If an audio rail supplies maximum +/- 67 volts, what is the maximum
power that can be delivered to an 8 ohm load (nominal)? Is there a
convenient formula?

Thanks!

"Michael" wrote in message
. com...

If an audio rail supplies maximum +/- 67 volts, what is the maximum power
that can be delivered to an 8 ohm load (nominal)? Is there a convenient
formula?

Thanks!


I believe it is Volts squareddivided by load =wattage

Michael wrote:

If an audio rail supplies maximum +/- 67 volts, what is the maximum
power that can be delivered to an 8 ohm load (nominal)? Is there a
convenient formula?

I'm not going to do your homework for you.

BUT, if V=IR and W=IV, then by substitution, W=V^2/R and W=I^2*V, right?
You can plug and chug.
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."

Scott Dorsey wrote:
Michael wrote:
If an audio rail supplies maximum +/- 67 volts, what is the maximum
power that can be delivered to an 8 ohm load (nominal)? Is there a
convenient formula?

I'm not going to do your homework for you.

BUT, if V=IR and W=IV, then by substitution, W=V^2/R and W=I^2*V, right?
You can plug and chug.
--scott

O.k... (67)^2 / 8 = 561 watts. The amplifier I am using is rated at 200
watts and the schematic shows the 67 voltage supply.

Michael wrote in
news
Scott Dorsey wrote:
Michael wrote:
If an audio rail supplies maximum +/- 67 volts, what is the
maximum power that can be delivered to an 8 ohm load
(nominal)? Is there a convenient formula?

I'm not going to do your homework for you.

BUT, if V=IR and W=IV, then by substitution, W=V^2/R and
W=I^2*V, right? You can plug and chug.
--scott

O.k... (67)^2 / 8 = 561 watts. The amplifier I am using is
rated at 200 watts and the schematic shows the 67 voltage
supply.

That's because the voltages are equivalent to the peak values of
the output, not the average voltages(see note 1). For a sine
wave, multiply the voltages by .707, So (67*.707)^2 / 8 = 275 W

Note 1. That is also assuming that the output transistors can
actually allow the total supply voltage to the load. There is
also some loss in emitter resistors, the output transistors and
internal wiring. The load may never see more that 60 Volts
unless the output shorts to the rail.

--
Bob Quintal

PA is y I've altered my email address.

Michael wrote:
Scott Dorsey wrote:
Michael wrote:
If an audio rail supplies maximum +/- 67 volts, what is the maximum
power that can be delivered to an 8 ohm load (nominal)? Is there a
convenient formula?

I'm not going to do your homework for you.

BUT, if V=IR and W=IV, then by substitution, W=V^2/R and W=I^2*V, right?
You can plug and chug.
--scott

O.k... (67)^2 / 8 = 561 watts. The amplifier I am using is rated at 200
watts and the schematic shows the 67 voltage supply.

No, it's a good bit more than that, since you can swing TWICE the 67V into
the load. Note that this is peak power... RMS power on a sine waveform
will be a lot smaller. Also the series resistance of the power supply
and the output transistors drops the maximum output swing to a good bit
less than the 134V in reality.
--scott


--
"C'est un Nagra. C'est suisse, et tres, tres precis."

If you are talking about RMS power for a sinusoidal waveform, its

P = (Vp^2)/(2R)

Where Vp is the peak voltage.However, as Vp approaches the rail
voltage, distortion becomes severe. With a few volts of headroom,
however, the output should be pretty clean. If you look at a data sheet
for a part like theTDA2009A, you can see how rapidly distortion
increases as Vp is increased near the rail voltage.


George Gleason wrote:
"Michael" wrote in message
. com...

If an audio rail supplies maximum +/- 67 volts, what is the maximum power
that can be delivered to an 8 ohm load (nominal)? Is there a convenient
formula?

Thanks!


I believe it is Volts squareddivided by load =wattage

Scott, are you talking about a bridge amplifer here?

jwvm wrote:
Scott, are you talking about a bridge amplifer here?

No, one that is complementary. You can swing the output rail up to +60V
by turning on one side of the output stage, or down to -60V by turning the
other one on. Total swing is from rail to rail.

I don't know who came up with the original idea but it was probably one
of the analog computer guys in the fifties. It's a great one, though.
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."

Michael wrote:

O.k... (67)^2 / 8 = 561 watts. The amplifier I am using is rated at 200
watts and the schematic shows the 67 voltage supply.

Well, actually, if it's running + and - 67 volts, the total voltage
swing is 134 volts. If we assume a sine wave which has an RMS value of
..707 (1/2 sqrt 2) times the peak, we have [(134 x .707)^2]/8 or 1122
watts. But there's more to dissipate the power in than just the
speakers, so 200 watts is probably a reasonably honest power rating.

How old is it and how much does it weigh?

Michael wrote:


O.k... (67)^2 / 8 = 561 watts. The amplifier I am using is rated at 200
watts and the schematic shows the 67 voltage supply.

OK, working backwards: 200 Watts sine wave power = 400 watts peak
P = V^2/R
V = sqrt(3200) = 56 Volts peak required to develop specified power...

The actual voltage of 67V is to allow for voltage drop in the outputs
transistors and variations in the AC supply voltage, so that the 200W
figure is guaranteed for a variation of (say) +/- 10% in supply voltage.

--
Anahata
-+- http://www.treewind.co.uk
Home: 01638 720444 Mob: 07976 263827

"Michael" wrote in message
. com

If an audio rail supplies maximum +/- 67 volts, what is
the maximum power that can be delivered to an 8 ohm load
(nominal)? Is there a convenient formula?

One problem with estimating amp power from supply voltages taken off a
schematic is that power supply regulation becomes an issue. Almost all power
amps use an unregulated power to supply their output stages. The voltage on
the schematic is usually taken with no signal. It could drop by from 5% to
20% under full load.

There are a couple losses in the output of a power amplifier. The amp
saturates at something less than the peak supply voltage and the unregulated
power supply droops. This could represent 10% or more in the total peak
voltage output.

Just for ease on numbers, assume that the output loss is a total of 7 volts
so that the peak output voltage is +/-60 volts.

The peak of a sinewave is 1.414 * RMS value of the sinewave.
The RMS value of a sinewave is the (peak voltage)/ 1.414 or 0.707 * peak
voltage.
The maximum RMS power in this example is 60* 0.707 = 42.4 Vrms.

Amplifier manufacturers use the continuous RMS to specify their output
power.

The RMS power from the output of an amplifier with a +/-60 Volt peak output
is:

Po = V^2/R
Po = (60 * 0.707)^2/8 = 1800/8 = 225W (with rounding)


John Phillips




George Gleason wrote:
"Michael" wrote in message
. com...

If an audio rail supplies maximum +/- 67 volts, what is the maximum
power
that can be delivered to an 8 ohm load (nominal)? Is there a
convenient
formula?

Thanks!


I believe it is Volts squareddivided by load =wattage

56 V is I believe the correct answer here. However, note that in order
to get this value, the formula should be

P = (Vp^2)/(2R) so that Vp = sqrt(2RP) = sqrt(2*8*16) = 56V

anahata wrote:
Michael wrote:


O.k... (67)^2 / 8 = 561 watts. The amplifier I am using is rated at 200
watts and the schematic shows the 67 voltage supply.

OK, working backwards: 200 Watts sine wave power = 400 watts peak
P = V^2/R
V = sqrt(3200) = 56 Volts peak required to develop specified power...

The actual voltage of 67V is to allow for voltage drop in the outputs
transistors and variations in the AC supply voltage, so that the 200W
figure is guaranteed for a variation of (say) +/- 10% in supply voltage.

--
Anahata
-+- http://www.treewind.co.uk
Home: 01638 720444 Mob: 07976 263827

I think that we are on the same page here. The .707 factor squared
becomes the 1/2 in my equation so the answers are the same.

I must be missing something here but the peak voltage across the load
for a standard amplifier with split supplies cannot exceed the rail
voltage. The peak-to-peak swing will be double the rail voltage but the
power output must be (Vp^2)/(2*R). A bridge amplifier will double the
peak voltage that appears across the load and increase the power output
by nearly a factor of 4. Where am I going wrong?

jwvm wrote:
I must be missing something here but the peak voltage across the load
for a standard amplifier with split supplies cannot exceed the rail
voltage. The peak-to-peak swing will be double the rail voltage but the
power output must be (Vp^2)/(2*R). A bridge amplifier will double the
peak voltage that appears across the load and increase the power output
by nearly a factor of 4. Where am I going wrong?

Okay, let's say that we have 20V rails. You use the output normally, and
it can go from +20V to -20V, so you have a total of 40V range.

Now, we bridge it. On positive going peaks, one channel is at +20V, the
other is at -20V, so there is +40V across the load. On negative-going
peaks, the first channel is now at -20V and the second at +20V, so there
is -40V across the load. The total output swing is now 80V.

It really is ingenious as hell. I don't know who came up with the idea but
he deserves a toast.
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."

Yes, we are on the same page! A bridge amplifier is indeed a very
clever design and works well even with a single supply. Since both
outputs should be biased to Vcc/2, output capacitors can be eliminated
if there are no offset errors between the outputs.

Scott Dorsey wrote:
jwvm wrote:
I must be missing something here but the peak voltage across the load
for a standard amplifier with split supplies cannot exceed the rail
voltage. The peak-to-peak swing will be double the rail voltage but the
power output must be (Vp^2)/(2*R). A bridge amplifier will double the
peak voltage that appears across the load and increase the power output
by nearly a factor of 4. Where am I going wrong?

Okay, let's say that we have 20V rails. You use the output normally, and
it can go from +20V to -20V, so you have a total of 40V range.

Now, we bridge it. On positive going peaks, one channel is at +20V, the
other is at -20V, so there is +40V across the load. On negative-going
peaks, the first channel is now at -20V and the second at +20V, so there
is -40V across the load. The total output swing is now 80V.

It really is ingenious as hell. I don't know who came up with the idea but
he deserves a toast.
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."