How to measure amplifier watages by useing volt meter.

phd500

What is the formula for converting volts to watts.
I have heard about audio reviewers doing this, but I have no idea how to convert the volts to watts, thanks for any help you can give me.

juice21

P=V2/R
The units for V, R and P are volts, ohms and watts respectivly.

Aaron

What are you trying to figure out with these formulas?

Aaron

jcaut

Somebody correct me if I'm wrong, but wouldn't this only be accurate if the output was at or near 60 Hz? You'd get the best results by using a true-RMS reading voltmeter and leaving a speaker connected to the amp during the measurement, for loading. And a speaker presents a dynamic impedence instead of a simple resistance. It might give you a ballpark estimate of an amp's power. This is probably the way the power meters work, on amps that have them.

2.83Vrms into an 8Ohm load would be 1Watt
20Vrms output into an 8 Ohm load would be 50Watts; Current in this example would be 20/8=2.5A

Like I said, somebody set me straight, if I'm wrong.

Jason

phd500

Ok say I measure 97 volts output from the amp when I run it with two 5 ohm subs wired in paralel. Can you put this in the formula to show me how to figure it out. I am kinda slow with this stuff:) thanks

phd500

Aaron, I have two JBL 15 power seriese subs, and I want to see how many ohms they will present an amp with when wired in parelel.

jcaut

phd500,

If you wire two speakers in parallel, the impedence presented to the amp would be half of what a single driver would be (assuming the impedence of each speaker is the same).

If you're wiring two 8 Ohm speakers in parallel, you'd have a 4 Ohm load. Two 4 Ohm subs wired in parallel would be a 2 Ohm load.

If you've got two 5 Ohm subs wired to the amp in parallel, then you're running at 2.5 Ohms, nominal. If you measured 97 V (AC) across the output terminals, with these speakers connected, then the current in the voice coil(s) is 97/2.5=38.8 Amps. The power would be 97 x 38.8=3763.6 Watts

Of course, this doesn't sound realistic to me. But that's what plugging those numbers into the formulas yield.

sgtgto

Hi:

Rather than make a statement I want to ask someone to do something for me. Turn on your stereo, turn it up, get a digital volt meter, measure the voltage across the terminals of one of your main speakers. Let me know what you read.


Gary;)

phd500

jcaut,

So if I measure 16 volts into a 4 ohm load then 16/2=8 amps. Then 16x8=128 watts?

jcaut

I know what you're getting at, Gary. I thought the same thing.

I'm just doing the math.:D

phd500,

According to Ohm's law, E=IR, where E is electromotive force measured in Volts, "I" is current in the system, measured in Amperes, and "R" is resistance in Ohms. As I said earlier, a speaker presents a dynamic impedence, rather than a simple resistance. If you use the nominal impedence of the speaker, which is kind of an "average" impedence, you could expect it to behave somewhat like a simple resistance for the purpose of the calculation. Therefore, if you measure 16 Volts and you assume the resistance (R) is pretty close to the 4 Ohms that your speakers are rated, then you can use Ohms law to calculate current: 16/4=4 Amps. It's not really necessary to calculate current, as I'll show you in just a minute.

The formula for power, in Watts is: current (in Amps) times the Voltage. So for your example, 16 x 4=64 Watts.

You can calculate "power" withouth going through the "current" calculation, with the formula Voltage squared, divided by resistance. For your example, (16 x 16)/4 = 64.

That's all I can tell you. As far as actually measuring this stuff with enough accuracy to be worth your time, well, that's up to you! And like I said earlier, I wouldn't tell you this stuff if I knew I was wrong, but I won't guarantee that I'm right, either!

Jason

sgtgto

HI:

I'll repeat my previous post. Take a digital volt meter and turn your stereo on and loud, measure the voltage across the terminals of your mains. Turn the stereo off and connect the volt meter in series with the positive lead and measure amps. Turn your stereo on and loud, let me know what you read.


Gary

phd500

Ok guys thanks for your help. I really appreciate it.

jcaut

Gary,

I was going to leave this topic alone, but then I saw your second post.

I thought you were referring to the difficulty in actually taking a measurement due the fluctuating readings. After I read your second post, I began to think that you know something that I don't (which wouldn't be unusual ).

I attempted to take some measurements this morning on my little stereo that I have at work. This is an old, Realistic receiver, rated something like 80wpc at 8 ohms. I played a CD-R containing some computer-generated sine wave test tones. Playing a 60Hz tone at a relatively loud volume, I measured: 14.0 Volts and .80 Amps
The power meters on the receiver (LED type) were showing 12 watts.

I haven't tried this on any of my "good" equipment yet. Am I missing something? Enlighten me!

Thanks,
Jason

sgtgto

Hi:

What I was getting at was that the measurements aren't as large as some people think. That is why I wanted you to measure it to avoid arguement. I believe some people believe you have as much as 100V at 50A across the speaker terminals and that is just not true. A speaker is not a current drawing device like a motor or electric stove. A speaker is really a transducer that only reacts to the signal sent to it from the receiver. A transducer changes electrical energy into mechanical motion and that is what a speaker does.

The ohms rating of a speaker will fluxuate during operation because of the change in signal input and the speaker operation itself. Does this help?


Gary

jcaut

Thanks for the clarification. Do you know if this method of measuring power output is sound? Other than the fact that you don't generally know the "real" impedence of the speaker at the frequency you use, I can't see anything wrong with it. If you actually measure both voltage and current, and multiply them to get watts, it's bound to be pretty close.

I knew the numbers wouldn't be large. You had me scared that I was missing something, and I didn't want to be giving out bad information! Thanks.

Jason

P.S. Could I infer, from my measurements, that the impedence of my speaker in the above example is about 16 Ohms at 60 Hz? That's probably near the resonance freq for the speaker system. The impedence will peak at the resonant frequency, due to the counter EMF generated by the moving voice coil, correct?

sgtgto

Hi:

I get 17.5 Ohms. Did you use a digital VM and is it calibrated? That seems kind of high.

The resonance frequency occurs when the resistance is smallest and maximum current and voltage can be obtained. Actually, impedance is the total resistance in an AC circuit and has to consist of capacitance, inductance, and resistance.

This is quiet involved and it has been a long time since I studied it. I don't really know if you could associate a speaker with resonance frequency. Maybe at one frequency but not overall.


Gary

jcaut

Gary,

Actually I just calculated it in my head and saw that it was about double the impedence I was expecting. I'll buy 17.5.

My VM is a cheapie Radio Shack. Calibration???

You sound as if you're very knowledgable in electronics. I personally know just enough to be dangerous, as they say.

I don't know about a multiple driver speaker system, but when dealing with an individual driver, I thought the peak impedence would occur when the driver was operating at its resonant frequency. But that seems to be the opposite of what you said.

I just thought of this as well: Impedence would rise as the frequency gets higher, due to inductive reactance (voice coil is an inductor), so the real "peak" would probably be at the highest frequency you feed it.

I have a feeling this has gone way beyond what phd500 wanted to get in to when he asked the original question. Thanks again for the info!

Jason