# Measuring output impendance

#### HEPBO3AH

##### Member
I gave it a go using the method where you derive a output impedance by finding the following values:
V1 - open circuit voltage while playing the 60Hz sine tone
V2 - voltage of the loaded circuit that I got by connecting IEMs and playing the same 60Hz sine tone
R - resistance of the load which in my case is 16 Ohm.

I used the calculator to avoid mistakes.

I got a result of around 13.71Ohm but I was confused by the Voltage readings.
V1 = 1.3mV
V2 = 0.7mv

I'm not sure if that is correct, all the online tutorials that talk about this in general have values that are in V and not mV.

Do my measurements sound plausible?

#### sergeauckland

##### Major Contributor
Forum Donor
I got the same answer using your numbers, but multiplied by 1000 to get volts, not mV.

I wonder how you measured 0.7mV reliably, and why it's so low.

An output impedance of 13.7 ohms is reasonable, and anyway, sengpielaudio's calculators I've always found reliable, so assume this one is too.

One further question, how did you measure the load impedance of 16ohms, or is that what the manufacturers rate the IEMs at? At 60 Hz, I would expect the impedance to be very close to the DC resistance, so it might be an idea to measure that, and use that number instead of 16 ohms.

S.

OP
H

#### HEPBO3AH

##### Member
I doubt the reliability as well as the values are so small that they are reaching the resolution of the multimeter.

The IEM is Mee M6 pro. They declare 16 Ohm at 1KHz so I didn't measure the impedance.
When measured at the 1KHz tone, the V2 was 0.7mV 0.8mV resulting in the 10 Ohm end calculation.
The 1KHz tone did not change the V1 voltage.

I measured using the multimeter digitech qm1529.

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#### pma

##### Major Contributor
Forum Donor
I use current injection method. One channel (amp) injects current via a defined resistor Ri to the output of the second channel or amp. The Ri divided by ratio of measured voltages tells the output impedance of the channel to which the current is injected. Here 0.094 ohm. Sweep the frequency to get output impedance frequency response.

#### solderdude

##### Major Contributor
When determining output R of a headphone amp source one should not use headphones and certainly not IEM's because of their low power rating.

Measure the output voltage at 100Hz and adjust to say 100mV or so.
Then apply a load of 15 Ohm or 33 ohm or so and measure the voltage drop when applying the load.

The method pma used is fine for power amps and could be used with headphone amps as well but would not go much beyond 0.1V and loads below 15ohm.
The reason for that is headphone outputs generally have current limits. When using a higher measurement voltage the current limit could be reached which drops the output voltage as well leading to an incorrect output R measurement. The actual value could be much lower than the calculated one in that case.

Measuring with mV and earphones is not recommended.

OP
H

#### HEPBO3AH

##### Member
I'll see what I can try. I do not really have any equipment besides the multimeter and wires. I can get some resistors, but that will have to wait until tomorrow.

Measure the output voltage at 100Hz and adjust to say 100mV or so.
This is interesting. I noticed that the unloaded output does not change at all no matter the volume or the frequency. Not sure what might be the cause.

I saw somewhere that some devices do not product voltage at all unless loaded by something and that there is method I can use through measuring the 2 different impedance loads.
But, I'd expect to not get the result I got if that is the case.

#### solderdude

##### Major Contributor
I saw somewhere that some devices do not product voltage at all unless loaded by something and that there is method I can use through measuring the 2 different impedance loads.
But, I'd expect to not get the result I got if that is the case.
I would expect some 'smart' devices to lower the max output voltage when a low impedance load is detected but have not heard of devices not outputting any voltage unless a load is detected below a certain impedance.