(OT a bit) Question on amp wattage calculations...

[Aurin]

Hello,
I'm new to this forum and amp/fx building in general and I think it's quite a nice place -- alot of good information lying around. Maybe I don't know what to put into the search or there aren't any topics with the "right" answer but.... I have a few questions that may or may not be somewhat silly.

I'm putting together a modified Ruby amp from ROG (Ruby + Fetzer valve + a lil' distortion mod I cooked up)  and I've got a few questions related to that...

Knowing the Ruby is based off of the Little Gem I go to the Little Gem page and the schem says it's a 1/2 W Amp .

How exactly are these things calculated? I know you can get Wattage by voltage squared over resistance  (Watts = Voltage * Amps; Amps = Voltage / Resistance) but it doesn't seem to add up for amps? Assuming a perfect amp running off 9V the max would, I assume, be 9V * 9V / 8Ohms ... which ends up being 10.125 Watts... definitely NOT the 1/2 W listed. Or am I completely off on these calculations?

But as I understand it speaker ratings are not quite the same as normal resistors.  so this may explain things but ... well, How? lol.

Next question is probably easier... If I like the sound of the ruby amp could I make a "bigger" version by adding a power transformer to step up the voltage (After the output capacitor, for instance)? Would it be more helpful to draw out a schematic to explain my meaning?

Thanks in advance to anyone who answers; hopefully they're easy questions =D

[R.G.]

You're very close to the answer. What you're missing is AC versus DC power.

If you have 9V across 8 ohms, you do indeed get 10.125 W. However, if you use the 9V to power an amplifier and the amplifier swings the 9V peak to peak across the resistor, the power is less. If you think about it, that makes sense - the voltage is only 9V at the very peak.

Worse, it's got to be real AC into a speaker, not 0 to 9V. So the speaker is decoupled from the DC by a capacitor or by using +/- 4.5V for a power supply, and now the peak voltage across the speaker is 4.5V. When the first guys working on electricity a couple of centuries ago tried to figure out what a volt was, they wanted the same measurement for DC (battery) and AC (alternator) power. What they came up with was to make the power out the same for the same volts. DC volts are easy, but how do you measure AC volts so that an AC volt across a given resistor gives the same power as a DC volt, when the AC volt changes?

What they came up with was the root-mean-square voltage. I won't type all the math in, but an AC volt RMS is the voltage that will cause the same heating in a resistor as the same numbered DC volts. If you have 9Vdc across an 8 ohm resistor you get 10.125W, so 9Vac RMS is the sine wave voltage that will cause the same heating. It turns out that the peak voltage of that sine wave must be 1.414 times bigger than the "DC" number to get the same heating, so that 9VacRMS squared divided by 8 ohms is 10.125W.

So if we power an amplifier from a total of 9V, its peak can only go to +/-4.5V, and then the RMS ac voltage that it puts out is 4.5V/1.414 = 3.18Vac RMS. Now we can compute the power as 3.18*3.18/8 = 1.266W. Still bigger than 1/2 W but in the ballpark.

Real amplifiers running from 9V can't actually pull the load up to 9.0000V or down to 0.0000V. There is some voltage lost because the output transistors can't saturate to zero volts across them. They can usually get to under 1V, though, so that makes our plausible output power be 4.5-1.0= 3.5V peak or 2.47Vrms, and the power 0.76W. So 1/2W is being conservative, but reasonable. Then on top of that you have the impedance variation of speakers, so being conservative is the proper approach.

If I like the sound of the ruby amp could I make a "bigger" version by adding a power transformer to step up the voltage (After the output capacitor, for instance)?

Yes, but it's going to be difficult to do this in a practical manner. An amp is properly thought of as a DC power supply that happens to have a tiny amount of other junk on it (the whole electronics!) that lets some of the power out in a carefully controlled manner. The DC power supply IS the amp for all questions involving power output.

Your question is a bit confusing. You could make a bigger version by using a power transformer, rectifiers, and filter capacitors to make a bigger voltage and current capable power supply , then choosing an amplifier (whether chip or discrete) which will work on that level of voltage and power.  If by "after the output capacitor" you meant adding a transformer on the output of the existing amp, no, that won't help because the amplifier can't drive the transformed load any more than it can get out of the existing power supply.

[Aurin]

Wow! Thank you for such a detailed explanation - that's quite helpful ... I asked mostly because I want to run the circuit at 12 V (Fet+IC are rated for it so it should be OK) and I wanted to figure out what the wattage would be when it's run @ 12V .... so... let's see...  12V halved = 6V - 1V (Our swing loss) = 5V / 1.414 = 3.53 VacRMS ... Wattage then would be (3.53 ** 2 ) / (8 Ohms) = 1.5 Watts (Rounded down to nearest half watt) . That sound about right?

I figured the power transformer question would be strange -- Basically I wanted to know if I could be lazy by taking the ruby circuit's output and running it through a transformer to increase the voltage by however much .... No worries on that, I won't play with things I don't understand until I understand them better =D.