I have a youtube link that shows different ways to do different tests. This is how I learned to measure and calculate uA leakage and hFE readings for choosing Ge Trannies.
https://www.youtube.com/watch?v=5UdMaB9C6BY&t=1262s

New guy here! I built a Keen tester on a breadboard, and it's powered by a rechargeable lithium-ion 9v battery. Problem is, it puts out closer to 9.4v rather than the needed 9v.

People had mentioned using a pot to reduce the battery voltage to exactly 9v before it hits the breadboard, but I'm not sure how to do that exactly (if/where to connect the pot to ground, etc...).

I'm only about three months in to building, so if I sound like a dum-dum for asking... I am! Ha...

Thanks so much!

Did I made confusion

Quote from: antonis on April 28, 2023, 04:28:07 PMYou'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..!!

Quote from: Ginsly on April 01, 2024, 12:24:39 PMPeople had mentioned using a pot to reduce the battery voltage to exactly 9v before it hits the breadboard, but I'm not sure how to do that exactly (if/where to connect the pot to ground, etc...).

Wellcome, I assume you have read master Keen tutorial,

for real leakage transistor no needed, just series 2k472 resistor with 1k resistor and measure voltage between 1k resistor,

I = V / R

if you get 0.1V

leakage wouldbe = 0.1V / 1000 ohm = 0,0001 A

or 0.0001 A * 1000 = 0.1 mA  (1A = 1000mA)
or 0.1 mA * 1000 = 100 uA (1mA = 1000uA or 1A = 1000000uA)

Quote from: zbt on April 01, 2024, 02:04:36 PMDid I made confusion

Quote from: antonis on April 28, 2023, 04:28:07 PMYou'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..!!

Quote from: Ginsly on April 01, 2024, 12:24:39 PMPeople had mentioned using a pot to reduce the battery voltage to exactly 9v before it hits the breadboard, but I'm not sure how to do that exactly (if/where to connect the pot to ground, etc...).

Wellcome, I assume you have read master Keen tutorial,

for real leakage transistor no needed, just series 2k472 resistor with 1k resistor and measure voltage between 1k resistor,

I = V / R

if you get 0.1V

leakage wouldbe = 0.1V / 1000 ohm = 0,0001 A

or 0.0001 A * 1000 = 0.1 mA  (1A = 1000mA)
or 0.1 mA * 1000 = 100 uA (1mA = 1000uA or 1A = 1000000uA)

I appreciate you chiming in! You might have misunderstood my question - I'm asking about the 9v power supply that the Keen tester needs in order to function and produce results. I'm using a 9v battery that measures higher than that (9.4v), and I'd like to find a way to lower it to 9v, as is required in the Keen test. Arcane Analog mentioned using a simple potentiometer to do that before it connects to the Keen tester, but I'm not sure how exactly to do that...

Ooops, for this question How to Calculate Transistor Leakage?  is not needed

as for leakage to calculate real hfe also not needed, just measure voltage between 2k472

0.4V different for me also not problem

IB = (VCC - VBE) / 2M2

9.0V - 0.3V (Ge) / 2M2 = 0,000003955
9.4V - 0.3V (Ge) / 2M2 = 0,000004136

it so small, so its no needed to exactly 9V

If you do need to use a pot, one outside leg goes to 9V+, the other outside leg goes to ground. The wiper (middle) will be your adjustable voltage.  This makes a variable voltage divider, a concept to learn more about and get used to Also understanding power and power ratings will be needed when you do this, at times...

Welcome to the forum!

Quote from: zbt on April 01, 2024, 02:53:59 PMit so small, so its no needed to exactly 9V

Ah, ok- so 9.3v or so wouldn't really affect results in a drastic way? Good to know!

@GibsonGM - Gotcha, thanks much! It seems like adding a resistor somewhere helps protect the pot too... does that sound familiar?

Be careful for hfe, once I use dying battery, the result would be different,
because Ib depend on VBatt

Vbatt 9.4 9.3 9.2 9.1 9 8.9 8 7
RB 2200000 2200000 2200000 2200000 2200000 2200000 2200000 2200000
VBE 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3
VB 9.1 9 8.9 8.8 8.7 8.6 7.7 6.7
IB 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
RC 2472 2472 2472 2472 2472 2472 2472 2472
leak 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001
V leak 0.2472 0.2472 0.2472 0.2472 0.2472 0.2472 0.2472 0.2472
Hfe 200 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001
V tr 2.292 2.270 2.247 2.225 2.202 2.180 1.978 1.753
hFE 2.045 2.023 2.000 1.978 1.955 1.933 1.730 1.506
% 204 202 200 197 195 193 173 150


I use buck converter for 12v supply


https://www.amazon.com/LM2596-Converter-Module-Supply-1-23V-30V/dp/B008BHBEE0

Quote from: Ginsly on April 01, 2024, 06:56:38 PMIt seems like adding a resistor somewhere helps protect the pot too...

Maybe but you'll have to count for voltage drop across that resistor..

In brief: A pot wired as voltage divider (like Sir Mike said above) has a "dual" operation..

1. It lowers the input voltage, according to wiper setting..
2. It further lowers the input voltage, according to the current flowing out of its wiper..

e.g. for 9V power supply, a 1k pot set at 50% rotation should exhibit y Volts at its wiper, where y = 4.5V - (500R * I) and I = current flowing out of wiper..

In general, Vout of a R1/R2 resistive voltage divider is set by: Vin * [R2/(R1+R2)] - [(R1*R2)/(R1+R2)] * I

(that's the reason for making a stiff(*) voltage divider, the current flowing through pot's body must be MUCH LARGER than the one coming out of its wiper - in other words, powered circuit's resistance should be MUCH BIGGER than divider's one..)
(*) immune to current variations..

P.S.
The above is the simplest implementation of Thevenin theorem

Quote from: Ginsly on April 01, 2024, 06:56:38 PM@GibsonGM - Gotcha, thanks much! It seems like adding a resistor somewhere helps protect the pot too... does that sound familiar?

Yes! I didn't want to confuse you, seeing you've had 4 forum posts, ha ha.  I almost added (and should have) - "the way I would do this is to connect the pot to the circuit before you apply power"...the circuit itself can be the current limiter.  In other cases, I set the pot to some calculated min. value and don't turn it any lower (ok, that's a hack but has worked).

Note the 2.472k resistor in the 'tester'.  That's a limiter on what will be drawn thru the 'top of the pot' where R is small.   I did quick math - LT Spice is your friend - and thought in this case you wouldn't burn a 10k pot. It is easy to do at other times, so a small limiting R is a great idea and might save a pot.  Here, oh, 47 ohms or something in that zone. You still need to get low enough R to drop enough of the excess voltage.   This is a quick and dirty method   A power diode might be another good way but isn't adjustable.

Pots are typically rated at 1/2 watt.  In a 9V circuit, 50mA or so is pushing up close to that. We're not near that here. 

Since it's a topic again...I have a bunch of Ge's I might as well test.  Does using a 1.5k resistor in series with a 1k trimmer seem reasonable to dial in the goal of 2.47k?  I have no need for total accuracy, just 'close enough' to see if any of them are in the ballpark would be fine!  Not sure since we're dealing with uA.

use trimpot for more precission

Quote from: Ginsly on April 01, 2024, 12:24:39 PMNew guy here! I built a Keen tester on a breadboard, and it's powered by a rechargeable lithium-ion 9v battery. Problem is, it puts out closer to 9.4v rather than the needed 9v.

People had mentioned using a pot to reduce the battery voltage to exactly 9v before it hits the breadboard, but I'm not sure how to do that exactly (if/where to connect the pot to ground, etc...).

You can put a silicon diode or one or two Schottky diodes in series with the power supply.

Or, if you are using a 2472 Rc resistor you should multiply by about 96 instead of 100 to get hfe.

Or use a 2365 Rc resistor, multiply by 100 for hfe and use (Vc -Vc*)/2365 (V -Vc*)/2365 without 2M2 for leakage.

mac

ZBT - ok, so 2.5k is close enough then?  I can easily make 2.5k (I literally just measured a 2.2k and 330R and it comes exactly to that).  Just that it is not 2.472k 

Quote from: mac on April 02, 2024, 10:26:06 AM

Quote from: Ginsly on April 01, 2024, 12:24:39 PMNew guy here! I built a Keen tester on a breadboard, and it's powered by a rechargeable lithium-ion 9v battery. Problem is, it puts out closer to 9.4v rather than the needed 9v.

People had mentioned using a pot to reduce the battery voltage to exactly 9v before it hits the breadboard, but I'm not sure how to do that exactly (if/where to connect the pot to ground, etc...).

You can put a silicon diode or one or two Schottky diodes in series with the power supply.

Or, if you are using a 2472 Rc resistor you should multiply by about 96 instead of 100 to get hfe.

Or use a 2365 Rc resistor, multiply by 100 for hfe and use (Vc -Vc*)/2365 for leakage.

mac

Ah- very interesting! Let's say I stick with the 9.4v battery and multiply by 96 instead of 100 to get hfe; in the case of a Ge transistor, how would I calculate leakage? Normally it would be something like say, .37(volts, without the 2.2m resistor)/2.472= 149 ua leakage, for instance. Would this stay the same in the 9.4v scenario?

if we measure with 1K resistor then 149uA * 1K = 0,149V  no need calculation 

at the same temperature, resistance (2472) would be constant,
leakage would be constant, change would be voltage value of C and E (VCE) of transistor

if you have 4 AAA 1.5V battery in series make it 6V, just try it.

Quote from: Ginsly on April 02, 2024, 12:51:45 PMin the case of a Ge transistor, how would I calculate leakage? Normally it would be something like say, .37(volts, without the 2.2m resistor)/2.472= 149 ua leakage, for instance. Would this stay the same in the 9.4v scenario?

Quote from: Ginsly on April 02, 2024, 12:51:45 PM(Vc -Vc*)/2365 for leakage.

Sorry, I typed it without the James Webb telescope...

for leakage (V - Vc*)/2365
where (V - Vc*) is the voltage across the 2365 resistor without the 2M2 resistor.

Don't get mad trying to get accurate results... tomorrow all readings will change! 

mac

Quote from: mac on April 02, 2024, 02:28:55 PMDon't get mad trying to get accurate results... tomorrow all readings will change! 

Somebody MUST create a leakage current function of temperature and phase of the moon coordinates..

Quote from: mac on April 02, 2024, 02:28:55 PMfor leakage (V - Vc*)/2365
where (V - Vc*) is the voltage across the 2365 resistor without the 2M2 resistor.

Don't get mad trying to get accurate results... tomorrow all readings will change! 

Haha... totally, I get that! With Ge it changes hour to hour... yikes!

Instead of using a 2364 resistor, I meant that I was going to leave the 2472 resistor in place and instead use your method of calculating hfe by multiplying by 96 instead of 100 when using a 9.4v battery in the tester. Great advice!

With the 2472 resistor still in place, omitting the 2.2m resistor (and again, using a 9.4v power source), would I still use the standard (V-Vc*)/2472 to calculate leakage or would I also altar that slightly because of the 9.4 volts being fed into the tester?