Technology of Fuzz tranny check math stuff

[Noplasticrobots]

http://www.geofex.com/Article_Folders/fuzzface/fftech.htm#Picking%20transistors

Ok, I'm a little confused on this so hopefully someone can give me a little insight. I have the above tester breadboarded and am fooling around with a 2N169 with gain of hfe 48. To start things off, do I measure the mA leakage across the 2.4k resistor? I assumed I did and used those numbers in this post.

My resistance across the 2.4k is 2.470k and 2.195 across the 2.2M

leakage in uA is 1.37 (0.00000137 amps?)

I read the DC voltage across the 2.4k resistor and get 0.88 volts.

I flip the switch and measure 0.889 volts across the 2.4k resistor.

"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."

So with my numbers, and the preceding paragraph as my instructions, I get the following:

(0.00000137)*(2470)=0.0033839

So how do I get voltage from 0.0033839? When using the articles numbers I actually get 0.22971, so moving the decimal point to the right one place gives us voltage?

So that means I have .00033839 volts?

Ok, so switch is flipped and I'm reading 0.889 volts.

So 0.889V-0.00033839V=0.88866161V

So my gain is .88?

And yes, this circuit is correct on the board. 

[bioroids]

How did you measure the leakage current?

The test is made so you only have to measure the voltage at the 2.4K resistor (with and without the base bias - the 2M2 resistor), substract them and voila you get the real gain.

The voltage you measure at the 2.4K resistor without the bias is not leakage current, but "false" gain caused by the leakage current.

Hope this helps!

Miguel

[Noplasticrobots]

Well, I understand that but I'm still a lttle confused on finding the leakage. The article goes from saying how much leakage is too much to saying "Let's assume we have a leakage of 93 uA".

So how do I find total leakage? Subtracting the voltages across the resistors from the gain? I think this formula gives you leakage, but I don't understand how you come out with voltage at the end:

(4uA)*(100)*(2472) =  0.9888V

If anyone has any experience witht his little device, could you please post the steps you take without example numbers.

[bioroids]

The leakage would be the voltage read across the 2K4 resistor divided by the resistor value, measured without the 2M2 bias resistor.

If you want to read leakage the easiest way is to use a 1K resistor instead of 2K4, then the voltage read is numerically equal to the leakage current in mA. So, if you read 0.100v that means you have 0.1mA of leakage, or 100uA.

Luck

Miguel

[Noplasticrobots]

Ok, so I replaced the 2k4 with a 1k resistor and I have a resistance of 2469 ohms.

Thus:

read voltage at 1k = 0.556V which equals 0.000556 A of leakage

(2469)*(0.000556) = 1.374432V      
            
read voltage at 2M2 = 0.57V         
            
So:

1.374432 - 0.57 = 0.804432V

Correct?

[bioroids]

Mmm I think I'm not getting something. Where do you have the 2468 ohms resistance?

Miguel

[Noplasticrobots]

Whoops! Thanks for pointing that out. ok, I've entered my correct resistor value of 1001 ohms into my spreadsheet and got this:

read voltage at 1k = 0.556 V = 0.000556 A of leakage

resistance in ohms = 1001         

0.000556 * 1001 = 0.556556 volts of leakage      
               
read voltage at 2M2 = 0.57         

0.556556 - 0.57 = -0.013444

So true gain = 0.013444

How's that look?

Also, can this circuit be used to accurately check Si transistors, or is it geared only to Germs?

[bioroids]

I think my explanation confused you more 

If you want to measure the gain, the best thing would be to put back the 2K4 resistor and do the procedure as R.G. says, but you can do it anyway.  Here's how:

You have to remember that the gain is the ratio between the output current and the input current. The leakage would be output current that is present even when there is no input current.

Lets use your circuit values.

You have to make two measures between the 1K resistor pins: (A) one without the base resistor and (B) one with it. It is best if you also measure the voltage between the pins of the 2M2 resistor when you put it. Or you can asume 4.4uA as RG article says.

From (A) you get 0.556volts. The resistor is 1K, so (because of Ohms Law) 0.556v/1001ohms gets you 0.555mA of current. As there is no input current at this time, this current is all leakage (wich is kinda high by the way).

Now you connect the 2M2 resistor and take measure (B). That is the part that got you confused.
Suppose (B) gives you 1volt. Then, having 1v/1001ohms that's 0.999mA of output current. But you know that 0.555mA are leakage, so you substract 0.999mA - 0.555mA that's a total of 0.444mA of real output current.

But, you know that the input current is 4.4uA=0.0044mA, so the ratio 0.444mA/0.0044mA gives you 100. That would be the real gain (supposing 1volt for measure (B) )

I hope this is more clear. It's a little harder than R.G.'s way because you got rid of the 2472ohm resistor that did the correct scaling, but the results are the same.

It is probably easier if you put back the 2K4 resistor and work by R.G. and then calculate leakage current by dividing the (A) value by the 2472ohm.

Luck

Miguel

[bioroids]

Oh my God I'm sure that explanation only mixes up things more.  I have to learn how to explain things.

By the way yuo can use this to test silicon transistors too, they work the same way, but there is no need to do it because they dont leak, and then the DMM figures are quite right.

Luck

Miguel

[Noplasticrobots]

I realized within minute sof posting that Si trannys don't leak, so I don't know what I was thinking!

Your explanation does make sense! I have trouble following R.G.'s article for some reason. It's easier for me to see the formulae laid out like you did. It's obvious that R.G.'s article has helped many people so I'm definitely the odd one out who doesn't get it. Thanks again Miguel! I'll redo my spreadsheet now!

[bioroids]

Cool!

If you have more doubts, just let me (us) know!

Miguel

[Noplasticrobots]

One more time for the world:

resistance in ohms= 1001   

read voltage at 1k= 0.556 V
output current at 1k = 0.556 / 1001 = 0.00055544456 Amps
         
read voltage at 2M2 = 0.57 V
output current at 2M2 = 0.57 / 1001 = 0.00056943057 Amps

real output current= 0.00056943057 - 0.00055544456 = 0.00001398601 Amps
         
True Gain = 0.00001398601 / .0044 = 0.00317863954

Thus true gain = 0.003? (that seems very low)

By the way, these transistors are super old (30 years maybe?)

[bioroids]

resistance in ohms= 1001   

read voltage at 1k= 0.556 V
output current at 1k = 0.556 / 1001 = 0.00055544456 Amps
         
read voltage at 2M2 = 0.57 V
output current at 2M2 = 0.57 / 1001 = 0.00056943057 Amps

real output current= 0.00056943057 - 0.00055544456 = 0.00001398601 Amps
         
True Gain = 0.00001398601 / .0044 = 0.00317863954

Thus true gain = 0.003? (that seems very low)

By the way, these transistors are super old (30 years maybe?)

When you put the 2M2 resistor, do you measure the voltage at the 1K resistor or at the 2M2 resistor?

Miguel

[Noplasticrobots]

I measured at 1k because I get 0 volts at the 2M2...

I've got an On-On SPDT switch in the circuit because I have no SPST. I wired the Tranny base into the center teminal of the switch and one of the end poles to the 2M2. Could that be the problem?

[bioroids]

Mmm strange. Are you sure you got the pinout of the transistor correct?

You can try with a 1M resistor instead of 2M2. You can assume about 8.4uA of input current in that case (or better, measure the voltage at the 1M resistor pins and divide that by the 1M ohms to get the exact input current)

You should get about 0.1 or 0.2 volts between the transistor base and gnd (in fact it will be -0.1v or -0.2v because of the positive ground) with the biasing resistor. 0v means its turned off and you're not suplying input current.

If the pinout is correct, I'm starting to think the transistor is bad. More than 500uA of leakage is too much.

Miguel

[Noplasticrobots]

I think you responded as I edited the last post but nevermind. I just now realized that the switch is actually an SPDT on/off switch. D'oh!! Must I ask: Why is it always the simplest mistakes!? 

[bioroids]

I dont think the switch is a problem if it is working right. Make sure it opens and closes correctly and you're fine with that.

Check my previous post. If you have other transistors, even silicon, you can try this stuff to see if the readings make more sense.

I'm going to a birthday party now! Keep posting your results and I'll check when I come back.

Luck

Miguel

[George Giblet]

Firstly, couple of things:
- There's no reason to change the collector resistor.   It doesn't matter what the value is too much.  What RG has done is rigged things so you get about 1V across the collector resistor/meter output when the transistor has gain 100.   The 1V choice ensures the circuit won't produce any funny results and the scaling lets you roughly read the gain on the voltmeter.
-  Your original posts quotes a current "leakage in uA is 1.37".  You shouldn't need to measure any currents.  With this jig you measure voltages.  If you want currents you convert the voltages to currents using ohms law I = V/R.

The way you should use the unit is: (assuming 2.4k resistor)
- The voltages are measured across the 2.4k resistor Rc.
- Switch open:  Measure voltage, call this V1;  this voltage relates to the collector current due to leakage
- Switch close:  Measure voltage, call this V2;  this voltage relates to the collector current due to leakage + the known base current through the 2M2 resistor.
- The actual gain (with the leakage removed) is  (V2 - V1) / (Rc * 4e-6) which is approximately  100 * (V2 - V1) with Rc=2.4k, and 250 * (V2-V1) with Rc = 1k.
- The leakage (refered to the collector) is  Ic = V1/Rc = V1/2.4k