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If I wanted to measure the gain that a power amplifier provides to a signal with a multimeter, do I simply do the following:

  1. Measure the input signal's voltage e.g. 1 Vrms
  2. Turn the amplifier up to full.
  3. Measure the output voltage of the amplifier e.g. 20Vrms
  4. Output / input = amount of gain that the amplifier provides.

Is it that simple, or do I need to add a resistor to the output which is a similar load to what a speaker would be and then measure the voltage drop across that resistor?

What if I wanted to determine the watts of the amp? Then would I need a resistor of a similar value to a speaker? Or would it be better to measure the resistance of the speaker with an ohmmeter, measure the amps output voltage and then I = V/R?

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  • \$\begingroup\$ It's worth making measurements both with and without a load, as while the amplifier itself is probably 'stiff' (very low impedance), its power supply will vary under load. The max unloaded output is what would be available for music peaks, the loaded output for steady power (like wot music isn't). Amplifier manufacturers who raise engineers' ire by rating 'music power' at the peaks are not all bad. \$\endgroup\$ – Neil_UK Feb 4 '17 at 2:24
  • \$\begingroup\$ So what you're saying is that if I measure it unloaded, then the Vrms result I get will be the equivalent to "music power"? I want to be able to say with some certainty: "This amp has a gain of ..." therefore I suppose it's best to measure it with a simulated speaker load such as 8 ohms. \$\endgroup\$ – Ryan Ashton Feb 4 '17 at 3:58
  • \$\begingroup\$ As the other answerers point out, you need to make sure your measurements don't take you into distortion, which would be all too easy to miss with a meter. Check several increasing amplitude levels, until the gain drops, then come back down to constant gain, crude, but better than nothing. If measuring gain, then maybe into a resistive load. \$\endgroup\$ – Neil_UK Feb 4 '17 at 9:16
  • \$\begingroup\$ I'd use a sine wave. Make sure that your meter can measure true RMS, and that it can can handle the frequency you are using to make your measurements. \$\endgroup\$ – JRE Feb 4 '17 at 11:22
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Yes, what you describe is a simple way to measure the gain of a power amplifier.

However, I would add a typical load to the amplifier. Audio power amplifiers have very low output impedance, but other amplifiers might not. In that case, the output voltage could vary significantly with load impedance.

In the case of a audio amplifier, connect a 8 Ω load. The speaker you plan to use might be good enough, but a 8 Ω power resistor would be better.

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  • \$\begingroup\$ What kind of wattage do you reckon the resistor should be? I mean, I know amps can vary considerably, let's say for a big amp like this (if I shoot big then that will cover a large range: rdn.harmanpro.com/product_documents/documents/2521_1430417955/… \$\endgroup\$ – Ryan Ashton Feb 4 '17 at 4:05
  • \$\begingroup\$ The data sheet says 2100W @ 8 ohms \$\endgroup\$ – Ryan Ashton Feb 4 '17 at 4:05
  • \$\begingroup\$ This is just an example, but it's quite powerful so whatever resistor covers this will also cover smaller amps. Right? \$\endgroup\$ – Ryan Ashton Feb 4 '17 at 4:06
  • \$\begingroup\$ I can generate a signal - ith my iPhone. And I can measure the Vrms with my multimeter. And I can measure the Vrms coming out of the amplifier with my multimeter. \$\endgroup\$ – Ryan Ashton Feb 4 '17 at 11:02
  • \$\begingroup\$ @Ryan: If you're talking about a audio power amp, the specs should tell you directly the maximum power it can put out. Get a resistor that can handle the power. Or, you can put multiple smaller resistors in parallel or series to get the 8 Ohms at higher power. For example, four 33 Ohm 10 W resistors in parallel get you close enough to 8 Ohms at 40 W. \$\endgroup\$ – Olin Lathrop Feb 4 '17 at 14:18
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Most amplifiers will have a gain that varies very little with output load so this is optional but probably a good idea because some amplifiers may go unstable on no-load and you might get rail-to-rail oscillations no matter what input signal you apply.

If your meter is good for the whole audio range then the method you propose is fine but, make sure that when on AC, the meter rejects dc levels because these could produce a big error in your estimations.

Also, make sure that the output isn't clipping. If it clips, the output level is restricted so, take two measurements at 50% levels to see that as a ratio, the two numbers are the same.

For measuring power, just use the equivalent resistance of the speakers impedance i.e. if the speaker is 8 ohms then use an 8 ohm resistor. Using the DC resistance value of the speaker coil is incorrect because that isn't the full picture of the impedance presented by an 8 ohm speaker at 400 Hz (a typical standard): -

enter image description here

Picture stolen from here.

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  • \$\begingroup\$ How do I make sure that my meter rejects D.C. when measuring? Will all this information about frequency, etc be in the meter documentation? (Fluke 179) \$\endgroup\$ – Ryan Ashton Feb 4 '17 at 4:19
  • \$\begingroup\$ Another question I have is: Should I use pink noise or a 1kHz sine? Pink noise makes more sense to me. \$\endgroup\$ – Ryan Ashton Feb 4 '17 at 4:23
  • \$\begingroup\$ Pink noise needs to be band limited to the amplifiers flat gain area to give you a meaningful number. Use 1 kHz is my advise or test across a range of frequencies. \$\endgroup\$ – Andy aka Feb 4 '17 at 9:19
  • \$\begingroup\$ But then I need to know what the flat gain area is. I'm just looking for a method that will allow me to measure gain on practically any amp and compare it to any other amp I have already measured. The easiest, most broadly-applicable method. \$\endgroup\$ – Ryan Ashton Feb 4 '17 at 9:43
  • \$\begingroup\$ Do you have a signal generator that can produce say 1 kHz. an android app will likely fit the bill plus you can also use an android to measure amplitude and some apps show an audio spectrum using the microphone but I'd bet you could wire an input to the Mic input. Maybe you have a Sig generator. \$\endgroup\$ – Andy aka Feb 4 '17 at 10:21

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