good day guys. i just found the circuit that I can use for my report but the problem is I'm am struggling finding the input and output impedance of this circuit. can someone help me?

 

Since this is your 'homework' you are expected to post your best effort so far at solving the problem; then we can provide guidance where needed.

 

You found a circuit that has errors and it does not work.
Its input transistors are wrongly PNP but they should be NPN.

Also the TIP41 transistor has a max allowed voltage of 40V but the circuit gives it up to 90V. Use a TIP41C instead.

Input and output Impedances are a range of numbers because the hFE of the transistors are a range of numbers. Maybe your assignment uses "typical" transistor hFE numbers.

 

good day guys. i just found the circuit that I can use for my report but the problem is I'm am struggling finding the input and output impedance of this circuit. can someone help me?
View attachment 295048

Hello there,

I did not check Q1 and Q2 as AG did because i did not get that far. I noticed first that the two output transistors Q4 and Q5 were Darlington transistors but there were only TWO diodes in the biasing scheme. That means each transistor only gets one diode to match its base emitter diode drops, which because there are TWO diode drops for each transistor BE diodes, both output transistors are starved for current. This means the output impedance, driving Q3 directly, would either be very high or not work at all.
So at the very least, two more diodes have to be added to the D1 and D2 string so the bias is closer to being correct.
There are better biasing schemes but if you want to stick with this one that might be ok, but only if you add two more diodes.
Note that two diodes, D1 and D2, are used when the output transistors are just regular single transistors not Darlington transistors.
If you like you could instead change the two output transistors to regular transistors, but you'd also have to change the biasing scheme anyway to make up for the need for higher bias currents, so maybe stick with the Darlington transistors.
I would guess when this is fixed the output impedance, using the direct Q3 drive, would be very low, which seems to be what the original designer was after.
I would also guess that once the input stage is fixed, if it needs to be, the output impedance will still be very low.

 

@Audioguru again identified a real problem. Q1 and Q2 are PNP and in the schematic the emitters are swapped with the collectors.

So..change the collectors to PNP emitters and the emitters (lower electrodes) to collectors. The part number for the transistors indicate the right kind of PNP for this use.

 

Its input transistors are wrongly PNP

I agree with Dick. They've just been drawn upside down.

 

Assume the schematic was corrected; What is the input impedance? Start at the input and you have C1 & R1. That gets you close to the right answer. Ony because this is a school project, you should go beyond that and find the answer is slightly lower than that. So agentperry15 please tell us what is the input impedance?

 

Don't forget the two more biasing diodes or this will NEVER work.

 

Sorry, I was wrong about the input transistors.
I should have looked at the similar PNP input transistors on the old LM324 opamp with correct emitter and collector connections.

 

Assume the schematic was corrected; What is the input impedance? Start at the input and you have C1 & R1. That gets you close to the right answer. Ony because this is a school project, you should go beyond that and find the answer is slightly lower than that. So agentperry15 please tell us what is the input impedance?
View attachment 295188

Looks much better now.

 

thank you for your answers guys. but still don't know to calculate this properly. do you have any circuit that has 25 watts output power composed only with transistor?

 

thank you for your answers guys. but still don't know to calculate this properly. do you have any circuit that has 25 watts output power composed only with transistor?

Hello again,

As mentioned previously, the first thing you need is a circuit that is drawn correctly. Since the two output transistors are Darlington transistors the bias arrangement with the two diodes is not correct so the output impedance would be very high if it worked at all. You need a working circuit first.
If you add two more diodes you can probably get some reasonable operation from that circuit.

To calculate the actual output impedance with a multiple transistor amplifier you have to go through some work to calculate the various nodes and such, and with a lot of transistors it's going to be a lot of work i am not sure if you willing to go through that so i have not given you any details yet. Since the calculation will be an estimate, you can probably simplify this circuit by driving Q3 and eliminating the two input transistors to start with. If you assume that Q3 is driven in a reasonable way then you don't have to include the first two to get an estimate. Feedback is normally imperative to get low output impedance, but that would also assume that Q3 is driven in a reasonable way. We can always test for any differences in spice without too much trouble, once the circuit is modified to be correct.

I can describe the procedure using a simplified transistor model which lends itself to this kind of calculation but even then i have a feeling you will not want to do it that way because you have to consider a lot of nodes even when you drive Q3 and eliminate the first two transistors.
What most people do these days, if you are allowed to that is, is use spice to get an estimate of the output and/or input impedances.

The main procedure is to drive the input with a signal that provides some output, let's say 2 volts peak for example, with an open circuit output.
Next, choose some load, maybe 10 Ohms but this could vary widely. The idea is to find a load that causes the output to decrease by one-half, and in this case it would mean the output would go down to 1 volt peak. When you find that resistance, the output resistance is equal to the output impedance. This assumes that the output impedance is purely resistive which is often ok to do with audio circuits.
It is probably better however to use two resistor values, starting with one maybe 20 Ohms then another 10 Ohms, and compare and calculate the output impedance from the two different output voltage levels. That way the output is never completely unloaded.

Another variation though is that since this is an AC amplifier you may have to use a very large capacitor in series with the output and the load resistance. In theory the capacitance has to be infinite, but you can usually get by with a very large value that drops very little voltage at the test frequency which is usually 1kHz.
This capacitor allows the output to assume it's normal DC level while still loading it properly for the test.
It's the same idea though, with no load if you get 2 volts peak and with a 10 Ohm resistor you get 1 volt peak, then the output impedance is 10 Ohms.
This comes from the theory that any voltage source that has an output resistance R will put out only half of the open circuit voltage when it is loaded with that same value R.

Does this make sense and help any?
If you really need to do the calculations by hand and can not use spice then we have to talk about this test some more.

Yes, we can show you some transistor amplifier circuits that are known to work well, but would that really help you?