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.
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
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.
Quote from: Arcane Analog on March 23, 2013, 12:22:16 AM
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!!!! :icon_biggrin: :icon_biggrin: 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?
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.
Thanks!! Now, I'll be making a rig to test a bunch of OC125s. :icon_biggrin:
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:
(http://img.photobucket.com/albums/v89/Troublestarter/Tester_zps7ff18474.jpg)
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.
Quote from: Arcane Analog on March 23, 2013, 01:41:25 PM
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:
(http://img.photobucket.com/albums/v89/Troublestarter/Tester_zps7ff18474.jpg)
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.
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.
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?
Quote from: Henry89789 on May 04, 2013, 12:15:14 AM
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
Quote from: Henry89789 on May 04, 2013, 12:15:14 AM
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.
Quote from: Henry89789 on May 04, 2013, 12:40:09 AM
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
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?
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.
Quote from: RG/GEOfex
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 (V
1, Base Current = 0), and when the switch is closed (V
2, 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 = (V
2 - V
1) ÷ (( 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:
(http://i.imgur.com/jqoX8nL.png)
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.
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.
(http://i1126.photobucket.com/albums/l603/henry7895/IMG_0253.jpg) (http://s1126.photobucket.com/user/henry7895/media/IMG_0253.jpg.html)
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.
Quote from: Henry89789 on May 05, 2013, 05:54:14 PM
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 :icon_twisted: 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.
Quote
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! :D
Quote from: Henry89789 on May 05, 2013, 05:54:14 PM
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.
(http://i1126.photobucket.com/albums/l603/henry7895/IMG_0253.jpg) (http://s1126.photobucket.com/user/henry7895/media/IMG_0253.jpg.html)
Look at your setup. Where is the 9V being applied to the transistor? Where is the ground?
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.
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?
Arcane:
Well, you either have a place to apply the probes on the side of the box that is not visible in the pic. Or, you insert the probes into holes in the perf board showing through the opening on the top of the box.
Try again. Both places to connect the DMM are visible from the top. The schematic really could not be more basic. Look at where the +9V and -9V connect.
Building pedals is not just about painting by numbers. Understanding the basics and critical thinking are key skills.
(http://www.geofex.com/FX_images/xstrtest.gif)
Kwai Chang Caine: Old man, how is it that you hear these things?
Master: Young man, how is it that you do not?
I think I figured it out. You use the transistor's "E" leg test "socket" and that black ground lead you got coming out the top. I tested it on mine. Works.
Okay, that was a fun exercise. I started another thread about getting a breadboard fuzz face circuit to work right so I can use it to manipulate resistor values to bias Q2C. How about giving me some clues on that one to see if I can figure it out?
Yes - you figured it out which is much better than just throwing you the answer I think.
> The gain ...95 when cool and 98 when warm.
> The leakage was 0.497 when warm and 0.332 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?
I can't know if your observations are correct.
However it is *generally* true that current-gain rises, slightly, with temperature; and leakage rises _significantly_ with temperature.
Leakage will roughly *double* for every 10 degrees C of temperature. Same basic physical/chemical law that says paint dries faster in summer, hot transformers die sooner, "cold-blooded" (really no-thermostat) reptiles are sluggish in the morning and active in the afternoon.
The reason current gain rises is more complex, or rather I don't have a quick explanation on hand, and it may be the difference of two phenomena so it may only *tend* to rise, but could fall.
If you are curious:
> the air conditioner cycled on and off
Don't fool around with little temp swings. Make a cup of hot tea and a glass of ice-tea. Tack long leads on the transistor, then stick it in a small plastic bag. Make your measurements in the hot and cold tea. Allow time for the hot/cold to soak through the bag and into the transistor.
Silicon can stand more heat/cold than you can. (Soldering-iron on the case is a traditional quick-check for heat troubles, though can be a bit hotter than is really wise.)
Germanium, especially vintage, you should limit the hot to tongue-safe. If you spilled it in your lap, would you be warmly annoyed or _scalded_? Likewise don't try dry-ice; regular ice-box ice is plenty cold for vintage Germanium.
From 1 deg C ice-tea to 65 deg C hot-tea, you should see very clear change of leakage. If double per 10 deg C, over 65 deg you figure 2*2*2*2*2*2 = 60 to 90 times the leakage (instead of 1.5 times in your very-small temp swing).
I've had a couple of queries from people wanting the spreadsheet I made for calculating leakage.
So here's a link. (http://cabbages.orconhosting.net.nz/graphics/joes_rubbish/fx/Germanium%20Transistor%20calculator.ods.zip)
This is an openoffice document so probably can't be opened in Excel. You will need something which can read them, like Openoffice, from www.openoffice.org (it's free!).
Openoffice spreadsheets will save as an excell (.XLS) spreadsheet. Do "Save as" and pick the appropriate file format.
However, I still philosophically object to using a nuclear spreadsheet to squash a fly. It's very nearly the level of arithmetic that can be quickly done in your head.
I was going to write an app, but hey it's only two equations. :icon_redface: I was a bit loathe to share the calculator because they always seem to end up with people relying on the numbers that come out but without understanding the implications of them.
I agree though, if you do the maths beforehand it's a pretty simple matter to calculate what voltage readings the various gain/leakage bands fall into. Then all that needs to be done is measure and sort.
On the other hand, if you have a pile of transistors to sort and match, spreadsheets do make sense.
You can number each transistor in the pile with those little sticky inventory dots from an office supply store. Then you run each one through the tester, entering only the device number in one column and its reading in the cell next to it in the spreadsheet. When the pile is entered, then you enter the calculation in the next column and drag-copy it for the whole column. With that done, sorting is done just by highlighting the rows and sorting on the column of interest.
After that, sorting for matching or where to cut off your selections gets really easy. You can have the spreadsheet construct graphs of it, etc.
I have a HTML page in my gallery that does the maths for you :icon_biggrin:
RG Test maths:
http://www.diystompboxes.com/smfforum/index.php?topic=45481.0
mac
I didn't want to waste time getting the breadboard to work so I built a kind of Biasing Pedal. It has sockets for Q1, Q2, the 470, the 2.2, and trim pots for the 33k and 8.2k. I test the transistor for gain and leakage with the tester discussed earlier and then bias them with this "Biaser." There's just one more thing I don't understand. You can see from the pic that the numbers on the meter show that each transistor is within the ranges considered desirable for fuzz pedals. These readings were obtained by adjusting the trim pots. So the question is :
Once you adjust the trim pots to get the desired readings how do you determine the ohms to which each trim pot was adjusted so you can install those transistors with the appropriate resistors "pre-biased" in another pedal?
(http://i1126.photobucket.com/albums/l603/henry7895/IMG_0259.jpg) (http://s1126.photobucket.com/user/henry7895/media/IMG_0259.jpg.html)
Quote from: Henry89789 on May 15, 2013, 07:49:24 PM
You can see from the pic that the numbers on the meter show that each transistor is within the ranges considered desirable for fuzz pedals.
that's mainly just for this specific fuzz pedal...and for some others based on this topology
Quote
Once you adjust the trim pots to get the desired readings how do you determine the ohms to which each trim pot was adjusted so you can install those transistors with the appropriate resistors "pre-biased" in another pedal?
you can only measure the ohms out of circuit, so you need to disconnect one leg of the trim. In my Fuzz Face test box, I have a switch on the trimmer so I can take it in and out of circuit
Luciferstrip:
Thanks for the reply. How do you connect such a switch? Or is it two switches? One on each side of the trim pot, so you disconnect the trimmer from the circuit on each side and then measure the resistance between the two switches ?
Socket the trimmers.
Quote from: Henry89789 on May 15, 2013, 09:26:33 PM
Thanks for the reply. How do you connect such a switch? Or is it two switches? One on each side of the trim pot, so you disconnect the trimmer from the circuit on each side and then measure the resistance between the two switches ?
you need only one spst switch on one side of each trimmer you want to remove from the circuit. a spst has only 2 lugs, so you should be able to figure out how to wire it. a socket would work if you're not putting it in a sealed enclosure.
I finally got Arcane Analog's Germanium Hfe/Lkg Tester put into an enclosure. I didn't want to use up a nice pedal box and I found this old kaput lantern lying around ....
(http://i1126.photobucket.com/albums/l603/henry7895/IMG_0343.jpg) (http://s1126.photobucket.com/user/henry7895/media/IMG_0343.jpg.html)
Not very compact but it works and its cool, even has a sort of hidden message about a mad Diogenes carrying a lantern looking for an honest germanium transistor. Thank you Arcane Analog for your assistance.
Hi Hey Hello. I know this is old, but is anyone still here? I am having trouble understanding this but think I do understand based on this thread. I'm building old school tone bender circuits with germanium transistors. 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?
And YES. I DID use the geofex link instructions, I DID build the tester with the resistors and used a power regulator for 9v on the nose. It is THAT link that everyone wants to share as the almighty answer but the language in it is SO confusing. I want to know if my numbers and way of getting them are right as it's never really simply explained. It's not! Nothing against Sir Keen, I'm pretty sure he started this whole website and forum we are on. But why does everyone act like it's explained in simple english, it's not. Thanks
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.
QuoteThe 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.
I use a 1k resistor from V to Collector and a 1meg resistor from V to Base to measure Ge or Si.
Twice base current, easy to read leakage.
OTOH RG tester makes gain readings easy.
Easy leakage reading, pay with gain. Easy gain reading, pay with leakage... sounds familiar? :icon_lol:
The model:
ic = hfe*ib + iL
V: battery
iL: leakage
Rc: collector resistor
Rb: base resistor
(V - Vc*) : voltage across Rc with Rb not connected.
(V - Vc) : voltage across Rc with Rb connected.
iL = (V - Vc*)/Rc
ic = (V - Vc)/Rc
ib = (V - Vbe)/Rb
Vbe: Base - Emitter voltage drop
hfe = (ic - iL)/ib
or
hfe = [(V - Vc)/Rc - (V - Vc*)/Rc]/(V - Vbe)*Rb
hfe = Rb/Rc*(Vc* - Vc)/(V -Vbe)
Vc*, Vc: collector voltages respect to GND
RG makes Rb/Rc/(V - Vbe) close to 100.
V - Vbe = 9v - 0.1v (typical) = 8.9v
2.2meg/2472/8.9v = 100/volt
so
hfe = 100*(Vc* - Vc) easy
iL = (V - Vc*)/2472ohm
What I do is to set Rc = 1k and Rb = 1meg so
hfe = 112*(Vc* - Vc)
iL = (V - Vc)/1000ohm easy
The eqs can be used to measure Silicons by making iL = 0.
And to know the gain under different working conditions.
For example, let's say you have a nice AD139 and want to build a Class A amp @ 100ma idle collector current. If a priori you suppose a gain of 100 then base current is about a milliamp, way higher than 4 microamps used in fuzzes.
Since hfe is a function of ic you can expect gain is going be different at 1ma than at 4ua.
You can choose Rc and Rb to test the transistor under a base current of 1ma, for example, Rb = 9kohm, Rc=10-100ohm
mac
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.
Quote from: mozz on April 26, 2023, 12:01:45 PM
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...
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.
Quote from: RobertJay on April 28, 2023, 07:02:28 AM
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.. :icon_wink:
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.. :icon_redface:
Quote from: antonis on April 28, 2023, 07:36:50 AM
Quote from: RobertJay on April 28, 2023, 07:02:28 AM
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.. :icon_wink:
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.. :icon_redface:
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?...
Quote from: Arcane Analog on May 04, 2013, 12:41:41 AM
Quote from: Henry89789 on May 04, 2013, 12:15:14 AM
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
Quote from: Henry89789 on May 04, 2013, 12:15:14 AM
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.
Quote from: Arcane Analog on March 23, 2013, 12:22:16 AM
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.
Quote from: Rob Strand on April 27, 2023, 06:36:35 AM
QuoteThe 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....
Quote from: RobertJay on April 28, 2023, 08:34:25 AM
I think the leakage is calculated by the lower voltage reading of the collector without base divided by.002472.
No.. :icon_wink:
https://www.diystompboxes.com/smfforum/index.php?topic=45481.0 (https://www.diystompboxes.com/smfforum/index.php?topic=45481.0)
Quote from: antonis on April 28, 2023, 09:32:41 AM
Quote from: RobertJay on April 28, 2023, 08:34:25 AM
I think the leakage is calculated by the lower voltage reading of the collector without base divided by.002472.
No.. :icon_wink:
https://www.diystompboxes.com/smfforum/index.php?topic=45481.0 (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?
Quote from: RobertJay on April 28, 2023, 11:03:23 AM
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
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!!!
You're free to not trust your power supply, your trasistors, your items values or your DMM but NOT Ohm's Law.. :icon_wink:
If you measure a voltage drop of 60mV across a 2k472 resistor then the current through this resistor indeed is 24μA..!!
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.
QuoteIt 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 ;D.
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.
QuoteAnother 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.)
Quote from: Rob Strand on April 28, 2023, 05:26:32 PM
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..)
QuoteOr 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.
QuoteI 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
Quote from: mac on April 29, 2023, 01:03:24 PM
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.
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
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.. :icon_wink:
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.. :icon_wink:
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://m.media-amazon.com/images/I/61KrlI7SqtL._AC_SL1500_.jpg)
https://www.amazon.com/LM2596-Converter-Module-Supply-1-23V-30V/dp/B008BHBEE0 (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.. :icon_wink:
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..) :icon_wink:
(*) immune to current variations..
P.S.
The above is the simplest implementation of Thevenin theorem (https://www.youtube.com/watch?v=jpO37lVBe-o)
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.
(https://i.postimg.cc/62zLjhS6/RGKeen-Tester.png) (https://postimg.cc/62zLjhS6)
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 :icon_mrgreen:
Quote from: mac on April 02, 2024, 10:26:06 AMQuote 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 :icon_mrgreen:
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! :icon_lol:
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! :icon_lol:
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! :icon_lol:
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?
Quote from: antonis on April 02, 2024, 04:02:43 PMSomebody MUST create a leakage current function of temperature and phase of the moon coordinates..
Ileakage(ic, T, a) = IRG(ic, T)*[R + r*cos(a/2)]
where
ic is the collector current
T the temparature
IRG is the RG leakage current
R is the average Sun to Earth distance
r is how much the Earth drifts from the average distance as it circles around the Sun
a is the angle of the orbit in radians
This Eq should be corrected to include moon phase and Mojo.
mac
Quote from: Ginsly on April 02, 2024, 04:07:07 PMwould 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?
Just measure the voltage drop across the 2472 and divide the result by 2472.
mac
mac, don't forget the minus sign - that lot are in the northern hemisphere.
@GibsonGM, OK :icon_mrgreen:
Due to different VBE even my lovely OC44 has 0.25V, also Si and Darlington,
this draft come to mind!
(https://i.postimg.cc/sM2DKcJQ/Match-Vbe-Hfe.png) (https://postimg.cc/sM2DKcJQ)
combine with Ian Fritz method for match vbe, maybe we can also match hfe,
using current constant source.
Thanks, I just 'approximated' the R.G. method, and managed to turn up 2 BJTs out of a bunch I have that seem like they'll work in a FF! 8) I need good germanium transistors approximately once every 20 years, ha ha :icon_mrgreen:
Yup for germanium, just measure real value of resistor, and voltage, a little counting is enough for precision.
I need a versatile one, also for superfuzz match.
Just bb the keen tester using a series of resistors to get 2.2m and 2.472k, of course the dmm has an inaccuracy, but good enough. I'm gonna think like a dog, "my ears" will be the ultimate judge, I won't "chase my tail" with irrelevant minute differences, else I will never get a "muff fuzz".
But I hope to sort the good, bad and ugly.
From Freestompboxes, a table:
2.472v = 1mA
1.236v = 500 uA
0.989v = 400 uA
0.741v = 300 uA
0.494v = 200 uA
0.247v = 100 uA
0.124v = 50uA
^ Voltage across 2.472K resistor with 2.2M lifted.
And RG says:
"How much leakage is too much? 100uA is common, 200 happens pretty often. More than 300uA means the device is suspicious, and more than 500uA I would say is bad."
Interesting thread,
How many decimal places should I calculate the leakage to ?
(At my age leakage is observed in drops per hour)
QuoteHow many decimal places should I calculate the leakage to ?
If you want that Hendrix tone as many as Pi! :icon_lol:
mac
Quote from: Phend on June 08, 2024, 03:12:54 PM(At my age leakage is observed in drops per hour)
https://lunderg.com/confidence-clamp/
Humm..
I am finding that my Elcheapo Tester and Cheapo DMM and the my state of the art Keen BB differ.
Case 1 1T308V
Tester 72
DMM 75
Keen 42
Case 2 2N518A
Tester 153
Keen 154
Case 3 (unknown)
Tester 102
DMM 166
Keen 89
I will use RG's test , since he has shown the math's and explanations.
The Chinese units I have are good for sorting resistors.
Those don't account for than darn leakage.
Apparently to make the holy grail FF these things need to be known.
Quote from: Phend on June 07, 2024, 04:26:55 PM"How much leakage is too much? 100uA is common, 200 happens pretty often. More than 300uA means the device is suspicious, and more than 500uA I would say is bad."
If you are vetting devices then it depends on what you expect. You might not be overly concerned with an AC128 at 300uA but you would certainly be thinking something is wrong if it was an OC44. Most devices are around the values RG quotes.
As for using the correct leakage for a given circuit, that's a different story. Maybe some AC128 work and some AC128's don't because the leakage is too high.
Apologies for raising this again!!
Given it all aligns to ohm's law, does it really matter about strictly 9v supply, 2.2m and a 2.472K resistors ?
Can I just use whatever voltage my battery puts out and whatever resistors I have to hand, say a 1M and 2k or something, measure voltages and plug them into a spreadsheet ?
Is there anything special about the RG ones apart from the ease by which you can visually convert the voltage readings to leakage current ?
Edit: ok, I might have found answers myself which is always a good sign! I can see that the choice of 2M2 makes the q1 base current similar to that in a fuzz face at operating conditions, 4uA. The choice of 9v and 2K472 them makes the calculations easier, but can be anything reasonable.
QuoteCan I just use whatever voltage my battery puts out and whatever resistors I have to hand, say a 1M and 2k or something, measure voltages and plug them into a spreadsheet ?
Rewind some pages, I posted the maths for any [Vcc, Rc, Rb] set.
mac
:-\ may be we can use VU meter?
I'm always surprised and amused at myself when this comes back up again. The leakage tester was made at a time when test-everything boxes were not available. I actually thought that by specifying a weird resistor value to indicate the leakage currents by just moving a decimal point was going to be useful, kind of a pre-computer/analog-computer way to get the numbers.
In retrospect, I maybe should have just stuck in a standard value and told everyone to get out their calculators. :icon_lol: It wasn't supposed to be confusing or complicated, honest.
I think John Why used a pencil ;D
Nice to meet you, Sir.
The VU just for the fun.
get out their calculators
(https://i.postimg.cc/mPnRQFvj/Slide-Rules-2022-42.gif) (https://postimg.cc/mPnRQFvj)-(https://i.postimg.cc/gLnBPdqK/Slide-Rule-1956.gif) (https://postimg.cc/gLnBPdqK)
Quote from: R.G. on January 10, 2025, 10:47:38 AMI'm always surprised and amused at myself when this comes back up again. The leakage tester was made at a time when test-everything boxes were not available. I actually thought that by specifying a weird resistor value to indicate the leakage currents by just moving a decimal point was going to be useful, kind of a pre-computer/analog-computer way to get the numbers.
In retrospect, I maybe should have just stuck in a standard value and told everyone to get out their calculators. :icon_lol: It wasn't supposed to be confusing or complicated, honest.
That's good to hear :) thanks!
Quote from: PRR on January 11, 2025, 01:23:11 AMget out their calculators
(https://i.postimg.cc/mPnRQFvj/Slide-Rules-2022-42.gif) (https://postimg.cc/mPnRQFvj)-(https://i.postimg.cc/gLnBPdqK/Slide-Rule-1956.gif) (https://postimg.cc/gLnBPdqK)
There is still a Log-Log Duplex in my center desk drawer in its leather case. It's a veteran of my first three undergrad years. As a senior, I saved up enough to get a four-function electronic. :icon_biggrin:
Quote from: PRR on January 11, 2025, 01:23:11 AMget out their calculators
(https://i.postimg.cc/mPnRQFvj/Slide-Rules-2022-42.gif) (https://postimg.cc/mPnRQFvj)-(https://i.postimg.cc/gLnBPdqK/Slide-Rule-1956.gif) (https://postimg.cc/gLnBPdqK)
There is still a Log-Log Duplex in my center desk drawer in its leather case. It's a veteran of my first three undergrad years. As a senior, I saved up enough to get a four-function electronic for about US$100.Yikes!! :icon_biggrin:
In retrospect, no superposition of Qubits can beat RG test, a sheet of paper, a pencil and a green display fx Casio ;D
mac