Measuring Audio Amplifier Power - Page 1
Forum section: electronics / beginners

author: prwiley
author_rank: Contributor
timestamp: June 14, 2020, 01:20:34 pm
content: I am trying to measure the power output of a simple audio amplifier I just built. The amp works great, driving a Jensen 35W 8ohm speaker just fine. In fact it can be quite loud.As I understand it, the procedure to measure the power is to connect a non-inductive load, provide a 1kHz signal, observe the output with an oscope, and then increase the amp volume until clipping starts. The voltage at which clipping starts is used to calculate power output. I am using a 8ohm 50W power resistor for the load. I am using a 60MV RMS 1k sine wave test signal.The trouble I am having is that when I attempt to connect my scope ground to the load, the signal on the scope goes flat goes flat. When the scope probe is connected to the + speaker lead without the ground connected I do see the signal at the right freq., although it's jittery as if a lower freq. singal is imposed upon it (ac noise, I guess).Using my DMM to measure voltage across the resistor, I get -40.42mv without the probe ground connected. With the probe ground connected I get 22.12mVBeing a bit new to electronics, I am sure I am missing something basic and I don't know how to troubleshoot. Can anyone point me in the right direction as to what the problem might be or what's going on? Thanks.

author: bdunham7
author_rank: Super Contributor
timestamp: June 14, 2020, 01:40:14 pm
content: Is this amplifier powered by a single voltage source, like a battery? Can you post a schematic of it?

author: TimFox
author_rank: Super Contributor
timestamp: June 14, 2020, 01:46:47 pm
content: This sounds like a "bridged" output, where neither side is grounded; one swings positive when the other swings negative. This allows generating an output voltage larger than the DC power in the amplifier.The floating voltmeter should read the correct voltage, but you can't accurately determine when the amplifier clips. You could infer it by when the output grows slower than proportional to the input.The usual answer is to use a differential connection: real differential probes are expensive, but many oscilloscopes have a pseudo-differential capability of subtracting one input from the other in "A-B" mode. Be careful of the common mode and proper grounding.You state "-40.42 mV": is that DC? Does your DMM do "true rms" on AC volts?One pedantic point: if you get this to work properly, and measure the rms voltage across a good 8 ohm load resistor, the power P = V2/R is themeanpower,notthe rms power, even though the audio industry perpetrates this vulgar error.

author: prwiley
author_rank: Contributor
timestamp: June 14, 2020, 02:49:25 pm
content: @bdunham7 12V wall wart. No path to mains ground. As much as a schematic as I have is in this document:http://www.diyguitarpedals.com.au/shop/boms/10%20Min%20Amp.pdf@TimFox. You are exactly right. I looked up the chip used: TDA7266M (https://www.st.com/resource/en/datasheet/tda7266m.pdf) and it is indeed a mono bridge amplifier, not a concept I understand so I have some reading to do.I am using the EEVBlog BM235 "True RMS DDM". The voltage I report is measured in the AC mode.So, the issue and procedure is described properly in the following video?I tried it out using the math function of my RIGOL DS1054 Because the scope is so slow, it's hard to adjust to the point where clipping seems to start, but I get a MEAN power of 4.7 to 6.7W. Which seems reasonable?

author: TimFox
author_rank: Super Contributor
timestamp: June 14, 2020, 03:10:11 pm
content: That power agrees with the data sheet for the amplifier working from 11 V DC. The 10% distortion level quoted indicates clipping of the waveform. The old rule of thumb was that 5% distortion was visible on an oscilloscope.In Figure 4 of the data sheet, note the sharp rise in distortion past roughly 5 W output.The article you cited is quite explicit about how to connect the sleeve of the two 1/4" phone jacks, since the output is not grounded.

author: prwiley
author_rank: Contributor
timestamp: June 14, 2020, 05:31:43 pm
content: TimFox, thanks a lot for your help with this.The part provided in the kit for the output jack of the nylon variety. I have to admit I paid more attention to the build instruction video that to the document up to now.Thanks again.

author: bdunham7
author_rank: Super Contributor
timestamp: June 14, 2020, 05:35:15 pm
content: Quote from: TimFox on June 14, 2020, 01:46:47 pmNot to pick nits, but with these type of amps it is more correct to say that both sides are grounded but not at the same time. During a cycle period one side of the amp will supply Vcc to one side of the output and the other side of the amp will ground the second side. No part is ever negative relative to the power ground. Vref in the diagram is similar to 'virtual ground' found in similar single-supply devices. This whole setup is much different than bridging two separate mono amps, each with their own bipolar supply, buy connecting the amps in series and feeding one an inverted input.

author: bdunham7
author_rank: Super Contributor
timestamp: June 14, 2020, 05:40:15 pm
content: Quote from: prwiley on June 14, 2020, 02:49:25 pmThe noise or jitter you noticed may be AC leakage, but the loss of signal has to be ground leakage somewhere. If your amp was truly isolated, the ground on your scope wouldn't matter. What are you using for the input signal? Is that grounded?Using the math function on a low-end DSO sort of works, as you've seen, but only crudely. You probably have a lot of AC-leakage common mode noise as Tim mentioned.

author: TimFox
author_rank: Super Contributor
timestamp: June 14, 2020, 05:51:58 pm
content: The usual definition of “grounded” is “connected to ground”. I disagree with the above nit-pick; one should not connect either side of this device’s output to the ground side of the input jack.

author: bdunham7
author_rank: Super Contributor
timestamp: June 14, 2020, 05:56:21 pm
content: Quote from: TimFox on June 14, 2020, 05:51:58 pmI'm not disagreeing with your statement about not grounding the outputs to either the power or input ground--it's shorting something in either case. However, if the power supply and input are both isolated from actual earth ground, then the scope ground shouldn't bother it.

author: TimFox
author_rank: Super Contributor
timestamp: June 14, 2020, 06:54:47 pm
content: The OP is testing the amplifier with a signal generator and oscilloscope. The generator certainly grounds the input (sleeve of the input 1/4” jack). The active device is a class-AB amplifier, where both output terminals swing with respect to “PS Gnd” (pinand “S Gnd” (pin 9) which are to be connected together (see data sheet). Therefore, one should probe the terminated output differentially (or quasi-differentially), or probe only one side with respect to ground (i.e. S-Gnd) and double the measured voltage.

author: bdunham7
author_rank: Super Contributor
timestamp: June 14, 2020, 07:22:18 pm
content: Let's call this type of bridged amp Type A and the type I described (both amps have bipolar supply) as Type B.On Type B, if you scope from GND to both outputs with two channels and display both, you will indeed see that each has a voltage of half the total output and that they are opposite in phase. If you add them, you get the full output. This would be found on more powerful, expensive audio equipment either consumer or commercial and is not that common in run-of-them mill audio equipment.On this one, Type A, if you do the same, you will see 'half the signal' on each output relative to ground, but it will be half of each cycle, like half-wave rectified AC, and again they will be out of phase. If you add them, you again get the full output. This is a much cheaper type of amplifier and is typically found in televisions, car radios and the like.OP--if you are still reading and want to know more about how your amp works, try this: Connect your scope ground to your power ground and probe both outputs like you just did. Don't use the math function, just display both channels at 5V/div and 1mS/div at full volume, and send us the photo. You'll be able to see your clipping more clearly as well as the crossover points.

author: prwiley
author_rank: Contributor
timestamp: June 14, 2020, 07:31:16 pm
content: @bdunham7 I am using a Rigol DG1022 Function Generator so it's certainly grounded. Scope and generator are plugged into the same outlet. Things are hooked together with alligator clips on long wires etc. so there are many entry points for noise.BTW my scope is a Rigol DS1054 not 52. . . typo above is now corrected . . . but still low end I suppose.The A-B calculated waveform is quite noisey.

author: bdunham7
author_rank: Super Contributor
timestamp: June 14, 2020, 07:43:01 pm
content: OK, so the signal generator is your ground. Can you try it the way I suggested just now? I think you'll get a much clearer picture, albeit in two parts.

author: prwiley
author_rank: Contributor
timestamp: June 14, 2020, 07:58:06 pm
content: Here are pictures.The first just before clipping, the second just after. Connecting the scope ground to the power supply ground made no difference I could see. What is interesting, and I didn't notice before, is that the scope voltage and the meter voltage don't match. The meter says 6.2V in AC mode just before clipping.

author: TimFox
author_rank: Super Contributor
timestamp: June 14, 2020, 08:13:19 pm
content: to bdunham7: From the datasheet (cited above by the OP). At 11 V supply, the amplifier can put out a noticeably distorted 7 W into 8 ohms, which is 7.5 V rms, or 21 V pk-pk.The circuit diagram shows two amp-shaped objects driving the two output pins. With 11 V, each amplifier can put out < 11 V pk-pk ideal maximum. Therefore, both outputs must swing between roughly +11 and 0, while the other swings between 0 and +11. This sounds like your "Type B" device, which is not restricted to expensive gear but used to deal with low supply voltages.With zero input signal (quiescent), each output pin should be close to one-half the supply voltage (+5.5V) with respect to "PS Gnd", and will swing from there to approximately +11 and 0. It is not specified on the data sheet, but I assume that "Vref" is half the supply voltage, with the resistive feedback shown internally.As I understand your "type A" model, the quiescent voltage at both pins should be close to zero--easily checked with a voltmeter.In the post above, the waveforms are as expected by me and your "Type B" description.to prwiley: can you post the positive and negative extrema of the clipped sine waves on both sides?

author: bdunham7
author_rank: Super Contributor
timestamp: June 14, 2020, 08:15:03 pm
content: That's interesting--it looks a little different than I anticipated. Apparently they avoid crossover distortion by always maintaining both sides a bit positive. As for grounding the scope, it really shouldn't have made that much difference--both the amp and scope are already grounded. It just reduces noise a bit.Still, you can see that neither side ever goes negative. Your scope and meter won't agree on this as your meter is measuring the AC component of the difference between the sides. The actual voltage that your load or speaker sees is always thedifferencebetween each trace.

author: TimFox
author_rank: Super Contributor
timestamp: June 14, 2020, 08:16:35 pm
content: Of course neither side goes negative: the active devices are connected between +11 V and zero. There is no transformer included.

author: bdunham7
author_rank: Super Contributor
timestamp: June 14, 2020, 08:29:45 pm
content: Quote from: TimFox on June 14, 2020, 08:13:19 pmAs the OP has posted pictures I guess we can see. You're right, this is not exactly what I described as my Type A, so we'll call it Type C. It is not the Type B I referred to--those types will swing the outputs BOTH WAYS from ground on their own. It is also not type AB because neither amp ever 'shuts off'. A better description would be two class A amps driven out of phase.

author: TimFox
author_rank: Super Contributor
timestamp: June 14, 2020, 08:50:13 pm
content: This is designed to work on a single supply, so the quiescent point is near half that supply voltage. Were it connected between +6V and -6V, the quiescent point would be near 0 V. Since the load is floating, it does not care where the quiescent common-mode voltage is, only the differential voltage.Class AB vs. Class A depends on how much quiescent current flows through the two internal transistors on each output pin (two transistors on each of the two amplifiers drawn on the data sheet): the data sheet shows 65 mA max total quiescent current, so the output stage is class AB on each pin. The 22 V pk-pk across the 8 ohm load will draw 2.75 A pk-pk, or 1.4 A on the positive or negative peak. With a large enough quiescent current, none of the transistors would cut off during the cycle, but the power dissipation at 0 V output would be huge. If the quiescent current through these transistors were zero, that would be Class B. These terms date back to the invention of amplifiers, before I was born.During most of a full cycle, one of the two transistors on each pin will shut off, but the other conducts. Let us call the four transistors Q1 (+ out to Vcc), Q2 (+ out to ground), Q3 (- out to Vcc), and Q4 (- out to ground).When Q2 cuts off and Q1 pulls up, Q4 pulls down and Q3 cuts off. The current then flows from Vcc through Q1 through the load through Q4 to ground.On the other half-cycle, Q1 cuts off, Q2 pulls down, Q3 pulls up, and Q4 cuts off. The current flows from Vcc through Q3 through the load (in the opposite direction) through Q2 to ground.This similar to a switching "H bridge", except that some quiescent current must flow to avoid crossover distortion for the audio application.With zero input, all four transistors are conducting moderately, but negligible current flows through the load.Note that most general-purpose op amps (with less power capability) also run the output stage in Class AB (sometimes called Class B, but the difference is only quantitative) for the same reason. In some audio applications, a resistor from the output to the negative supply is used to force the output stage into Class A and avoid crossover distortion.

author: bdunham7
author_rank: Super Contributor
timestamp: June 14, 2020, 09:27:42 pm
content: Quote from: TimFox on June 14, 2020, 08:50:13 pmI hadn't even considered the internal operation of the op-amp sections themselves, so I'll assume you're correct thateach sideruns in some sort of equivalent AB. Still, I'm not sure you can call it an AB amplifier, because if you connected the load to either side and ground, you would have a working Class A setup, would you not? Internal operation aside, the only reason you don't have high quiescent current through the load here is because the output is opposed by another similar output. You could substitute an ouput capacitor for the other side, just like old-school solid state amplifiers. I think I understand how it works now, so no point in wrangling over the semantics.I've not looked at an amplifier design like this in many years, but the one that I was basing my assumptions on used discrete components arranged in the H-bridge configuration with bias, just as you've stated here. IIRC, it did not 'float' to half of the supply voltage (car audio application) but I could be wrong about that. The difference in looking at an op-amp design is that the op-amp cansinkcurrent whereas the individual sides of a typical AB amplifier cannot, they can at most shut off. Thus an amplifier using two op-amp sections behaves a bit differently

author: TimFox
author_rank: Super Contributor
timestamp: June 14, 2020, 09:46:08 pm
content: If you connected the load through a large capacitor from one side (e.g., "+ Out") to any constant potential (including PS ground), you would still have a Class AB amplifier, since the quiescent current is through the amplifier chip, not through the load. It can't be a Class A amplifier unless none of the transistors cut off, which means that the quiescent current is very large and does not contribute to the load current. If you connected the load resistor between + Out and groundwithout a capacitor, and the amplifier forced the quiescent output voltage anywhere near the normal Vcc/2 point, you would have a very large quiescent power dissipation in both the device and the load (DC power).The reason why this device forces the two outputs to Vcc/2 is found in the data sheet diagram: there is an internal reference voltage and the resistor feedback drives the DC output voltages there.In a "normal" single-ended class AB or B amplifier, for example with split supplies, one side pulls up and the other pulls down, the first one sources and the second one sinks, and there may be an intermediate region where they both pull together but not through the load. This is the quiescent current. If run from a single supply, the bias circuit forces the quiescent point to near Vcc/2 and the output capacitor charges up to prevent DC current through the load. This is why audioneveruses a single device in class AB or B, but could use one in class A if the load can tolerate a large DC quiescent current (such as through a transformer). RF power amplifiers can use a single tube running in AB, B, or even C, where the tuned circuit output network deals with the distorted current waveform (no sink). Again, the standard nomenclature for amplifier classes A, AB, and B refers to the fraction of the cycle where each active device conducts.