It is somewhat confusing.
RMS is short for "Root Mean Squared" which is short for "the square root of the average value of the square of the amplitude".
Put simply, the RMS voltage was originally used to tell you what equivalent dc voltage you would need to produce the same average heating power in a resistive load as an ac voltage. It is like a convenient way to equate ac and dc systems for the purpose of calculating average power.
This is handy when dealing with mains electrics - if your ac mains is 230Vac RMS then you know that you would need 230V dc to heat a resistor the same amount or make a light bulb (incandescent) the same brightness.
The term RMS can only apply to an ac voltage or current. It has no meaning for a dc voltage or current.
To convert between peak V or I and RMS V or I you divide the peak by the square root of 2 = 1.414
When you measure 70mV across Re I assume this is a dc voltage. So RMS does not apply.
I personally find it easiest just to deal with peak V and peak I. Just work out the peak power using P = (V*V)/R or P = (I*I)*R, then just divide by 2 to get average power.
RMS is short for "Root Mean Squared" which is short for "the square root of the average value of the square of the amplitude".
Put simply, the RMS voltage was originally used to tell you what equivalent dc voltage you would need to produce the same average heating power in a resistive load as an ac voltage. It is like a convenient way to equate ac and dc systems for the purpose of calculating average power.
This is handy when dealing with mains electrics - if your ac mains is 230Vac RMS then you know that you would need 230V dc to heat a resistor the same amount or make a light bulb (incandescent) the same brightness.
The term RMS can only apply to an ac voltage or current. It has no meaning for a dc voltage or current.
To convert between peak V or I and RMS V or I you divide the peak by the square root of 2 = 1.414
When you measure 70mV across Re I assume this is a dc voltage. So RMS does not apply.
I personally find it easiest just to deal with peak V and peak I. Just work out the peak power using P = (V*V)/R or P = (I*I)*R, then just divide by 2 to get average power.
traderbam said:It is somewhat confusing.
RMS is short for "Root Mean Squared" which is short for "the square root of the average value of the square of the amplitude".
Put simply, the RMS voltage was originally used to tell you what equivalent dc voltage you would need to produce the same average heating power in a resistive load as an ac voltage. It is like a convenient way to equate ac and dc systems for the purpose of calculating average power.
This is handy when dealing with mains electrics - if your ac mains is 230Vac RMS then you know that you would need 230V dc to heat a resistor the same amount or make a light bulb (incandescent) the same brightness.
The term RMS can only apply to an ac voltage or current. It has no meaning for a dc voltage or current.
To convert between peak V or I and RMS V or I you divide the peak by the square root of 2 = 1.414
When you measure 70mV across Re I assume this is a dc voltage. So RMS does not apply.
I personally find it easiest just to deal with peak V and peak I. Just work out the peak power using P = (V*V)/R or P = (I*I)*R, then just divide by 2 to get average power.
Yes it was somewhat confusing but thanks everyone for the effort to show me the light.
My conclusion is, My mistake was, I confused RMS DC measurement be equal to AC RMS but of course the fact that I measure RMS DC says nothing about the RMS AC voltage or current.
So I got wrong in taking the Peak to Peak current in my formula.
Now I calculate the RMS (average) current by Irms= 0,71xI
And then like roender and others indicates my correct formula is:
P rms = R * (0.71 * I)^2
I think I get it now.
Thanks again.
With best regards,
Bas
Yep, I think you've figured it out.
Just be a little careful with the terminology...
No. RMS current is not the same as average current. The average ac current is zero. That's why the RMS formula is used.
Average power, on the other hand may be calculated using RMS voltage or RMS current, but this is not to be confused with the literal definition of RMS power (the root mean squared value of the power waveform) which is not the same thing at all.
Just be a little careful with the terminology...
No. RMS current is not the same as average current. The average ac current is zero. That's why the RMS formula is used.
Average power, on the other hand may be calculated using RMS voltage or RMS current, but this is not to be confused with the literal definition of RMS power (the root mean squared value of the power waveform) which is not the same thing at all.
roender said:You are correct Andrew .... but many amplifier data sheets specify the RMS power capability, which is a nonsense form mathematical POV
Power RMS is just a language term, noting more
Please see the attachment, Power RMS for Dummies
RMI-FC100 specs:
Class A power: 2W rms into 8ohm with Iq=0.6A (maximum Iq for optimal B class bias)
Class B power: 90W rms into 8ohm
stinius said:
RMI-FC100 specs:
Class A power: 2W rms into 8ohm with Iq=0.6A (maximum Iq for optimal B class bias)
Class B power: 90W rms into 8ohm
I was drunk when I've done that
I forgot to do the math for push-pull OPS (not SE) and the power is in 6ohm, the impedance of my MA gold GS20 speakers
Sounds about right for working out the RMS output going into a load by measuring current, but remember that it will probably be lower than that for Class A as quiescent current is almost 100% of the time higher than the corresponding voltage swing on the output needed to drive the load. I would use this equation:
P = (0.707*V)^2/4*Z
Where P is the RMS power, V is the maximum peak to peak voltage swing of the amplifier's output and Z is the loudspeakers impedance.
Hope that helps
P = (0.707*V)^2/4*Z
Where P is the RMS power, V is the maximum peak to peak voltage swing of the amplifier's output and Z is the loudspeakers impedance.
Hope that helps
Vizion, calculation for single-ended cl. A differs from push-pull topology.
Ideally, push-pull amp stays in cl. A as long as current through the load (loudspeaker) is equal to or less than twice the bias current. With current through the load higher than that the amp will keep working in class AB.
For single-ended amp the bias current equals the maximal current through the load. If the load tries to draw current higher than that the current clipping will occur. There are workarounds about that (Aleph, the modulated current source) but it's no longer pure singlle-ended topology.
Ideally, push-pull amp stays in cl. A as long as current through the load (loudspeaker) is equal to or less than twice the bias current. With current through the load higher than that the amp will keep working in class AB.
For single-ended amp the bias current equals the maximal current through the load. If the load tries to draw current higher than that the current clipping will occur. There are workarounds about that (Aleph, the modulated current source) but it's no longer pure singlle-ended topology.
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