Measuring DAP Output Resistance
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Supremevegbeef

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Is this the right procedure?????

Measure resistance of male to male 3.5mm stereo cable.
Plug into DAP with nothing playing.
Measure resistance between ground and either signal side.
Subtract previously measured resistance of cable.

Is this correct???
 
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castleofargh

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probably not. first the resistance isn't always very interesting in our world of sine waves. instead we usually care about impedance, and that can give a different value depending on the frequency being played.
and second, a DAP when playing nothing, could very well "open" some circuit(to save battery, to hide how much it's actually hissing when playing,...), and let you measure something irrelevant.

I would suggest instead to find a known load. like say a resistor or an IEM with a sure impedance value at a given frequency(often 1kHz). then you play a 1kHz sine you've put on the DAP if that's what you're looking for, and measure the voltage at the DAP unloaded. then repeat with the load. from the voltage deviation, you should be able to devise the impedance using some Ohm's law witchcraft. might need someone to confirm if I'm not wrong because it's been a while since I played with that, but I think it was something like: load's impedance*(V1-V2)/V2= impedance of the DAP.

ps: I remember something about cases where an output unloaded will be a problem and then there is a need to use 2 known loads and work out a tiny bit more of Ohm's devil magic. But in practice the one load method always delivered for me, so I only mention this because I remember being warned about the possibility in another life. ^_^
 
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dazzerfong

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No: you need to measure voltage twice, once without a load, then repeat once more with a small-ohm (like 10-20 ohm) resistor shorted from +ve to -ve. During this whole time, a low-amplitude constant AC sine wave (preferably 1 kHz) should be playing (don't want it to high because you might deliver too many amps and cook your source).

The reason you need to measure twice is that you need to find what the source voltage is as well as the scaled voltage. Using Ohm's law, if you know what the voltage drop is, you can work backwards and find out what your resistance is.

The ratio between your voltages is the same as the ratio between your resistor and the output impedance. For example, say you measure a 1V signal with no load, and then you measure 0.5V @ 10 ohms. Assuming Z = the output impedance, R is the resistor you use, V1 = 1, V2 = 0.5:

V2 = V1 * R/(Z+R) (voltage divider)
V2/V1 = R/(Z+R)
(V2/V1)(Z+R) = R
Z+R = R/(V2/V1) = R*V1/V2
Z = R*V1/V2 - R
Z = 10*2 - 10 = 10

In this example, your output resistance is 10 ohms.

Cables are generally a non-issue, but to truly factor it out requires you to know what the impedance the cable is at 1 kHz.

The problem comes from your measuring equipment. Most multimeters aren't well-equipped to measure AC signals at kHz range, especially electrician's multimeters. If that's the case, use a 60Hz sine wave instead: it won't be as indicative as a 1kHz measurement, but better than nothing.

probably not. first the resistance isn't always very interesting in our world of sine waves. instead we usually care about impedance, and that can give a different value depending on the frequency being played.
and second, a DAP when playing nothing, could very well "open" some circuit(to save battery, to hide how much it's actually hissing when playing,...), and let you measure something irrelevant.

I would suggest instead to find a known load. like say a resistor or an IEM with a sure impedance value at a given frequency(often 1kHz). then you play a 1kHz sine you've put on the DAP if that's what you're looking for, and measure the voltage at the DAP unloaded. then repeat with the load. from the voltage deviation, you should be able to devise the impedance using some Ohm's law witchcraft. might need someone to confirm if I'm not wrong because it's been a while since I played with that, but I think it was something like: load's impedance*(V1-V2)/V2= impedance of the DAP.

ps: I remember something about cases where an output unloaded will be a problem and then there is a need to use 2 known loads and work out a tiny bit more of Ohm's devil magic. But in practice the one load method always delivered for me, so I only mention this because I remember being warned about the possibility in another life. ^_^
Your last paragraph is correct: there are cases where a zero-ohm measurement will cause a problem. This is because there's nothing to 'regulate' the amount of current delivered. Generally, however, this is not an issue these days.
 
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