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

Thanks enter image description here

  • 2
    \$\begingroup\$ 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 \$\endgroup\$ – Voltage Spike Jan 31 '18 at 19:52
  • \$\begingroup\$ Its not homework you moron! \$\endgroup\$ – John Feb 1 '18 at 12:19
  • \$\begingroup\$ 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. \$\endgroup\$ – Voltage Spike Feb 1 '18 at 16:13
  • \$\begingroup\$ 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. \$\endgroup\$ – John Feb 2 '18 at 21:26
  • \$\begingroup\$ 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. \$\endgroup\$ – 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.

The above assumes that RL isn't connected.

I have also assumed that the circuit is for an audio amplifier with mid-band frequency around 1 kHz.

| improve this answer | |
  • \$\begingroup\$ Thanks. If the load is an 8 ohm speaker, then the output impedance would be R5 or R6 || RL = 10Ω || 8Ω = 4.4Ω? \$\endgroup\$ – John Jan 31 '18 at 17:31
  • \$\begingroup\$ If the load is an 8 ohm speaker it isn't part of the amplifier and doesn't contribute to the amplifier's output impedance. \$\endgroup\$ – Andy aka Jan 31 '18 at 17:34
  • \$\begingroup\$ If RL is a resistor and part of the amplifier then yes, the output impedance is 10||8 = 4.44 ohms. \$\endgroup\$ – Andy aka Jan 31 '18 at 17:36
  • \$\begingroup\$ @John Thanks is not needed but maybe wait a while before you formally accept the answer in case someone does a better job. \$\endgroup\$ – Andy aka Jan 31 '18 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.