How to measure amplifier current (amperage) output to speakers in realtime? And some other questions about power measurements.

[Hatto]

I was recently loaned a nice digital oscilloscope to play around with my stereo setup. I've sucessfully set it up to measure stereo amplifier voltage output under load (8ohm speakers) in realtime. My objective is to find the clipping point of my stereo amp under real-world conditions. I determined the voltage output is capped at +-9V (18Vpp). I tried measuring the current output in real-time with a nice multimeter, but I'm guessing I wasn't succesful in doing so considering most I could measure was 100mA.

My first question is, since P = V^2/R: Can one say the amplifier can provide 10.125W (9^2/8) peak power?

I'm trying to find a case where the difference between Vpeak and Vrms is maximum and it looks like this in one of the extreme cases:

In this case, Vpeak (8.60V) = 5x Vrms (1.72V) which means there's 25x power differential between RMS and peak power for this track.

I'm trying to find out, at what point clipping might occur due to a current cap. Now, I know that I can calculate the instantaneous current by using instantaneous voltage and impedance, but I don't have the impedance curve of my speakers.

My second question is: What is a conveniently accurate way to measure the current output of the amplifier in real-time?

As far as I know, impedance rarely dips under the nominal rated value (8ohms) and current draw is maximum when the impedance is at minimum. So I'm assuming the amplifier should be able to supply enough current to support 9V over 8ohms.

Therefore: Is it safe to assume that the amplifier can provide a peak current of 1.125A (9V/8ohms)?

In general I would be delighted if anyone can enlighten me about the error of my ways.

 

[voodooless]

My second question is: What is a conveniently accurate way to measure the current output of the amplifier in real-time?

Use a .1 Ohm resistor in series with the load and measure the voltage over it. I = V/R, Bob’s your uncle. Just make sure you have a differential probe.

If the second channel can measure voltage, and you can do some math in the scope, you can get power out of it in real-time as well.

 

[Hatto]

Use a .1 Ohm resistor in series with the load and measure the voltage over it. I = V/R, Bob’s your uncle. Just make sure you have a differential probe.

If the second channel can measure voltage, and you can do some math in the scope, you can get power out of it in real-time as well.

Hmmm, but then the amplifier will be powering a 9ohm load, which would lead to a different power output than what it is rated for 8ohms.

EDIT: Missed the point (literally)

 

[DonH56]

Use a .1 Ohm resistor in series with the load and measure the voltage over it. I = V/R, Bob’s your uncle. Just make sure you have a differential probe.

If the second channel can measure voltage, and you can do some math in the scope, you can get power out of it in real-time as well.

Thanks for the suggestion, the reason I'm trying to measure current output in real-time is because I'm also listening to my speakers for any signs of distortion as I'm measuring. Besides the power output of the amplifier differs with impedance so I'm not sure how I can correlate the current value I got under constant 1ohm load to the real-world speaker under load.

Read @voodooless' answer again. Use a 0.1-ohm resistor in series with the speaker wire (the speaker is the load), measuring across the resistor with a differential probe (not a regular probe that grounds one side!) The voltage across the resistor divided by the resistance (0.1 ohms) gives you the current. Use another probe to measure the voltage at the speaker terminal (or anywhere after the resistor) with reference to the negative side of the speaker (which may or may not be ground; again, safest to use a differential probe) to get the voltage across the speaker. Now you know the current into the speaker and the voltage across it, so the power is their product (V * I).

HTH - Don

 

[Hatto]

Read @voodooless' answer again. Use a 0.1-ohm resistor in series with the speaker wire (the speaker is the load), measuring across the resistor with a differential probe (not a regular probe that grounds one side!) The voltage across the resistor divided by the resistance (0.1 ohms) gives you the current. Use another probe to measure the voltage at the speaker terminal (or anywhere after the resistor) with reference to the negative side of the speaker (which may or may not be ground; again, safest to use a differential probe) to get the voltage across the speaker. Now you know the current into the speaker and the voltage across it, so the power is their product (V * I).

HTH - Don

Gotcha!

 

[DVDdoug]

Just make sure you have a differential probe.

Or put the resistor on the ground-side. (Assuming it's not a bridge amplifier and there is a ground to the speaker.)

I tried measuring the current output in real-time with a nice multimeter, but I'm guessing I wasn't succesful in doing so considering most I could measure was 100mA.

Multimeters are not good for "dynamic" measurements. They measure continuous RMS or DC. On old style electro-mechanical meter movement does a pretty good job of averaging (because of the inertia). But, you shouldn't run continuous test-tones into a speaker, especially high frequencies into the tweeter. They ARE designed for dynamic signals.

I'm trying to find out, at what point clipping might occur due to a current cap.

If the impedance drops and the amplifier can't supply the current, the voltage will also drop.

Usually the short-term current is supplied (or "assisted") by capacitors in the power supply so the voltage & current won't sag immediately on program dynamics. It might take a little more time for current to limit output-power. (The amplifier may be able to supply higher short-term peaks than it can supply continuously.)

Therefore: Is it safe to assume that the amplifier can provide a peak current of 1.125A (9V/8ohms)?

Yes, and it's HEAT that burns-up transistors/MOSFETs (or kicks-in the thermal protection) so even if the speakers occasionally draw "excess" current intermittently with peaks at certain frequencies, it's usually OK.

As far as I know, impedance rarely dips under the nominal rated value (8ohms)

Often it does. But power amplifiers usually have some safety margin (the amp isn't going to burn-up with 7.9 Ohms) and the dynamics & variability of music helps... You aren't playing the "worst" frequency at full-power continuously. And that kind of thing is usually more dangerous for the speaker (depending on the speaker's power ratings.)

P.S.
If you can easily get to the amplifier's internal power supply voltage you can check to see if it holds-up with loud music or during program peaks.

 

[Hatto]

Thank you @voodooless, @DonH56 and @DVDdoug for the answers!

P.S.
If you can easily get to the amplifier's internal power supply voltage you can check to see if it holds-up with loud music or during program peaks.

Nice. My amp has an external PS so it is fairly easy. What do you mean by "holding up with music"? As in "not flatting out"?