I'd like to know how to measure output impedance on my (and friend's) devices, specifically for headphone gear, and the headphone output on various receivers. I think I have all of the tools that I would need, but if someone could give the down low on what I would need as far as equipment, and what tools I would need to measure the output impedance as accurately as possible, that would be amazing. Also, I assume that I would need to actually be playing something through the headphone out, so what would be the best sound to play?
Thanks!
A quick search for:
measure output impedance
measure output impedance of amplifier
Will give you many sets of complete instructions, some with videos.
Thank you! I was just a bit worried, as I recently saw another thread here that said that measuring output impedance by hand can be inaccurate, and I'd rather be as accurate as possible.
Thank you! I was just a bit worried, as I recently saw another thread here that said that measuring output impedance by hand can be inaccurate, and I'd rather be as accurate as possible.
https://en.wikipedia.org/wiki/Output_impedance (https://en.wikipedia.org/wiki/Output_impedance)
You solve for Zs (output impedance) and Vs (open circuit voltage) in that diagram by using two different load impedances so that you have two knowns and two unknowns and can do the algebra. FWIW, I'd measure the open circuit voltage (no load at all, which gives you Vs), and then a relatively large load (16 ohm for headphone amp, 4 ohm for power amp). The accuracy will be limited only by how well you can measure the voltage and impedance, which with even a $25 multimeter should be very accurate. For best results, use a resistor, not a pair of headphones as the load.
You need: 60 Hz sine test signal, digital multimeter, dummy load.
Measure the voltage unloaded and with dummy load.
Calculate:
Zout = (Rload * (Vnoload - Vload)) / Vload
Example:
We measure 1V unloaded, 0.9V loaded and our dummy load is 30 ohm ==> (30 * (1 - 0.9)) / 0.9 = 3.33 ohm output impedance at 60 Hz.
I'd like to know how to measure output impedance on my (and friend's) devices, specifically for headphone gear, and the headphone output on various receivers. I think I have all of the tools that I would need, but if someone could give the down low on what I would need as far as equipment, and what tools I would need to measure the output impedance as accurately as possible, that would be amazing. Also, I assume that I would need to actually be playing something through the headphone out, so what would be the best sound to play?
We just lately covered that here but what thread, what forum?
Here's a good general article:
NWAVGUY headphone amp measurements (http://www.innerfidelity.com/content/nwavguys-heaphone-amp-measurement-recommendations)
and another:
Westhost (Elliot) article on measuring source impedance (http://sound.westhost.com/impedanc.htm)
"By measuring the output voltage with no load, and with a known load, you can calculate the output impedance. This online calculator (http://www.raltron.com/cust/tools/voltage_divider.asp) makes it easy. The no load voltage is the "Input Voltage", R2 is the known load resistance (don't use headphones), the Output Voltage is the loaded voltage. Click Compute and R1 is the calculated output impedance. This can be done using a 60 hz sine wave file (Audacity can create such a file), a Digital Multi Meter (DMM), and a 15 – 33 ohm resistor. Most DMMs are only accurate around 60 hz. Play the 60 hz sine wave file and adjust the volume for about 0.5 volts. Then attach the resistor and note the new voltage. For example, 0.5 volts with no load, and 0.38 volts with a 33 ohm load gives an output impedance of about 10 ohms. The math is: Zout = (Rload * (Vnoload - Vload)) / Vload " - NwAvGuy (http://nwavguy.blogspot.com/2011/02/headphone-amp-impedance.html)
@Arnold B. Krueger
So Arny, you strike me as the kinda guy who has a 15 - 33 ohm resistor right on hand; might you please do us a big favor and measure your current Yamaha or Denon receiver's headphone output impedance? I know lots of people besides myself would be curious to know the results even though they would only be specific to that particular AVR. Thanks.
"By measuring the output voltage with no load, and with a known load, you can calculate the output impedance. This online calculator (http://www.raltron.com/cust/tools/voltage_divider.asp) makes it easy. The no load voltage is the "Input Voltage", R2 is the known load resistance (don't use headphones), the Output Voltage is the loaded voltage. Click Compute and R1 is the calculated output impedance. This can be done using a 60 hz sine wave file (Audacity can create such a file), a Digital Multi Meter (DMM), and a 15 – 33 ohm resistor. Most DMMs are only accurate around 60 hz. Play the 60 hz sine wave file and adjust the volume for about 0.5 volts. Then attach the resistor and note the new voltage. For example, 0.5 volts with no load, and 0.38 volts with a 33 ohm load gives an output impedance of about 10 ohms. The math is: Zout = (Rload * (Vnoload - Vload)) / Vload " - NwAvGuy (http://nwavguy.blogspot.com/2011/02/headphone-amp-impedance.html)
@Arnold B. Krueger
So Arny, you strike me as the kinda guy who has a 15 - 33 ohm resistor right on hand; might you please do us a big favor and measure your current Yamaha or Denon receiver's headphone output impedance? I know lots of people besides myself would be curious to know the results even though they would only be specific to that particular AVR. Thanks.
Can I cheat?
I have the service manuals for both AVRs, and they both have 470 ohm resistors, 1 per channel running from the power amp outputs to the headphone jack. Bad, bad news.
The good news is that I use neither jack. My headphones are Sennheiser digitals, with line level inputs that are driven by the line level outputs of a dedicated 2-channel Rane equalizer.
Sure, cheating is fine, if your known resistor's value translates to a more specific headphone jack output impedance value you can then verbalize to me so I can start doing some numbers cruching on a specific value other than "it's bad". I can't do 1/8th rule calculations using the numerical value of "bad" now can I?
https://en.wikipedia.org/wiki/Output_impedance (https://en.wikipedia.org/wiki/Output_impedance)
The accuracy will be limited only by how well you can measure the voltage and impedance, which with even a $25 multimeter should be very accurate. For best results, use a resistor, not a pair of headphones as the load.
I have an analog multimeter from the 70s, does that count?
I also have a really nice digital oscilloscope. I watched a youtube video where the guy in the video was using an oscilloscope, would that be more accurate than the multimeter, or should I stick with the one I have/buy a digital one at the hardware store?
A scope or multimeter will both work. Anything that can measure ac voltage.
https://en.wikipedia.org/wiki/Output_impedance (https://en.wikipedia.org/wiki/Output_impedance)
The accuracy will be limited only by how well you can measure the voltage and impedance, which with even a $25 multimeter should be very accurate. For best results, use a resistor, not a pair of headphones as the load.
I have an analog multimeter from the 70s, does that count?
It should work provided you can get enough of an indication to have reasonable values to work with.
The potential problem is that general purpose multimeters only have accurate response up to 400 or 1,000 Hz. They may have needles or displays that fluctuate below 50 Hz. As long as you work with signals in that range, which can be useful for measuring the source impedance of headphone amps, they are good enough.
However, the seriousness of the potential problem is vastly reduced by the fact that the impedance calculation is based on the ratio of two voltages. If an inaccurate meter causes one voltage to be so many percent low, it will have the same effect on the other voltage, and the ratio of the two voltages will be correct even though the individual voltages have built-in errors.
I also have a really nice digital oscilloscope. I watched a youtube video where the guy in the video was using an oscilloscope, would that be more accurate than the multimeter, or should I stick with the one I have/buy a digital one at the hardware store?
I have an Owon USB digital oscilloscope and its numeric readout features seem to be sufficiently accurate. Yours may be just as good if not better. With response up to 10-20-50-100 MHz, the frequency response problems with using general purpose multimeters to measure audio would appear to be circumvented.
Sure, cheating is fine, if your known resistor's value translates to a more specific headphone jack output impedance value you can then verbalize to me so I can start doing some numbers cruching on a specific value other than "it's bad". I can't do 1/8th rule calculations using the numerical value of "bad" now can I?
I can confirm that the 470 series resistors are the overwhelmingly dominant influences relating to the source impedance of these two AVRs. The AVR power amps themselves have source impedances of under 0.1 ohm, and the source impedance at the headphone jack is the sum of that and the 470 ohm resistor.
Interesting coincidence that they were both the same.
More expensive AVRs may have separate stand-alone headphone amps, but there is no guarantee that their source impedance is down in the 1 ohm or so sweet spot that we like.
Isn't it favorable to have a (sine) function generator at hand to measure the impedance as a function of frequency instead of at a single frequency or DC? A single value will never give you the full picture what's happening. That's even more important for headphones.
Thanks Arny, so as I understand it the 470 ohm resistors are wired in series from the positive terminals of the AVRs speaker outputs and you estimate the output impedance for the headphone jack to be around 470.1 ohms.
Fixing this with an intermediate buffering amp like a Fiio connected to the headphone out to establish a nice low output impedance for one's headphones seems inelegant and like a visual eyesore to me. Have you ever explored building a simple, passive voltage divider network like this guy (https://diyaudioheaven.wordpress.com/tutorials/power-amp-adapter/) discusses on the lower half of the page underneath the simplistic method AVRs use which he first discusses, or have comments on it?
Thanks Arny, so as I understand it the 470 ohm resistors are wired in series from the positive terminals of the AVRs speaker outputs and you estimate the output impedance for the headphone jack to be around 470.1 ohms.
Fixing this with an intermediate buffering amp like a Fiio connected to the headphone out to establish a nice low output impedance for one's headphones seems inelegant and like a visual eyesore to me. Have you ever explored building a simple, passive voltage divider network like this guy (https://diyaudioheaven.wordpress.com/tutorials/power-amp-adapter/) discusses on the lower half of the page underneath the simplistic method AVRs use which he first discusses, or have comments on it?
I think that what we want is a headphone signal source that puts out about 2 volts RMS with a source impedance of no more than 1 ohm.
A 100 wpc AVR puts out about 32 volts RMS into 8 ohms. A simple voltage divider with source of impedance of 1 ohm and an input impedance of 8 ohms would have a gain of 1/8 which would cut the 32 volts down to 4 volts. So this is doable without violating the law of conservation of energy. The investment in reasonably stable and accurate non-inductive 8 resistors might be in the same range as a good cheap headphone amp. It might get hot if you crank it, but it would not require external power.
If you really wanted to use a power amp, you'd have to note that:
- it needs to have a common ground, or else it will blow up
- the noise floor will likely be audible
- resistors as attenuators will essentially turn a lot of power into heat
Thick film resistors of 1 ohm and 10 ohm, 35W, 1% tolerance can be had for less than $4 each. Note: the 35W rating requires a heat sink.
AVRs are a real work horse: they do so many things, for such a reasonable cost, so they are a dream come true for audio buffs who don't fall for the lies of the stereophool magazines, but why oh why must they almost all have this fatal flaw?
Sure we can fix it with an kludge outboard $28 (shipped) Fiio complete with its own dedicated power supply, lithium ion battery [possibly its most expensive part], volume control pot, tone control (bass boost switch), case, etc. which means if we strip it down to just the essential op amp and a few wires the material cost for a Japanese AVR maker is couple of bucks, tops, maybe even under $1, yet even when we buy a top level $2,000+ receiver from the big brands they won't spring for this for us?! ARGH!
This commonly repeated notion of "Well, some of the top level units use a dedicated op amp for the headphones instead of just the simple resistor method" is effectively untrue with current designs. I challenge anyone reading this to cite a specific, non-British AVR made within the past decade which does, based on manufacturer's claims, not anecdotal forum post claims.
What about instead of attenuating the unneeded power into heat, using resistors or pads, we use an autoformer? There are expensive volume control companies (http://www.intactaudio.com/atten.html) which do it this way which I can't justify the cost of, but then again there are also less expensive ways to buy an autoformer meant for tapping the speaker outputs of an audio amp which are more affordable. (http://www.nilesaudio.com/images/PDF/VCS50R-cutsheet.pdf)
It's still a clunky device. Audio performance will probably not be as good as a similarly priced headphone amp.
I see now they even make my autoformer idea as an assembled product. (http://www.russound.com/product_detail.php?i=2223)
But that's 3x as expensive as the "switch" ... probably just due to the case.
What about instead of attenuating the unneeded power into heat, using resistors or pads, we use an autoformer? There are expensive volume control companies (http://www.intactaudio.com/atten.html) which do it this way which I can't justify the cost of, but then again there are also less expensive ways to buy an autoformer meant for tapping the speaker outputs of an audio amp which are more affordable. (http://www.nilesaudio.com/images/PDF/VCS50R-cutsheet.pdf)
Transformers require careful engineering to avoid adding audible nonlinear distortion (particularly at low frequencies) and frequency response aberrations.
The cut sheet that you linked neatly avoids dealing with both issues, right? ;-)
Ditto for the Niles volume control, right?
Nice thing about resistors is that engineering products with zero nonlinear distortion and negligible frequency response aberrations is pretty simple.
Transformers require careful engineering to avoid adding audible nonlinear distortion (particularly at low frequencies) and frequency response aberrations.
Which is why it is so difficult to design a tube amplifier with very low distortion and flat response.
Transformers require careful engineering to avoid adding audible nonlinear distortion (particularly at low frequencies) and frequency response aberrations.
Which is why it is so difficult to design a tube amplifier with very low distortion and flat response.
Precisely. To that end any number of transformerless tubed power amps have been developed. Some of them are pretty good.
They typically run into severe economic and reliability problems because of the vast number of tubes required to deliver large amounts of power into low impedance loads.
Transformers require careful engineering to avoid adding audible nonlinear distortion (particularly at low frequencies) and frequency response aberrations.
The cut sheet that you linked neatly avoids dealing with both issues, right? ;-) Ditto for the Niles volume control, right?
The Niles link I gave says "
Frequency Response: 20Hz To 20kHz (+/-) 2dB " which I could probably live with especially since I expect it is more like +/- 1 dB over the more limited central range I'm more concerned with that doesn't include the frequency extremes of 20 Hz and 20 kHz.
Intact Audio only has an animated graphic, no written spec, showing their autoformer based volume control has a flat response with less than 1 dB alteration out to 100kHz (and beyond) for any attenuation level from 0 to -42 dB:
But that's for the line-level control. Also, these things are not pure inductances. There's also resistance that will change with position. So it might not satisfy a 1/8th rule with certain headphones either. And we've still ignored the nonlinearities of cheap speaker-level autotransformers.
As for resistors, you wouldn't just put 1 in series. You'd e.g. put 180 ohm || 30 ohm for a >250 ohm headphone.
But again, why not just get a headphone amp?
But that's for the line-level control. Also, these things are not pure inductances. There's also resistance that will change with position. So it might not satisfy a 1/8th rule with certain headphones either. And we've still ignored the nonlinearities of cheap speaker-level autotransformers.
I notice the more complete of the performance specs that have been shown for transformer based solutions only went down to 1 KHz. It is below 100 Hz where you separate the men from the boys in transformer land.
As for resistors, you wouldn't just put 1 in series. You'd e.g. put 180 ohm || 30 ohm for a >250 ohm headphone.
I think that the recommended solution might be something like an 8 ohm resistor in series with a 1 ohm resistor in parallel.
That follows the 1/8 rule for headphones with impedances as low as 8 ohms which seems to be pretty good coverage.
But again, why not just get a headphone amp?
Letsee, 2 of these for $10 shipped:
1 ohm NI power resistors (http://www.ebay.com/itm/1-POWER-FILM-RESISTOR-1-OHM-15W-1-TO126-MP915-1-00-1-CADDOCK-Kool-Pak-/321678873501?pt=LH_DefaultDomain_0&hash=item4ae58e139d)
Two of these for $5 shipped:
10 ohm NI power resistors (http://www.ebay.com/itm/2x-Genuine-Brand-New-Caddock-MP820-110-20W-10Ohm-10-Thick-Film-Power-Resistor-/261668925165?pt=LH_DefaultDomain_0&hash=item3cecaeeeed)
Put them on a scrap piece of aluminum or other good heat sink.
And you have a zero-added distortion, as good as possible frequency response solution for any AVR.
Due to the built-in attenuation, problems with hum and noise will be far less likely.
And that whole thing would probably go up in smoke with a bit over 1V output.
And that whole thing would probably go up in smoke with a bit over 1V output.
I've experimented with those little Caddock resistors, and when properly heat sinked, they seem to do quite well.
Proper heat sinking is not trivial. They are usually in TO-220 packages, and so are the output devices in many of the power amps that they are used with. Therefore, an old power amp heat sink might be in order.
This is a good time to remember that we are using them with music and dialog, not test tones, so their heat stress is vastly reduced by the crest factor of music and dialog.
With a 100 wpc receiver, my proposed design is limited to 3 volts by power amp clipping.
I'd still consider raising the parallel resistor value. It will reduce the power dissipation and/or you can get higher output voltage for hard to drive headphones, which I think are the targeted thing to drive here.
Can't imagine someone building such an adapter for in-ears or sensitive low impedance headphones.
Forty years ago I worked in audio equipment repair. One trick that we used to provide a high-wattage load on an amplifier was to take a 8 ohm ceramic resistor and insert it into a container of water. This kept the temperature of the resistor at a safe level while delivering many times its rated wattage. I don't recommend this in general, but it could be used as a temporary solution until you can supply a proper heat sink.
If you're already ordering parts anyway then just add proper heat sinks and maybe even a case and headphone jack etc. to build a nice small adapter box.
Two of these for $5 shipped:
10 ohm NI power resistors (http://www.ebay.com/itm/2x-Genuine-Brand-New-Caddock-MP820-110-20W-10Ohm-10-Thick-Film-Power-Resistor-/261668925165?pt=LH_DefaultDomain_0&hash=item3cecaeeeed)
Yes but those are 10% resistors. They could give a worst-case mismatch of +- 1.7 dB between channels, which could be easily audible.
Not water, use mineral oil (or maybe light motor oil). Ham radio operators do that to dummy loads for their radio transmitters. Heathkit even sold oil filled dummy loads.
22 AWG nichrome wire has a resistance of about 1 ohm per foot. A 25 foot spool (less than $5) would make two nice resistors, which you would tap at a point that gave you a 10:1 divider. Wind each around a non-conductive, non-flammable cylinder.