Input/output Impedance

Hi guys,

how to MEASURE and Calculate the INPUT/OUTPUT impedance for a circuit (consist of just transistors and or op-amps)?
How to you guys do so? Are you try it with a meter or just guess it? or...?

Thank in advance

It depends entirely on the specific circuit, generally for correctly designed opamp circuits it's obvious and simple.

Nigel Goodwin said:

It depends entirely on the specific circuit, generally for correctly designed opamp circuits it's obvious and simple.

Click to expand...

Here are 2 circuits for instance which I would like to know how to measure /calculate the input/output impedance for them?

Another question Which I have is how the designer noticed that a 10uF capacitor is a good chioce for the top of R1?
I know the formula which states F=1/(2xpixRxC), but?! And what The 'R' symbol standes for in this formula?

The opamp one is simple, input impedance is 23.5K, output impedance is pretty low (which is all you need to know).

The first one I would measure.

the first circuit you would need to measure, since it has a 1k output resistance, but there is also some negative feedback, which would lower it a bit more. the second one can be calculated from the output resistance of the op amp and the feedback ratio. there are two ways to measure output impedance. the first and most common is the indirect method. you feed in a signal at the frequency of interest (1khz is most often used). you then measure the output voltage open circuit. you then add a known load resistance to the output and measure the voltage again. for instance, if the first voltage measurement was 5Vrms, and the second voltage was 2.5Vrms with a 1k load resistor, then you know that Zo=Rload, so Zo=1k. the way this is calculated is:

Vopen= 5V (open circuit voltage)
Vloaded=2.5V (voltage across load)
Iload=2.5mA (current through load)
Vzo=Vopen-Vloaded=2.5V (voltage dropped across the output impedance is open circuit voltage minus load voltage)
Vzo/Iload=1000 ohms (divide output impedance voltage drop by load current to get output impedance)

the other method is the direct metnod, and it's something i have been tinkering with. it works very similar to the way a DC ohmmeter works, but uses an AC current source. you can find dtails about it here:
https://www.electro-tech-online.com/blogs/unclejed613/92-wouldnt-ac-ohmmeter-handy.html

Nigel Goodwin said:

The opamp one is simple, input impedance is 23.5K, output impedance is pretty low (which is all you need to know).

The first one I would measure.

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Thanks Nigel, but how could you calculate it? I read this website but could not get any method to meausre or calculate it:Non-Inverting Operational Amplifier Circuit :: Electronics and Radio Today

Besides, Can I consider this as a rule of thumb that the higher the input impedance the better result I will expect?

unclejed613 said:

the other method is the direct metnod, and it's something i have been tinkering with. it works very similar to the way a DC ohmmeter works, but uses an AC current source. you can find dtails about it here:
https://www.electro-tech-online.com/blogs/unclejed613/92-wouldnt-ac-ohmmeter-handy.html

Click to expand...

You can find AC ohmmeters on ebay. Here's a moderately priced one:

**broken link removed**

or, you could go the big bucks route:

**broken link removed**

or, you could roll your own multi-frequency version:

**broken link removed**

Electronman said:

Hi guys,

how to MEASURE and Calculate the INPUT/OUTPUT impedance for a circuit (consist of just transistors and or op-amps)?
How to you guys do so? Are you try it with a meter or just guess it? or...?

Thank in advance

Click to expand...

Determining properties of circuits such as voltage gain, current gain, input impedance and output impedance are done with network analysis.

For your first circuit, the 10K feedback resistor has no effect at midband (audio) frequencies because the right end is connected to a 220µF capacitor which goes to ground. This causes the feedback to only be effective at essentially DC.

The input impedance of this circuit is just the base input resistance (β+1)*re in parallel with 10K. You can determine re from the expression re = .026/Ie, where Ie is the emitter current.

The output impedance is approximately 22K/(β+1) || 1K, ignoring the collector resistance of Q1 which is probably much larger than 22k.

Had the 10k feedback resistor not had an AC ground at its right hand end, the calculation of input impedance would have been more complicated.

Nigel gave the correct input impedance for the opamp circuit. The feedback has no effect on the input impedance for this circuit.

The output impeadance if the 1st circuit is easy to calculate once you realize that the feedback is for bias only, as the cap in Q2's emitter keeps the sample point at AC ground. The impeadance them is simple the 1K resistor in parallel with Q2's base reflected resistance.

Zout = 1K || 22K/β(Q2)

In the second circuit, the output impeadance is the small amp's output impeadance parallel with the effect of the feedback resistors. The junction if R2 and R1 will remain at VCC/2, so the current through R2 due to a test voltage applied at the output is:

V(test) - VCC/2
------------------
R2.

The same current flows in R1. A KVL and a little algerbra and you have your answer.

After consulting Sedra and Smith, I have to change my answer for circuit #2. We define the inherent output resistance of the amplifier as ro, which can be from 50 ohms to several hundred ohms. Also, the open loop gain, we define as "A" Then, V- is the test voltage, Vt times the resistive divider:

V- = Vt*R1/R1 + R2. And the amplifier output voltage is Vo = -A*V-, which is: -A*Vt*(R1/R1 + R2), or --A*Vt*β,where β=R1/R1 + R2.

Now, the current due to the test voltage is: (Vt/R1 + R2) + (Vt - Vo)/ro, or using the equations above for Vo:

I = (Vt/R1 + R2) + (Vt + A*Vt*β)/ro

Finally, 1/Rout = I/Vt = 1/(R1 + R2+ + (1 + Aβ)/ro, and so

Rout = (R1 + R1)//[ro/(1 + Aβ)

The sedond term dominates so

Rout ~= ro/(1 + Aβ); where β is R1/(R1 + R2)

Hi and thanks for all inputs,
could you tell me why the input impedance of the noniverting op-amps in the second circuit is 23.5k??
Besides I asked a question about how to determine the value of the cap connected on the top of the R1 in the circuit? I want it to be short at AC so how determine the value of it?

Thanks

In impeadance analysis, set all DC sources to zero. Then, the iinput impeadance seen by the signal source is the two 47K Ohm resistors in parallel. Half of 47K is 23.5K. For the capacitor question, you want the impeadance of the cap ot be much less than the resistors connected to it, so since one reisitor is 1K, then use the formula:

Xc = 1/2ΠFC << 1K ohm, where C is the value of the capacitor, and F is the frequecy of operation. Solving for C gives:

C >> 1/(2ΠF*1Kohm)

BrownOut said:

In impeadance analysis, set all DC sources to zero. Then, the iinput impeadance seen by the signal source is the two 47K Ohm resistors in parallel. Half of 47K is 23.5K. For the capacitor question, you want the impeadance of the cap ot be much less than the resistors connected to it, so since one reisitor is 1K, then use the formula:

Xc = 1/2ΠFC << 1K ohm, where C is the value of the capacitor, and F is the frequecy of operation. Solving for C gives:

C >> 1/(2ΠF*1Kohm)

Click to expand...

Thanks but why R1 and R2 do not affect the impedance of the input too? As a rule of thumb, Can I consider this which the higher the input impedance the better result I will get in a circuit or...?

About the cap question, I read somewhere that for that cap the formula says: F-3dB=1/(2 pi RC). Can you tell me how much helpful that formula is really?

And the last question is how to calculate the cap values for an input or output of an amplifier like those 2 above circuits too? What value for "F" at your formula I must put when I want to design an amplifier? just 20Hz?

Thanks

Thanks but why R1 and R2 do not affect the impedance of the input too? As a rule of thumb, Can I consider this which the higher the input impedance the better result I will get in a circuit or...?

Click to expand...

Think about the input current. Does it flow through R1 and R2?

About the cap question, I read somewhere that for that cap the formula says: F-3dB=1/(2 pi RC). Can you tell me how much helpful that formula is really?

Click to expand...

Not much helpful for your earlier question. Use the equation I gave you.

And the last question is how to calculate the cap values for an input or output of an amplifier like those 2 above circuits too? What value for "F" at your formula I must put when I want to design an amplifier? just 20Hz?

Click to expand...

You need to know what the frequency of the signal will be that you're trying to amplify. If there is a range of frequencies, you need to use the lower limit when designing a coupling capacitor that you want to be a virtual short.

the + and - inputs of the op amp are separated by a theoretically infinite impedance. there are actually physical limits to this, and it's usually in the range of 10E+10 to 10E+12 ohms, so it's much higher than you need to worry about. so R1 and R2 are isolated from the input impedance of the + input. if this were an inverting amp, then R1 would be in series with the input and a "virtual ground" at the - input, and the input impedance would be the resistance of R1. R2 wouldn't matter because it would be dropping the whole inverted output voltage across it and ties to the same "virtual ground".

BrownOut said:

After consulting Sedra and Smith, I have to change my answer for circuit #2. We define the inherent output resistance of the amplifier as ro, which can be from 50 ohms to several hundred ohms. Also, the open loop gain, we define as "A" Then, V- is the test voltage, Vt times the resistive divider:

V- = Vt*R1/R1 + R2. And the amplifier output voltage is Vo = -A*V-, which is: -A*Vt*(R1/R1 + R2), or --A*Vt*β,where β=R1/R1 + R2.

Now, the current due to the test voltage is: (Vt/R1 + R2) + (Vt - Vo)/ro, or using the equations above for Vo:

I = (Vt/R1 + R2) + (Vt + A*Vt*β)/ro

Finally, 1/Rout = I/Vt = 1/(R1 + R2+ + (1 + Aβ)/ro, and so

Rout = (R1 + R1)//[ro/(1 + Aβ)

The sedond term dominates so

Rout ~= ro/(1 + Aβ); where β is R1/(R1 + R2)

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I know this will probably confuse our OP, but Zout of the op amp is complex (has an imaginary component) due to the gain roll off of the op amp. In other words, A is complex, so Zout is complex, and appears to have an inductive component. Zout will rise with increasing frequency, and the voltage will lead the current.

if the OP has LTspice, i have a paper i'm writing about amplifier output impedance that has experiments in it that can be run in LTspice.

Your opamp has an input capacitor, an output capacitor and a feedback to ground capacitor. Each one drops the output 3dB at the cutoff frequency.
Three capacitors cause the output to be -9dB at the cutoff frequency.
-9dB is almost 1/10th the power so if the cutoff frequency is 20Hz then it will be at a very low level and 100Hz will also be reduced.
So select a cutoff frequency of 7Hz or less for good bass.

BrownOut said:

Not much helpful for your earlier question. Use the equation I gave you.

Click to expand...

But I took a look at PRACTICAL ELECTRONICS FOR INVENTORS book and it used the formula I told (What that formula States really?) And when Is help full? What is F-3dB for an input cap which is working at 20Hz to 20Khz?