Continue to Site

# Amplifier Input Resistance

Status
Not open for further replies.

#### frozone45

##### New Member
Hello, I had a question about finding input resistance. The question is shown in the attached "Input Resistance Problem Picture.pdf" file.

My attempt at solving the problem was to disconnect the signal source from the original circuit, and then attach a test independent voltage source where the signal source originally was. Then, I was to find an expression for "Rin". This expression would be my answer. I obtained: Rin = 1000 [Ω], as shown in the attached "Problem Work.pdf" file. The circuit including my test source is shown in the "Test Source Picture.pdf" file.

The correct answer for the problem is -43 [Ω]. I cannot figure out why this the correct answer.

Would anyone have any tips or advice for how to fix my mistakes in attempting this problem?

#### Attachments

• Input Resistance Problem Picture.pdf
178.2 KB · Views: 168
• Problem Work.pdf
443.1 KB · Views: 154
• Test Source Picture.pdf
308.2 KB · Views: 169
Bump

Hello, I had a question about finding input resistance. The question is shown in the attached "Input Resistance Problem Picture.pdf" file.

My attempt at solving the problem was to disconnect the signal source from the original circuit, and then attach a test independent voltage source where the signal source originally was. Then, I was to find an expression for "Rin". This expression would be my answer. I obtained: Rin = 1000 [Ω], as shown in the attached "Problem Work.pdf" file. The circuit including my test source is shown in the "Test Source Picture.pdf" file.

The correct answer for the problem is -43 [Ω]. I cannot figure out why this the correct answer.

Would anyone have any tips or advice for how to fix my mistakes in attempting this problem?

That was a good move disconnecting the voltage source from the amplifier. The amplifier input impedance will be the same no matter what you hang in front of it. Instead, insert a 1 amp current source from ground to where the voltage source was before. Now write your node equations. They should look like the attachment below. You should also get the same solved answers

"vl" is the voltage on the left side of R680 and "vr" is the voltage on the right side of R680 and the 5 V source. So, vl is -43 volts. Dividing by the 1 amp source gives the input impedance of -43 ohms. I did not have the time and patience to go through your calculations.

Ratch

Frozone45,

You are welcome.

Ratch

That was a good move disconnecting the voltage source from the amplifier. The amplifier input impedance will be the same no matter what you hang in front of it. Instead, insert a 1 amp current source from ground to where the voltage source was before. Now write your node equations. They should look like the attachment below. You should also get the same solved answers

View attachment 111438
"vl" is the voltage on the left side of R680 and "vr" is the voltage on the right side of R680 and the 5 V source. So, vl is -43 volts. Dividing by the 1 amp source gives the input impedance of -43 ohms. I did not have the time and patience to go through your calculations.

Ratch

I was able to find the correct input resistance. Thank you so much for the help! Sorry for the late reply

Status
Not open for further replies.