obtain HFE from voltage

[Dimitree]

hi everyone
I have a particular question, it's a bit hard to explain for me since my english is not fine, but I'll try:

in a transistor stage like this:


knowing every value (power supply voltage, resistors and capacitors) and the voltages across the 3 pin of transistor, is it possible to calculate the transistor gain that we need in order to get the same voltages again (and so the same results?) in a "clone" of this stage?

I'm sorry if this is a stupid/obvious question but I'm not an expert at all, so thank you in advance.

[alanlan]

Yes, it is pretty much Rc/Re.  There are some "but's". 
- The higher the value of Re (up to a limit), the more negative feedback there is and the closer you will get to the stated equation. 
- Also, you have to take into account input source impedance and output load to properly calculate the gain of the stage when it is "in circuit".
- The input and output caps will affect the frequency response.
- The available voltage swing on the output will depend on the gain, the power supply voltage and the bias current (required by the transistor) and bias resistor.

The circuit isn't very good as a production circuit - it would be improved by having a resistor from the base to ground as well as from base to supply.  That way, you can set the base voltage and therefore reliably predict the biasing.  Your bias circuit should carry 5 to 10 X the worst case base current so it isn't significantly affected by individual component variations.

There are other considerations, I could probably go on if I thought about it a bit more.

[CynicalMan]

I think he was talking about the transistor hFE, not the stage's gain.

Edit: I wasn't thinking. hFE is just Ic/Ib. So, using ohm's law, that's ((Vcc - Vc) / Rc) / ((Vcc - Vb) / Rb).

[Dimitree]

thanks for the help guys.
so could you help me to apply those formula to my "schematic"? (i'm not that fast with electronic-math  )
for example, calling Vout the voltage on the second cap, and Vin as the base voltage, how can I do?

[Dimitree]

the point is that I had a fuzz circuit borrowed from a friend, I like the sound and I wanted to clone it, so I copied everything but it didn't soudn the same, the voltage across the transistor were different so I guess the problem is that I haven't used the same HFE. I couldn't check it 
now knowing the voltages and the values I'd like to understand what HFE would I need in order to get the same result, without trying tons of transistors.

[CynicalMan]

Vout and Vin aren't the information you need. You need the voltage on the base (Vb), the voltage on the collector (Vc), and the supply voltage (Vcc), as well as the resistor values. If you post the voltages here, I can show you the calculation.

Have you tried tweaking the bias? By changing the value of the base resistor so that the voltage on the base matches your friend's values, you might be able to make a closer clone.

[Dimitree]

no I didn't tried that
if I get the same voltages, it means the clone is 1:1 on the paper? (obviosly other parts of the circuit could change that)

btw another difficulty is that that NPN transistor is reversed 

supply voltage is 9.2V
so voltage on the base is 0,7V (base resistor 910k)
voltage on the collector 0,2V (going to ground with 3k resistor)
voltage on the emitter is 3V (connected to + supply) resistor is 180k

that's pretty mad but I like how it sounds

[CynicalMan]

if I get the same voltages, it means the clone is 1:1 on the paper? (obviosly other parts of the circuit could change that)

There are many other factors, even within the one stage, but that should be enough to make them sound alike.

btw another difficulty is that that NPN transistor is reversed  

If by that you mean that the collector and emitter are swapped, then the calculation I posted before doesn't apply. Normally the Emitter-Base diode is forward-biased and it controls the collector current. But in that case it would be reverse-biased. Now, you're dealing with the reverse hFE, Ie/Ib, which applies when the the Emitter-Base diode is reverse-biased and the Base-Collector diode is forward-biased. In your case it's  ((Vcc - Ve) / Re) / ((Vcc - Vb) / Rb), or  3.7.

[Dimitree]

Many thanks for the help Alex, I really apprecciate
yes the emitter and collector are swapped, as you can by the voltages collector goes to ground while emitter to + supply.
so the result is 3.7, and how that apply to HFE that I can read with my tester (or with Geofex hfe germanium transistor tester) ?

[Dimitree]

btw do you suggest me to adjust the bias to obtain that voltages, instead of looking for that transistors?

[CynicalMan]

Sorry, I missed your previous post. Considering the wide range of reverse betas that transistors have, I'd try to match the beta. If the circuit is socketed, I'd just measure them in there. The voltage over the 910k resistor should be around 1.4 times the voltage over the 180k resistor.

[Dimitree]

Considering the wide range of reverse betas that transistors have, I'd try to match the beta. If the circuit is socketed, I'd just measure them in there.

no unfortunately I don't have the chance to check the beta on the original, just the voltages, so I'm at the beginning again

The voltage over the 910k resistor should be around 1.4 times the voltage over the 180k resistor.

on the original it's 0.7V over the 910k (on the base) and 3.1V over the 180k (on the emitter)

[CynicalMan]

no unfortunately I don't have the chance to check the beta on the original, just the voltages, so I'm at the beginning again

I mean check on your clone if possible.

on the original it's 0.7V over the 910k (on the base) and 3.1V over the 180k (on the emitter)

The voltage over a resistor is the voltage on one side relative to the other. In other words, it's the difference between the voltages on either side. So, with a 9.2V supply, the voltage over the base resistor is the difference between 9.2V and 0.7V, or 8.5V. You can measure it by just putting your meter's probes on each side of the resistor.

The voltage over the base resistor is 8.5V and the voltage over the emitter resistor is 6.1V.

[Dimitree]

perfect, so that's right, it's 1.4 times
sorry if I'm so slow to understand those things

on my clone I can check everything, for example using a 2N5089 I have almost the same voltage on the base and collector, but really different on the emitter. Same thing using 2N5088, 2N3904, and some other. (on the original there's a 2N5089)
so what could I check?