Mesuring input impedance...

[O.Binette]

Hi,

I would like to know how I can calculate the input impedance and the output of a circuit, by hand or with a multimeter.

And since I don't really know what is input impedance, can we increase it by putting a resistor before the amplification stage? and is having the input connected to ground by a small (1000k-10k) resistor increasing the input impedance?

Thanks!

[R.G.]

I would like to know how I can calculate the input impedance and the output of a circuit, by hand or with a multimeter.

Theoretical calculation is a series of approximations.

You can measure the input impedance by (1) putting in a test signal of some known amplitude (2) measuring the output amplitude (3) inserting a resistor in series with the input and increasing the resistor until the output is halved. The inserted resistor is then equal to the input impedance. This only works if the circuit is not a compressor or a clipper, which changes the output amplitude on its own.

And since I don't really know what is input impedance,

Input impedance is literally the input voltage divided by the input current. If you put in one volt of input signal, and it requires one milliampere of input signal to get this to happen, the input impedance is 1V/1ma = 1K ohm.

can we increase it by putting a resistor before the amplification stage?

This increases the input impedance of the entire circuit, including the inserted resistor. However, it decreases the amount of signal getting to the circuit by the voltage divider effect of the added resistance and the actual input impedance. If your circuit has an input impedance of 100K and you put 1M in series with it, then the new input impedance of the whole thing is 1.1M. However, the actual signal getting to the circuit is 100K/1.1M = 1/9th of the input voltage at the front of the new resistor. Not to mention that the added resistor adds noise.

and is having the input connected to ground by a small (1000k-10k) resistor increasing the input impedance?

No. That decreases the input impedance because the input is paralleled by the new resistor and makes the input suck more current than it otherwise would have.

[Incubusguy]

I'll offer up a simple explanation and hopefully others will chime in with more/better explanations!

You can think of the connections between stages as potential dividers. Assuming you have a guitar, pedal and amp, you can look at the chain as follows:

Guitar into pedal: The output of the guitar can be thought of as R1 of the potential divider. Therefore, the output impedance of the guitar is a series resistance (ignoring inductance and capacitance). The input impedance of the pedal is R2 of the potential divider and is, therefore, a parallel resistance to ground.

Pedal into amp: The output of the pedal, in this case, is R1 of the potential divider and the input of the amp is R2 of the potential divider. Again, R1 and R2 are series and parallel resistances respectively.

Putting a series resistor between two stages will increase the source impedance/R1. Putting a parallel resistor between two stages will alter the load impedance (it may increase or decrease depending on the input impedance of the following stage). Using the potential divider and parallel/series resistor equations, you can calculate the effect of the added resistor if you know the source/load impedances.

Now, you can use the same equation in a different way to determine the input/output impedance of a circuit.

If you have a known voltage source, say a 1V peak sine wave, entering a circuit, then you can take 1V as the input to your potential divider. If both resistances in the potential divider are equal, you would expect to see 0.5V on the output. We can't change R1, the output impedance, but we can alter what load impedance, R2, it sees. Therefore, if we use different values of parallel resistors on the output, we can plot a graph of how the potential divider voltage output changes with respect to R2 (it will look like a parabola). When you have 0.5V on the output, you know that the value of your resistor, R2, is equal to the output impedance of the circuit.

I must point out that I have technically described resistances because impedance is a combination of inductance, capacitance and resistance. However, the same kind of principle still applies; it's just a bit more complicated.

EDIT: It would seem R.G. got there first! 

[PRR]

> This only works if the circuit is not a compressor

If, instead of measuring the box output, you measure the box INput, you avoid any monkey-motion inside the box.

The practical problem is that input signal may be smaller than your meter will read accurately. Also it is essential that meter loading be negligible (or accounted-for). While modern DVMs are perhaps more suitable than my old VTVM, guitar signals may be quite small and most popular-price DVMs have poorly specified AV-Volt impedance and response. 

Other problems:

Most inputs vary with frequency. To do a thorough job you should sweep all frequencies of interest (and a bit more to check for surprises; the 2-Q phono preamp tends to go wacko just below the audio and). (Again, beware the DVM which rolls-off above 400Hz.)

Input signal must be in the "linear zone" (which is a dubious thing for a fuzz/clipper). If you apply 10V signals to a high-gain 9V box, you won't get the answer you want. It may be best to observe "typical" signals first and set-up to measure with similar signals. 20mV may be the upper limit of "linear" for some pedals (though impedance in the non-linear zone may then be of interest). Cheap DMMs may reach 5KHz with full 200mV inputs but get very slow for smaller inputs.

Pre-calculating input impedance is a "simple" problem in circuit design. (By your questions, I suspect you have not traveled much along this long path.)

> can we increase it by putting a resistor before the amplification stage?

Putting in a series resistor increases input impedance, yes. If it increases significantly it also cuts the signal. This is useful when you have a 0.05V 200 ohm mike input and a 10V medium-Z signal. A 40K resistor in series with the 200 ohms will give a nice high 40.2K input and the desired 10V/0.05V attenuation. (I knew a mix-board which was all like that.) OTOH, in stage/pedal work, the "reason" you might care about impedance is to NOT have loss. So sticking resistors in series is usually bad. However a classic fuzz has a 6K transistor base impedance padded-out with a 39K resistor which is "essential" to how it all works well.

> input connected to ground by a small (1000k-10k) resistor increasing the input impedance?

That decreases input impedance. If you have a 100K input, and add 10K in parallel, it looks like 9K (9.09090..K). This is useful for a few cases where a box NEEDS a 600 ohm load, and the next box has a 100K input. You put a 601 ohm resistor across it. (That's 597 ohms; for some reason back in 600 ohm days we always used 601 ohm 1% termination resistors.) You still find it in video where several hi-Z TV monitors are connected to one line which must have 75 ohms at the end so the signal does not "bounce" and cause "ghosts". Monitors sometimes have a hi-Z/75 switch. You put the one on the end of the run at 75 and all the rest at hi-Z.

[Paul Marossy]

This gadget here will give you an approximate input impedance for an existing pedal: http://www.diyguitarist.com/PDF_Files/ImpedanceTester.pdf

[O.Binette]

Thanks for the good info, it's going to be useful!

O.B.