Been wondering about this topic, found a few (rare) mentions that have different opinions, wondering if anyone could clear this up:
Lets say 10V supply into 10 ohms, to simplify things, amp is ideal, so max output voltage = supply voltage,
Instinctively, P=V^2/R so max power = 10W.
But, 10V is +-5V, so my output voltage is +-5V max, wouldn't that mean power = 2.5W?
Help me clear this misconception pls.
Lets say 10V supply into 10 ohms, to simplify things, amp is ideal, so max output voltage = supply voltage,
Instinctively, P=V^2/R so max power = 10W.
But, 10V is +-5V, so my output voltage is +-5V max, wouldn't that mean power = 2.5W?
Help me clear this misconception pls.
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10V is the peak-to-peak voltage. You need to consider the RMS voltage to get the RMS power.
The formula is:
P[RMS] = Ec^2/(8*Rl) where:
P is the RMS power in watts
Ec is the supply voltage. In your case, is either 10V or +/-5V=10V (same thing)
Rl=Load resistance in ohms.
In your case, the RMS power is 10^2/(8 * 10) = 1.25W
a single polarity 10Vdc is equivalent to a dual polarity +-5Vdc supply.
An amplifier with an AC output from a theoretically ideal output stage can develop an output voltage of 5Vpk from the 5Vdc supply.
+-5Vpk is equivalent to 3.5Vac.
The maximum power into 10r0 is V^2 / Rload = 3.5^2 /10 = 1.25W
or P = Vpk^2 / 2 / Rload = Ipk^2 / 2 * Rload = Vac / Rload = Iac * Rload = Iac * Vac
The real maximum output from an amplifier operating on +-5Vdc driving a 10r load will be Pmax ~ 4^2 / 2 / 10 ~= 800mW
If the load is a reactive speaker then expect the maximum current that could be demanded ~= three times the equivalent resistor, i.e. a 10ohm speaker driven with 4Vpk may demand <=1.2Apk.
An amplifier with an AC output from a theoretically ideal output stage can develop an output voltage of 5Vpk from the 5Vdc supply.
+-5Vpk is equivalent to 3.5Vac.
The maximum power into 10r0 is V^2 / Rload = 3.5^2 /10 = 1.25W
or P = Vpk^2 / 2 / Rload = Ipk^2 / 2 * Rload = Vac / Rload = Iac * Rload = Iac * Vac
The real maximum output from an amplifier operating on +-5Vdc driving a 10r load will be Pmax ~ 4^2 / 2 / 10 ~= 800mW
If the load is a reactive speaker then expect the maximum current that could be demanded ~= three times the equivalent resistor, i.e. a 10ohm speaker driven with 4Vpk may demand <=1.2Apk.
Hi, a useful bit of info here:- http://www.bcae1.com/voltages.htm
Bottom line is that you have to convert your theoretical 5V peak sine wave into the equivalent DC heating value called the rms value which would be around 3.5V. You can then calculate what power would be dissipated into the particular load you were using by using the usual formula V (squared) divided by R (Load resistance). Assuming this was an 8 Ohm speaker, power dissipated with a continuous pure sine wave would be 3.5 x 3.5 = 12.25/8 = 1.53 Watts.
Les
P.S. Sorry if I have repeated some of the above posts which were not there when I started mine LOL!
Bottom line is that you have to convert your theoretical 5V peak sine wave into the equivalent DC heating value called the rms value which would be around 3.5V. You can then calculate what power would be dissipated into the particular load you were using by using the usual formula V (squared) divided by R (Load resistance). Assuming this was an 8 Ohm speaker, power dissipated with a continuous pure sine wave would be 3.5 x 3.5 = 12.25/8 = 1.53 Watts.
Les
P.S. Sorry if I have repeated some of the above posts which were not there when I started mine LOL!
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if you can remember that
I = V / R - Ohm's law
and that
P = I * V
then all the others follow exactly from those two equations.
No reason to be either lazy nor stupid.
But frequent repetition may well be required to commit this to permanent memory, for it to stick. Just like a PC:- RAM is volatile and HDD can be almost permanent, if we know where the index to the library/files is kept.
I = V / R - Ohm's law
and that
P = I * V
then all the others follow exactly from those two equations.
No reason to be either lazy nor stupid.
But frequent repetition may well be required to commit this to permanent memory, for it to stick. Just like a PC:- RAM is volatile and HDD can be almost permanent, if we know where the index to the library/files is kept.
Sorry, this answer/explanation is misleading to any newbie looking for clarification.
Thanks, I'd have to bookmark this. 
I understand much of the calculations, especially the electrical formalae, I'm just confused as to what other factors I have to include.
So what I understand is:
10V -> +-5V, so equivalent heating power divide by 4
Then 5V peak sine wave -> divide by root 2 to get RMS voltage, equivalent power divide by 2
So P[RMS] = Ec^2/(8*Rl) -> the 1/8 in the formula comes from above
I notice that everybody (and program) is considering RMS power with regard to sine waves, is that standard practice in this particular industry? (Because an amp can output square waves, like class-D, or digital buffer, would P[RMS]=Ec^2/(4*Rl) instead?)
I understand much of the calculations, especially the electrical formalae, I'm just confused as to what other factors I have to include.
So what I understand is:
10V -> +-5V, so equivalent heating power divide by 4
Then 5V peak sine wave -> divide by root 2 to get RMS voltage, equivalent power divide by 2
So P[RMS] = Ec^2/(8*Rl) -> the 1/8 in the formula comes from above
I notice that everybody (and program) is considering RMS power with regard to sine waves, is that standard practice in this particular industry? (Because an amp can output square waves, like class-D, or digital buffer, would P[RMS]=Ec^2/(4*Rl) instead?)
The problems start when you consider MOSFET output stages.
On my amp it takes 7 volts bias voltage so I have to subtract 3.5 volts from the peak voltage at least to get near the correct result.
Just to complicate things further the harder the MOSFET is driven the more volts it needs on the gate.
BJT's arent as bad they just need a volt or two bias.
Or at least that is my understanding.......
On my amp it takes 7 volts bias voltage so I have to subtract 3.5 volts from the peak voltage at least to get near the correct result.
Just to complicate things further the harder the MOSFET is driven the more volts it needs on the gate.
BJT's arent as bad they just need a volt or two bias.
Or at least that is my understanding.......
My memory is extremely volatile.
It took a long time to get this image burned into my EEPROM.....
An externally hosted image should be here but it was not working when we last tested it.
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My memory is extremely volatile.
It took a long time to get this image burned into my EEPROM.....
An externally hosted image should be here but it was not working when we last tested it.
Strange, I had P = V * I etched into my brain since I was a child. Must be my father showing me the utility bill.
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