Input impedance test

[Buffalo Tom]

Hi. Im trying to measure the input impedance on a couple of buffer pedals I have. So using this method seems really simple https://www.harmonycentral.com/articles/effects/how-to-test-input-impedance-for-guitar-effects-and-amps-r428/

Im using a test signal generator set to 100mV ac 1 kHz. The pot is 5M ohm linear.

But my problem starts here:

4. Rotate R1 until the meter reads exactly 50% of what it did in step 3.

I can't decrease the value to 50mV with my potentiometer. When I rotate the value only get higher.. several hundred mV.
Have tried a bunch of connections but I can only increase the value.. What Im I missing? 

[antonis]

Reverse pot wiring..?? 

Verify your wiring with a 1M & 500k resistors instead of pot, set alternatively in series with signal..
(you may use 1M & 2x1M in parallel.. in case you don't get any practically voltage difference reading, measured input impedance should be veeeery low..)

What you actually have to do is form a voltage divider consisting of pot's value and effect input impedance..
(signal generator's output impedance should be negligible..!!)

When you read on divider junction point 50% of previous got voltage (the one you've got with no upper divider resistance), you'll ideally have equal divider resistances..
(of course, the above only stands for DDM input impedance of value much higher than measured 'cause this impedance is also set in parallel with effects input impedance..)

You may, of course, get any other measurement (except 0% & 100%) at any pot setting (other than those on extreme ends) and calculate input impedance by Vmeasured = (Vsignal X effect input impedance) / (pot resistance + effect input impedance) 

[Buffalo Tom]

Reverse pot wiring..?? 

Verify your wiring with a 1M & 500k resistors instead of pot, set alternatively in series with signal..
(you may use 1M & 2x1M in parallel.. in case you don't get any practically voltage difference reading, measured input impedance should be veeeery low..)

What you actually have to do is form a voltage divider consisting of pot's value and effect input impedance..
(signal generator's output impedance should be negligible..!!)

When you read on divider junction point 50% of previous got voltage (the one you've got with no upper divider resistance), you'll ideally have equal divider resistances..
(of course, the above only stands for DDM input impedance of value much higher than measured 'cause this impedance is also set in parallel with effects input impedance..)

You may, of course, get any other measurement (except 0% & 100%) at any pot setting (other than those on extreme ends) and calculate input impedance by Vmeasured = (Vsignal X effect input impedance) / (pot resistance + effect input impedance)

Thanks! Problem solved... The test pedal was not connected properly.  The method works great.

[Buffalo Tom]

Trying to measure output Z now with the same equipment... Putting 500mV ac 1 kHz to pedals input. The 5M testpot is wired after the output of the pedal and I trim it to measure 250mV. So far soo god.. But when I measuring the pot it says around 1.6M ohm.. So that can't be the outputZ (I was expecting a couple of hundred ohms here..)  .To high value on the test pot? Or maybe this is not the way it can be done... Worked great for input Z still.

[antonis]

I presume you've just quoted my previous post without reading (or understanding) it..

When placing pot AFTER effect, you form a voltage divider on effects out & GND, so you have to take measurement on pot's lug 1..!!
(you've probably grounded pot's lug 2, didn't you..??)

BUT..

You have to be sure for your effect exact voltage amplification..!!
(e.g. in case of a humble single BJT Emitter follower, voltage output is calculated from: Vin X R_load/(r_e+R_load ), where r_e = 0.025/collector current & R_load is total Emitter resistance, including pot's effective resistance..)

BUT..
(again)

In the above case, output impedance is considered r_e..
(you don't have to measure anything..)

Now, in case op-amp buffer, you can estimate it from manufacturer datasheet or use a pot of 100R (or lower) value..

BUT..
(again)

You have to be sure you don't overload (low resistance -> high current) your specific effect output..

P.S.
To get things more clear, consider effect's output as a voltage source of zero impedance in series with an impedance which is to be measured..
That impedance forms toghether with pot's effective resistance a voltage divider and on divider's node voltage is simply
generator's Vout X pot's resistance / (effect's output impedance + pot's resistance)..
(e.g. for a buffer of 0.95 voltage amplification, Vin 100mV and an output impedance of 100R, you'll get 47.5mV when use 100R resistor as load..)

Sorry but I can't post a diagram to make it more clear..

[Buffalo Tom]

Thank you very much Antonis! Will dig in and do more testing tomorrow.

I presume you've just quoted my previous post without reading (or understanding) it..

When placing pot AFTER effect, you form a voltage divider on effects out & GND, so you have to take measurement on pot's lug 1..!!
(you've probably grounded pot's lug 2, didn't you..??)

BUT..

You have to be sure for your effect exact voltage amplification..!!
(e.g. in case of a humble single BJT Emitter follower, voltage output is calculated from: Vin X R_load/(r_e+R_load ), where r_e = 0.025/collector current & R_load is total Emitter resistance, including pot's effective resistance..)

BUT..
(again)

In the above case, output impedance is considered r_e..
(you don't have to measure anything..)

Now, in case op-amp buffer, you can estimate it from manufacturer datasheet or use a pot of 100R (or lower) value..

BUT..
(again)

You have to be sure you don't overload (low resistance -> high current) your specific effect output..

P.S.
To get things more clear, consider effect's output as a voltage source of zero impedance in series with an impedance which is to be measured..
That impedance forms toghether with pot's effective resistance a voltage divider and on divider's node voltage is simply
generator's Vout X pot's resistance / (effect's output impedance + pot's resistance)..
(e.g. for a buffer of 0.95 voltage amplification, Vin 100mV and an output impedance of 100R, you'll get 47.5mV when use 100R resistor as load..)

Sorry but I can't post a diagram to make it more clear..