I've never seen a chart, but that would be handy.
This Ohm's Law Calculator will teach you a lot. (the formula is there, too)
Remember though that will be peak power, not RMS. Divide your rail voltage by the square root of 2 (1.41) to get the RMS voltage that those rails can supply. For example:
50V rail/8 ohm load = 312 watts peak. But RMS value is lower. 50V/1.41 =35.46V or 157 watts RMS into the 8 ohm load.
You'll have to add a few extra volts to the rails to make up for losses in the output devices. And that should get you close enough. It's all Ohm's law.
Welcome to the forum!
This Ohm's Law Calculator will teach you a lot. (the formula is there, too)
Remember though that will be peak power, not RMS. Divide your rail voltage by the square root of 2 (1.41) to get the RMS voltage that those rails can supply. For example:
50V rail/8 ohm load = 312 watts peak. But RMS value is lower. 50V/1.41 =35.46V or 157 watts RMS into the 8 ohm load.
You'll have to add a few extra volts to the rails to make up for losses in the output devices. And that should get you close enough. It's all Ohm's law.
Welcome to the forum!
That is largely correct if you have positive and negative voltage rails. If it's single ended, divide the voltage by two.
In addition, realize that at maximum power the supply voltage will sag after a few milliseconds, so you must decide whether you want to know the maximum you can get in a pulse, or steady state. The filtering matters too, as the dips in the mains cycles will be visible, more or less, in the power supply output when loaded. You need to use the lowest momentary voltage in your calculations. To make it more complicated, the resistance of the transformer winding changes with temperature and so a hot transformer delivers lower voltage than a cold one. Further, the mains voltage must be held constant during these tests.
In addition, realize that at maximum power the supply voltage will sag after a few milliseconds, so you must decide whether you want to know the maximum you can get in a pulse, or steady state. The filtering matters too, as the dips in the mains cycles will be visible, more or less, in the power supply output when loaded. You need to use the lowest momentary voltage in your calculations. To make it more complicated, the resistance of the transformer winding changes with temperature and so a hot transformer delivers lower voltage than a cold one. Further, the mains voltage must be held constant during these tests.
Hi axperion2
Simple formula to calculate power into a given load resistance is;
Power(average) = V^2/(2*R)
This is the PEAK (rail) voltage squared, divided by twice the load resistance.
However, please remember that this is for ONLY one rail and applies ONLY to Sine waves.
Example for 100W into 8R requires a 40V rail but there are two rails, so +-40V.
Again, this is theoretical maximum and will need some headroom to give the power you are looking for.
Sandy
Simple formula to calculate power into a given load resistance is;
Power(average) = V^2/(2*R)
This is the PEAK (rail) voltage squared, divided by twice the load resistance.
However, please remember that this is for ONLY one rail and applies ONLY to Sine waves.
Example for 100W into 8R requires a 40V rail but there are two rails, so +-40V.
Again, this is theoretical maximum and will need some headroom to give the power you are looking for.
Sandy
Hi,
for that 100W into 8ohm example, you will find that it needs a +-48Vdc to +-51Vdc to get your 100W continuous into 8r0.
When doing this the supplies will have sagged to between +-44Vdc to 46Vdc.
Starting from the required output power, calculate the peak output voltage,
Then add between 8V and 11V to each of the dual polarity rail voltages to account for losses through the PSU and amplifier.
for that 100W into 8ohm example, you will find that it needs a +-48Vdc to +-51Vdc to get your 100W continuous into 8r0.
When doing this the supplies will have sagged to between +-44Vdc to 46Vdc.
Starting from the required output power, calculate the peak output voltage,
Then add between 8V and 11V to each of the dual polarity rail voltages to account for losses through the PSU and amplifier.
Hi again.
As an example of the overhead required for rail voltages, I'll cite the amplifier I'm designing at present.
I was looking for high-ish voltage to use as a class A unit possibly for a current dumping project that could be amended fairly easily to class AB for higher power.
my spec. was that the class A should be usable to the maximum rail voltage I would need for the power amp and obviously be as high quality as possible.
However, all was going welll 'till I found the simulation showed distortion creeping in when the output voltage was about 80% of my rail voltage, i.e. with rails of +-52V, I was only getting output of +-41V at 0.003% THD. Below this, I was getting consistant 0.001% THD.
The THD climbed rapidly above this but the virtual scope showed no clipping or any other obvious sign that would account for it.
So it would appear that around 15% - 20% overhead is needed for good performance.
Sandy
As an example of the overhead required for rail voltages, I'll cite the amplifier I'm designing at present.
I was looking for high-ish voltage to use as a class A unit possibly for a current dumping project that could be amended fairly easily to class AB for higher power.
my spec. was that the class A should be usable to the maximum rail voltage I would need for the power amp and obviously be as high quality as possible.
However, all was going welll 'till I found the simulation showed distortion creeping in when the output voltage was about 80% of my rail voltage, i.e. with rails of +-52V, I was only getting output of +-41V at 0.003% THD. Below this, I was getting consistant 0.001% THD.
The THD climbed rapidly above this but the virtual scope showed no clipping or any other obvious sign that would account for it.
So it would appear that around 15% - 20% overhead is needed for good performance.
Sandy
Hi,
your arithmetic is flawed. Add 25% to your 41Vpk and you get ~51Vdc for your supply. This is within the range of adding 8V to 11V
ok here is my contribution this is base on a proaudio amps with a regulated front end/driver stage which is 20% higher than the main rail voltage.
200W @ 8Ohms = 70VDC+/-
400W @ 8Ohms = 90VDC+/-
600W @ 8Ohms = 110VDC+/-
800W @ 8Ohms = low rail 65VDC+/- step 130VDC+/- class H
1000W @ 8Ohms = low rail 75VDC+/- step 150VDC+/- class H
200W @ 8Ohms = 70VDC+/-
400W @ 8Ohms = 90VDC+/-
600W @ 8Ohms = 110VDC+/-
800W @ 8Ohms = low rail 65VDC+/- step 130VDC+/- class H
1000W @ 8Ohms = low rail 75VDC+/- step 150VDC+/- class H
Distortion near power rails could be caused by the VAS. Even with a CCS load it is still feeding more current into the output stage at peaks. Most amplifiers begin to distort when asked to get too near their supplies. There may be advantages in running the VAS from a slightly higher supply voltage than the output stage, although many designs do the opposite. Need to be careful about output latchup, though.
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