After all this time making amplifiers, it still escapes me how to calculate exactly power output, I have some serious doubts. I have a 6 ohm test load of several wattage (300-400W). I have an oscilloscope too. This is my multimeter.
So my supposed correct procedure to measure power output is:
1.- Hook up the 6ohm load between the speaker output terminals.
2.- Hook up the multimeter to the speaker output terminals, set it to 200 AC volts scale.
3.- Hook up the oscilloscope between the speaker output terminals.
4.- With a senoidal 1Khz signal on the input, turn the volume up until I can see clipping on the oscilloscope. Then go back just a little.
Now the mess starts for me. Some talk about true RMS meters. With that setup, I can see 22vac on the multimeter. Is that the correct procedure to calculate power:
P=V^2/R
P=22^2/6= 80W
Something is escaping to me, as for this example (chipamp LM3886), the datasheet states that clipping occurs near 4 volts under PSU supply. If I am supplying it with +/-35vdc, how can it be possible that clipping occurs just at 22vac?
Please someone correct me in this procedure, I think I am messing up concepts 🙄
Best regards!
So my supposed correct procedure to measure power output is:
1.- Hook up the 6ohm load between the speaker output terminals.
2.- Hook up the multimeter to the speaker output terminals, set it to 200 AC volts scale.
3.- Hook up the oscilloscope between the speaker output terminals.
4.- With a senoidal 1Khz signal on the input, turn the volume up until I can see clipping on the oscilloscope. Then go back just a little.
Now the mess starts for me. Some talk about true RMS meters. With that setup, I can see 22vac on the multimeter. Is that the correct procedure to calculate power:
P=V^2/R
P=22^2/6= 80W
Something is escaping to me, as for this example (chipamp LM3886), the datasheet states that clipping occurs near 4 volts under PSU supply. If I am supplying it with +/-35vdc, how can it be possible that clipping occurs just at 22vac?
Please someone correct me in this procedure, I think I am messing up concepts 🙄
Best regards!
I have always wondered about this too, but lucky for me someone had posted in another post something about audio power and put in a Wikipedia link.
Audio power - Wikipedia, the free encyclopedia
Not sure if it helps, but it might be a start.
Audio power - Wikipedia, the free encyclopedia
Not sure if it helps, but it might be a start.
35 volt rail minus 4 volt for clipping point = 31V peak / 1.414 = 21.92V RMS as displayed by your meter.
Thanks epicyclic for the answer. So I should calculate peak value with 31v and constant output value with 22v?
So on a 6 ohm load, that is:
160W peak.
80W constant output.
So this will be qualified as an 80W pure sinuoidal output amplifier?
So on a 6 ohm load, that is:
160W peak.
80W constant output.
So this will be qualified as an 80W pure sinuoidal output amplifier?
P = I * V = V^2 / R = I^2 * R
This is the general rule that applies to DC and any AC waveform that has exactly the same heating effect as the DC. rms is applied to an AC waveform to indicate the heating effect equivalence.
Now let's move on to sinewave power. This is just one special case of an infinite number of AC waveforms.
P = Iac * Vac = Vac^2 / R = Iac^2 * R
In this case ac is the meter reading shown by an AC voltmeter that is calibrated to show sinewave equivalence to the DC voltage. The meter does not need to be rms.
If you read the same voltage on an oscilloscope, you will see the peak to peak voltage Vpp, where Vpp = 2 * sqrt(2) * Vac (as read on that voltmeter calibrated to read Vac equivalent to the DC voltage).
When using peak values of sinewave we usually deal with Vpk (the amplitude of the voltage waveform about Zero Volts). Vpp = 2 * Vpk
The power formulae for the peak voltage and peak current are:
P = Ipk * Vpk / 2 = Vpk^2 / R = Ipk^2 * R
Now to the general (covers all types of waveform including pulsed, square, triangular, sinewave, multiple sinewaves, DC and varying DC and all the other waveforms that exist) formulae for power:
P = Irms * Vrms = Vrms^2 / R = Irms^2 * R.
Irms and Vrms must be read on a meter calibrated for directly reading any waveform and displaying the heating effect as a constant and equivalent DC voltage.
rms reading meters are usually more expensive compared to average reading AC meters calibrated to read sinewaves.
Both average reading and rms reading meters are only accurate within the frequency range specified by the manufacturer and only to the accuracy of each scale specified by the manufacturer.
Notice the four groups of formulae are in exactly the same format.
This is the general rule that applies to DC and any AC waveform that has exactly the same heating effect as the DC. rms is applied to an AC waveform to indicate the heating effect equivalence.
Now let's move on to sinewave power. This is just one special case of an infinite number of AC waveforms.
P = Iac * Vac = Vac^2 / R = Iac^2 * R
In this case ac is the meter reading shown by an AC voltmeter that is calibrated to show sinewave equivalence to the DC voltage. The meter does not need to be rms.
If you read the same voltage on an oscilloscope, you will see the peak to peak voltage Vpp, where Vpp = 2 * sqrt(2) * Vac (as read on that voltmeter calibrated to read Vac equivalent to the DC voltage).
When using peak values of sinewave we usually deal with Vpk (the amplitude of the voltage waveform about Zero Volts). Vpp = 2 * Vpk
The power formulae for the peak voltage and peak current are:
P = Ipk * Vpk / 2 = Vpk^2 / R = Ipk^2 * R
Now to the general (covers all types of waveform including pulsed, square, triangular, sinewave, multiple sinewaves, DC and varying DC and all the other waveforms that exist) formulae for power:
P = Irms * Vrms = Vrms^2 / R = Irms^2 * R.
Irms and Vrms must be read on a meter calibrated for directly reading any waveform and displaying the heating effect as a constant and equivalent DC voltage.
rms reading meters are usually more expensive compared to average reading AC meters calibrated to read sinewaves.
Both average reading and rms reading meters are only accurate within the frequency range specified by the manufacturer and only to the accuracy of each scale specified by the manufacturer.
Notice the four groups of formulae are in exactly the same format.
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"160w peak "
I dont like that way of relating to peak power .
I prefer..........with an unregulated power rail you could get perhaps 1 or 2 volts extra over the voltage the rail droops to when supplying the amp at full output . Therefore for a music transient the amp might be able to supply say 32.5v peak = 22.98V RMS = 88W over 6 ohms .
I dont like that way of relating to peak power .
I prefer..........with an unregulated power rail you could get perhaps 1 or 2 volts extra over the voltage the rail droops to when supplying the amp at full output . Therefore for a music transient the amp might be able to supply say 32.5v peak = 22.98V RMS = 88W over 6 ohms .
For a sinewave 80W average and 160W peak are exactly what you would expect. Also, rarely mentioned, is 0W minimum power. The instantaneous power varies from 0 to 160 through each half-cycle. The average is 80W. In almost all cases the thermal time constants are such that all you see is the 80W average, but the power really is varying from 0 to 160W.
There is another sort of 'peak' power: this is the average for a short period, when the supply rails don't have time to droop. Amplifier makers sometimes deliberately confuse these two, or even combine them, so an amp which can give 40W (average) for long periods but 60W (average) for a short period is advertised as 120W peak. No lies are told, but people are misled (and confused).
There is another sort of 'peak' power: this is the average for a short period, when the supply rails don't have time to droop. Amplifier makers sometimes deliberately confuse these two, or even combine them, so an amp which can give 40W (average) for long periods but 60W (average) for a short period is advertised as 120W peak. No lies are told, but people are misled (and confused).
Is there a number where most amplifiers clip ( class-a/ab ) in relation to psu voltage or is it typically 2-3 volts below...
I use non inductive resistors as a load because loudspeaker impedance varies across the frequency band. To find the maximum clean power before clipping, I do the following:
With oscilloscope: Turn off amplifier and disconnect all inputs and speakers from the amp. Connect sine wave source to input and set frequency. Connect power resistor to output. Connect scope's probes across the resistor. Set volume to minimum and turn on the amp. Increase volume until you see the top and bottom of the waveform flatten out, back off until the wave is smooth again. Take the voltage reading. With a scope, peak to peak is the easiest. Use formula to find Vrms: Vrms = (Vpp/2)/(sqrt 2). So, if you measured 30v pp, the Vrms is 10.6. Square the Vrms value and divide by the load resistor's value. In this example, the amplifier is making 14 watts into an 8 ohm load.
If you don't have a scope, but do have a true rms meter, check its manual to see the frequency range that the true rms meter is accurate. Connect the source and load as described above. Connect the true rms meter across the resistor and be sure to set it for AC volts. Connect a speaker in series with a 330 Ohm 1 watt resistor across the output. This allows you to hear the clipping point without extra loading. I use a frequency of around 150 Hz as it is very easy to hear the clipping point. As you increase the level, the sine wave will start to sound harsh. The onset of clipping is indicated by a slight raspy sound. Back off the level just until the sound becomes smooth. Take the reading off the meter, square it and divide by the load resistance.
If testing a stereo amp, I like to determine the power with both channels driven. In this case, the test signal must be applied to both channels, balance and/or levels must be set even during the test. Don't forget to add a load resistor on each output channel.
With oscilloscope: Turn off amplifier and disconnect all inputs and speakers from the amp. Connect sine wave source to input and set frequency. Connect power resistor to output. Connect scope's probes across the resistor. Set volume to minimum and turn on the amp. Increase volume until you see the top and bottom of the waveform flatten out, back off until the wave is smooth again. Take the voltage reading. With a scope, peak to peak is the easiest. Use formula to find Vrms: Vrms = (Vpp/2)/(sqrt 2). So, if you measured 30v pp, the Vrms is 10.6. Square the Vrms value and divide by the load resistor's value. In this example, the amplifier is making 14 watts into an 8 ohm load.
If you don't have a scope, but do have a true rms meter, check its manual to see the frequency range that the true rms meter is accurate. Connect the source and load as described above. Connect the true rms meter across the resistor and be sure to set it for AC volts. Connect a speaker in series with a 330 Ohm 1 watt resistor across the output. This allows you to hear the clipping point without extra loading. I use a frequency of around 150 Hz as it is very easy to hear the clipping point. As you increase the level, the sine wave will start to sound harsh. The onset of clipping is indicated by a slight raspy sound. Back off the level just until the sound becomes smooth. Take the reading off the meter, square it and divide by the load resistance.
If testing a stereo amp, I like to determine the power with both channels driven. In this case, the test signal must be applied to both channels, balance and/or levels must be set even during the test. Don't forget to add a load resistor on each output channel.
nice ..... one may keep this in mind as a general rule for amplifiers that not feature VI limiters that work as SOA limiter devices ....
Amplifier will produce the specific amount of power on bench under the conditions described above at resistive load and 1KHZ sine but will eventually fail to produce the exact same amount of power in a reactive load or real life conditions ....
Under this conditions 2 pairs of 1943 5200 in a class AB amplifier will easily measure almost 300W ... will they ever produce this in real life ? .... i don't think so ....
Even a TDA 2030 will meet specs as described in the datasheet and produce the amount of power on a resistive load but it will never make it in relalife conditions for the same reason
I presume that 3886 will be one of the same family .....
Amplifier will produce the specific amount of power on bench under the conditions described above at resistive load and 1KHZ sine but will eventually fail to produce the exact same amount of power in a reactive load or real life conditions ....
Under this conditions 2 pairs of 1943 5200 in a class AB amplifier will easily measure almost 300W ... will they ever produce this in real life ? .... i don't think so ....
Even a TDA 2030 will meet specs as described in the datasheet and produce the amount of power on a resistive load but it will never make it in relalife conditions for the same reason
I presume that 3886 will be one of the same family .....
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