# How would you calculate the input and output impedance of this amplifier?

How would you calculate the input and output impedance of this amplifier? I have done a lot of research and found so many conflicting pieces of information. Not sure what is right.

C1 = 10Uf, C2 = 1000Uf R1 = 50KΩ, R2 = 7.5KΩ, R3 = 820Ω, R4 = 100Ω, R5 = R6 = 10Ω D1 = D2 = 1N4148 Q1 = 2N3904, Q2 = TIP31C, Q3 = TIP32C RL = 8Ω Or 16Ω

Supply Voltage = 12V

• I'm voting to close this question as off-topic because Homework (or a regular problem) without an attempt at a solution is off topic – Voltage Spike Jan 31 '18 at 19:52
• Its not homework you moron! – John Feb 1 '18 at 12:19
• Sorry, we get a lot of questions like this on SE.EE that are homework. I would be great if you could try and show what you have done to help people answer your question. – Voltage Spike Feb 1 '18 at 16:13
• Not to worry. I was on another site and the respondents were treating people like dirt, which was very frustrating! I am working on a procedure for designing such a circuit from scratch. I will post the it as soon as I have it complete. – John Feb 2 '18 at 21:26
• I did not mean a personal attack when I tried to close your question, but as a means for you to improve it so you can get some answers. We also try and improve the quality of questions to keep EE.SE looking nice. – Voltage Spike Feb 2 '18 at 21:30

If Q1 was an NPN transistor (as I suspect it should be) the input impedance at mid-band (no appreciable extra impedance due to C1) is approximately R1 || R2 and, if you wanted to be pedantic you would have that impedance in parallel with $\beta\cdot$R4 where $\beta$ is the gain of Q1.
The output impedance (mid band hence ignoring C2) is dependant on the current flowing through Q2 and Q3. If that current is small then the internal $r_E$ might be higher than R5 or R6 and will sway things. If you ignored $r_E$ then on positive half cycles the output impedance is R5 and on negative half-cycles it is R6. If R = R5 = R6 then the output impedance is approximately R.