) / 0.002472 (megaohms) = collector leakage current in microamps

The problem with this stuff is you either understand all the details of how the jig works, like mac posted,
OR, you accept the gods and have done the hard work and given formulas that work.   They are in fact all the same
thing but one requires understanding what is going on and one is based on faith.  Anything in between results in conflict  .

The potential for different units (A, mA, uA) creates some confusion but if the formulas say what the units are then you have to accept them.

Another inquiry on the DIYStompboxes FB page gave the same calculation. VC with no base voltage divided by 2472 then multiplied by 1000000. or just divide the vc by .002472
These are what the voltages are and what I see the calculation is, but my uA leakages are too low then for this Ge transistor. A transistor that should have a min leakage of 100uA are calculating to be 24uA.
.06 / .002472 = 24uA. Either this is the wrong way to measure and calculate, or the transistors are not as advertised.
I Just Don't Know!!!

24uA is certainly a reasonable value.   If a transistor stated a *minimum* of 100uA you might start to doubt how valid that is.  Often you don't care if the leakage is too low.   Most production transistors would quote a maximum leakage because large values are difficult to allow for in circuit designs.   Unfortunately some guitar pedal circuits need a magic amount leakage to work and that creates problems sourcing transistors, especially these days.  (You can often tweak the circuit to accommodate different leakage.)

[antonis]

Unfortunately some guitar pedal circuits needs a magic amount leakage to work and that creates problems sourcing transistors, especially these days.  (You can often tweak the circuit to accommodate different leakage.)

Or place the pedal inside an oven with adjustable temperature..
(it should be more practical than Shokley's equation implementation and surface leakage dissociation for not counting on temperature elevation..)

[Rob Strand]

Or place the pedal inside an oven with adjustable temperature..
(it should be more practical than Shokley's equation implementation and surface leakage dissociation for not counting on temperature elevation..)

Perhaps in this case winter temperatures.    I pondered that when I posted.  Going from 100uA to 24uA due to temperature is going to be in the order of 20deg below normal, so around 3 degC.   Seems too much drop.

I don't think 24uA leakage is unreasonable.

The leakage is difficult to predict accurately.   You can see how the I0 (or IS) in the Schottky equation also depends on temperature.   It's quite a strong effect, temperature cubed.
https://electronics.stackexchange.com/questions/35692/why-does-the-base-emitter-voltage-of-a-bjt-decrease-with-temperature

The post here has some details on leakage but it doesn't get into temperature effects,
https://www.diystompboxes.com/smfforum/index.php?topic=127535.msg1223527#msg1223527
[Just to be clear.  The leakage here is ICBO but the leakage measured by the jig is ICEO.  They are
related by the transistor gain. for example
https://electronics.stackexchange.com/questions/124668/iceo-icbo-physical-interpretation-in-bjt]

At the end of the day, if you want to use transistors with different leakages you have to tinker a bit to kick the circuit back on track.

[mac]

I don't think 24uA leakage is unreasonable.

I have a lot of Toshiba 2SA49, 52, 53 that leak less than 50ua.
In RG words they are "prime specimens", but too trebly compared with AC128.

mac

[mozz]

I have a lot of Toshiba 2SA49, 52, 53 that leak less than 50ua.
In RG words they are "prime specimens", but too trebly compared with AC128.

mac

2SA's are generally RF transistors, C-B capacitance is usually low, hence the better RF qualities. You can measure it, I can on a good benchtop type LCR meter. Maybe add 50-100pf across C-B would cut the trebly response, will have to try it next fuzz, which could be any day now. Generally the worse the transistor spec, the higher capacitance. Just used some AC151's that were 32-35pf. Just measured some genuine AC128 pulls, 130, 140, 145pf. I think my 2SB54,56's are about 50pf.

[mac]

I also have a bunch of Matsushita 2SA101, 102 that I use in Rangemasters.
Toshiba 2SB54, 56s can be used on any fuzz.
So Matsushita 2SB172, 175, 176 family and some others like 2SB475, etc.
IMHO best Japanese Germs for FF are Hitachi 2SB77, Sanyo 2SD72, and the King of Darkness Matsushita 2SD352

mac