1
\$\begingroup\$

This is one of the tasks in my BJT amplifier lab. I have to find impedance seen to the right from V1 which can easily be done by taking a frequency sweep and measuring \$\small \frac{V_1}{I_{C_2}}\$ at 1kHz. I also have to find the input and output impedances. Input impedance is also easy since I can just divide the input voltage by the current through it. But how to find the output impedance. I know I cannot use the traditional circuit theory method of killing the independent sources and adding a test source since that will change the biasing of the transistors. Also I need to measure it using LT Spice and not calculate it using small-signal analysis.

So what should I do here?

Any help would be appreciated! enter image description here

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Hint: Change R_L to7 Ohms to 9 Ohms. Since V_out won't change much... \$\endgroup\$ Nov 22, 2022 at 16:20

1 Answer 1

0
\$\begingroup\$

The biasing of this amplifier is always going to change depending on the input voltage, and therefore, the amp's output impedance varies too. You must set the input voltage (V1) to zero, otherwise there is no defined output impedance for you to even determine. Once you've done this, you can attach an AC current sink to the output (disconnect the load) and measure the AC voltage variation at the output.

\$\endgroup\$
2
  • \$\begingroup\$ Yes, but in order to find the two other impedances that I mentioned we do not have to kill the AC input source. Isn't there a way to do so for this case as well. Also the measured value matches the calculated value range for impedance measured to the right from node \$V_1\$ which is \$1k//1k//(\beta + 1)\cdot 8 = 400 \text{to} 500 \Omega\$. And the measured value is \$ 406 \Omega \$... \$\endgroup\$ Nov 22, 2022 at 17:19
  • \$\begingroup\$ It simply doesn't make any sense to leave the input signal running while trying to measure the output impedance. It's just going to interfere and ruin your measurement. You should always just have one AC source in the circuit when trying to determine its AC parameters. In this case, that source has to be an AC current source connected to the output. All other sources have to be DC only. \$\endgroup\$ Nov 22, 2022 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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