Opamp Zin et Zout measurement

[kgavionics]

hi
i'm trying to measure the LM741 input and output impedance and i'm facing a problem:

-Since the input impedance is very high and the output impedance is very low , i can't use the conventional method to measure the input impedance with a potentiometer.

can someone give me a clue please

[c4757p]

May I ask why? The LM741 isn't exactly the world's best op amp - what application are you using it for that actually requires a measurement of such a high impedance, and yet is perfectly OK with an LM741?

[kgavionics]

May I ask why? The LM741 isn't exactly the world's best op amp - what application are you using it for that actually requires a measurement of such a high impedance, and yet is perfectly OK with an LM741?

it's a lab.
I'm trying to verify the effect of negative feedback on a non inverting op amp.

[nitro2k01]

it's a lab.
I'm trying to verify the effect of negative feedback on a non inverting op amp.

Would you care to explain exactly what the assignment is about? Is it actually about measuring the input/output impedances, or is that something you thought of doing during the assignment? From what you're saying, "I'm trying to verify the effect of negative feedback on a non inverting op amp." it seems like they just want you to connect a signal on the positive input, and a potentiometer through negative feedback and see the signal vary in amplitude as you adjust the pot.

[kgavionics]

it's a lab.
I'm trying to verify the effect of negative feedback on a non inverting op amp.

Would you care to explain exactly what the assignment is about? Is it actually about measuring the input/output impedances, or is that something you thought of doing during the assignment? From what you're saying, "I'm trying to verify the effect of negative feedback on a non inverting op amp." it seems like they just want you to connect a signal on the positive input, and a potentiometer through negative feedback and see the signal vary in amplitude as you adjust the pot.

normallay, when we want to measure for instance the input impedance of an amplifier, we use a pot between the signal source and the amplifier input,after that we apply for signal to the amplifier and we adjust the the pot to have the half of the input signal.the last step is to measure the pot resistance that's equal to input inpedance of the amplifier and because the input impedance of the opamp is very high, i can't use this method.

[nitro2k01]

I see. The datasheet for the LM741 quotes an input impedance of min: 1M/typ: 6M for the LM741A variant and min: 300k/typ: 2M for the non-suffix and C variants. However, what you're actually measuring is the ratio of the voltage divider formed by the external resistor (in your case the potentiometer) and input impedance of the amplifier. You're using a ratio of 1/2 because it simplifies the calculations for you. But this can be done with any resistance (assuming it's close enough to the imepdance) and thus any voltage ratio.

The theory is that the resistor and the input impedance form a voltage divider. This ratio can be expressed as:

Z/(R+Z) = V_out/V_in

Z is the amplifier input impedance. R is the resistor value. V_in is the input voltage. V_out is the measured output voltage.

If R=Z, (which is what you're aiming for in your experiment) then obviously the left side becomes R/(2R) = 1/2. However, if you solve the equation for Z, you can easily realize that you can choose any value of R (within reason; should be close to the input impedance) This gives the following formula, assuming I didn't make any mistakes.

Z = R*V_out/(V_in-V_out)

Now go find a 1M (or so) resistor and try it out.

[NiHaoMike]

Connect an AC microammeter between the source and input. Combined with a voltage reading, the impedance can be calculated. Use a low frequency like 100Hz to avoid stray capacitance problems.

For the output impedance, ground the input and apply a signal to the output through a resistor (100R-1k or so), making sure to stay well within the output current rating. Then measure voltage and calculate impedance by thinking of it as a voltage divider.

[kgavionics]

thank you guys for your help,its very much appreciated.