How to measure the impedance

[zesla]

Hi there,

I want to know how to practically measure the input/output impedance of a biased transistor and an op-amp?

Thanks a bunch

 

[MikeMl]

Input impedance:

1. Set the frequency of a sine-wave audio signal generator to the freq at which you want to measure the input impedance of your amp.

2. Connect the generator output to the input of your amplifier. Note the rated output impedance of the generator. Use the output with the lowest output impedance. My HP gen has both 50Ω and 600Ω outputs.

3. Set the generator amplitude to the nominal level for your amplifier which makes an unclipped output from your amplifier. This is assuming that your amplifier is linear. Note the Peak to Peak amplitude at the output. Useful for computing the gain, too.

4. Insert various resistances between the gen output and your amplifier input until the output from your amplifier is half of the original amplitude. I would use a rehostat or a decade resistance substitution box. The resistance value that reduces the output to half is the input impedance at the set frequency.... Dont forget to account for the output impedance of the generator.

 

[MikeMl]

Output impedance:

1. Set the frequency of a sine-wave audio signal generator to the freq at which you want to measure the input impedance of your amp.

2. Connect the generator output to the input of your amplifier.

3. With no load connected to the amplifier (just a scope probe or high-Z voltmeter), set the generator amplitude to make an output from your amplifier which is about 1/10th of the clipping level.

4. Connect various load resistances to your amplifier output put until the output from your amplifier is half of the original amplitude. The resistance value that reduces the output to half is the output impedance of the amplifier at the set frequency.

Note: Output impedance from opamps with negative feedback is usually verrrry low. However, they have a finite current drive capability, which causes distortion of the output when you connect a low-impedance load. The distortion may go away at lower output amplitudes, so you must use a scope to observe what is going on...

 

[zesla]

Thanks Mike, but very strange!! specially the NO 4!

Why the HALF OUTPUT is the base parameter and is when the resistor is equal to the input/output impedance?!!

A very stange method to determine the impedances to me!

 

[audioguru]

Thanks Mike, but very strange!! specially the NO 4!

Why the HALF OUTPUT is the base parameter and is when the resistor is equal to the input/output impedance?!!

A very stange method to determine the impedances to me!

It was a strange description.
1) The signal from the low output impedance generator is adjusted for an output from the amplifier that is not distorted.
2) Insert a resistance between the generator and the input of the amplifier and adjust its value until the output level of the amplifier is half what it was without the resistor.
3) Measure the value of the adjustable resistor which has a value the same as the input impedance of the amplifier.

 

[zesla]

Thanks audioguru,

But Why the HALF is the key of the input/output impedance????
I am not able to see main relationship between the input impedance and the half output!? Is it due to something like 3dB freq or half output power which is used to describe F3dB?

 

[Nigel Goodwin]

Thanks audioguru,

But Why the HALF is the key of the input/output impedance????
I am not able to see main relationship between the input impedance and the half output!? Is it due to something like 3dB freq or half output power which is used to describe F3dB?

It doesn't have to be half, half just makes the maths very easy that's all.

 

[audioguru]

-3dB is half the power but is 0.707 times the voltage.
-6dB is half the voltage.

When the output drops to half the level with your added resistor then the input impedance of the amplifier is the same as the value of the resistor.

 

[zesla]

-3dB is half the power but is 0.707 times the voltage.
-6dB is half the voltage.

Is this the reason of what does happen regarding the equivalence resistor values and the input impedance?

When the output drops to half the level with your added resistor then the input impedance of the amplifier is the same as the value of the resistor.

Yeah but yet nobody told the reason?
Nigel told that the half value is not critical which this makes the understanding the reason much harder!

Nigel,
Which math are you talking about?

 

[Nigel Goodwin]

Is this the reason of what does happen regarding the equivalence resistor values and the input impedance?


Yeah but yet nobody told the reason?
Nigel told that the half value is not critical which this makes the understanding the reason much harder!

Nigel,
Which math are you talking about?

What you're doing here is making a potential divider - two resistors, one known, one unknown.

By measuring either the voltage, or the current, you can easily calculate the value if the unknown resistor - simple ohms law.

It's simple enough maths - BUT if you make the known resistor the same value as the uknown one, you don't have to calculate anything at all, just read it off.

Easiest way is to use a potentiometer, adjust it for half the original signal, then remove it and measure it's resistance with a meter,

 

[zesla]

What you're doing here is making a potential divider - two resistors, one known, one unknown.

By measuring either the voltage, or the current, you can easily calculate the value if the unknown resistor - simple ohms law.

It's simple enough maths - BUT if you make the known resistor the same value as the uknown one, you don't have to calculate anything at all, just read it off.

Easiest way is to use a potentiometer, adjust it for half the original signal, then remove it and measure it's resistance with a meter,

Thanks a lot Nigel, It clearly makes sense now. I took it just when you mentioned it as a potential divider. It is a very good trick to measure the input impedance.

But If my brain serves, I remember that there was another method to measure the impedances by measuring Vopen then Vfull load by a known resistor then use a formula to calculte the impedances?

 

[Nigel Goodwin]

Thanks a lot Nigel, It clearly makes sense now. I took it just when you mentioned it as a potential divider. It is a very good trick to measure the input impedance.

But If my brain serves, I remember that there was another method to measure the impedances by measuring Vopen then Vfull load by a known resistor then use a formula to calculte the impedances?

Like I said, the simple maths - work it out, it's only ohms law.

Two resistors in a potential divider, you know the voltage at the top, the voltage at the middle, and one of the resistors - what value is the second resistor?.

 

[aljamri]

But If my brain serves, I remember that there was another method to measure the impedances by measuring Vopen then Vfull load by a known resistor then use a formula to calculte the impedances?

I think what have been explained here is how to measure Output impedance. Input impedance has different procedure, may be thats what you remember.

 

[Nigel Goodwin]

I think what have been explained here is how to measure Output impedance. Input impedance has different procedure, may be thats what you remember.

It's the same procedure, the only difference is that the known and unknown resistors swap places.

 

[MrAl]

Hi,


There's just one catch to the "Add resistor until output goes to half of what it was" method. That is, some amplifiers are biased at a certain DC point so you have to be able to change the resistance (say for the load) without disturbing the DC bias point.
One way to accomplish this is to use a very large value capacitor in series with the load. The capacitor value is to be chosen to provide a very low impedance at the test frequency. The capacitor charges up to the bias level and then the resistance can be varied as needed.
It should be noted that some amplifiers, without this extra large capacitor, will shift bias and wont work as well, if at all.

The reason we want to use a resistance that gives us half output is because when we finally find the right resistance it has to be the same as the output impedance because of the voltage divider effect of two resistances that are exactly the same (the output is half of the input). You can choose any resistance you like up to a certain point but you really have to stay within range of the design point, and if you choose a larger resistance than the output wont drop as much.

 

[crutschow]

Using the resistor method to reduce the output voltage to 1/2 for a circuit with a lot of negative feedback, such as many op amp circuits and power amplifiers, will give an interesting result. The output voltage will initially have little variation with load change, since the closed loop output impedance is very low. Then the voltage will suddenly drop when you reach the current limit of the output. Thus the measurement to 1/2 output voltage basically gives you the open loop output impedance of the circuit, not the closed loop impedance during normal operation.

To measure the closed-loop output impedance you would just vary the resistance to a value above the current limit point, which will typically give only a small voltage change. Then use ohm's law to calculate the output impedance. This can be in the milliohm region for a circuit with significant feedback.

 

[Nigel Goodwin]

Using the resistor method to reduce the output voltage to 1/2 for a circuit with a lot of negative feedback, such as many op amp circuits and power amplifiers, will give an interesting result. The output voltage will initially have little variation with load change, since the closed loop output impedance is very low. Then the voltage will suddenly drop when you reach the current limit of the output. Thus the measurement to 1/2 output voltage basically gives you the open loop output impedance of the circuit, not the closed loop impedance during normal operation.

To measure the closed-loop output impedance you would just vary the resistance to a value above the current limit point, which will typically give only a small voltage change. Then use ohm's law to calculate the output impedance. This can be in the milliohm region for a circuit with significant feedback.

Measuring power amplifiers is a VERY, VERY different matter!

 

[audioguru]

I think you measure the output impedance of a power amplifier by injecting a signal into its output from another amplifier that has an extremely low measured output impedance, in series with a very low value resistor.

 

[Nigel Goodwin]

I think you measure the output impedance of a power amplifier by injecting a signal into its output from another amplifier that has an extremely low measured output impedance, in series with a very low value resistor.

Or you could measure the voltage drop at the output when you connect a 4 ohm load, but the voltage drop will be VERY small.

However, for a solidstate amplifer there's really no need whatsoever to measure (or indeed know) the output impedance.

 

[audioguru]

Some people still think that an amplifier's output impedance must be the same as the speaker's impedance in order to throw away 2/3rds of the power of the amplifier and so that the amplifier does not damp the resonances of a speaker.