I have found that the maximum unclipped output voltage on my RF 750s amp is 39V/channel for a bridged output of 78V.

It has a minimum impedance of 4 ohms bridged.

Using ohms law and doing the math (V^2/R=W), it capable of putting out 1,521 watts.

Now, since I don't like to play my stereo in my neighbourhood, I haven't measured the amp output with my sub connected.

Obviously current cannot flow without the sub in the circuit, so does that mean that the actual unclipped output of the amp is much lower with the sub connected because of the current flow?

Anyone with experience on this?

 

If you had no load, then R is at least 10,000 Ohms, so your amp put out0.1521 watts during the test.

 

It put out 0 watts with no load.... it put out a theoretical wattage though

but you will never see the no load wattage you derived since once the amp is driving a load that will lower terminal voltage. Power is a function of Voltage Current and any impedance factors (more of an issue with AC but still an issue for audio since the loads are all very inductive).

I had spent an hour writing a nice tech discussion on this and lost it so this is as good as I am going to give you

 

With no load on the amp, the voltage measured at the output is meaningless.
Don't bother using for any calculations.

The ONLY way that test is useful is with a REAL Resistive 4 ohm load RESISTOR. (linear) Of course, your resistor has to be big enough to take rated power.

Even with the sub hooked up, the test won't be real, because of the impedence changing for every frequency.

 

So you're saying you couldn't use and compare the theoretical output of two amps to determine which is more powerful despite their ratings?

Even if I measure the voltage with the sub hooked up at a specific frequency (60Hz for example) at least that gives me a pretty good idea of what output the amp will do at that volume level and frequency before clipping.

 

by testing without load it implies the ideal power the amp is capable of nothing more. I would not call it valid.

Testing with a resister is fine but you WILL burn up the resister unless you know what you are doing.

P=V x I x pf

Your testing ignores the ability of the amp to produce, and sink the heat of, current.

 

You miss to multiply the peak to peak voltage to a constant value 0.707 to get the actual wattage of the amp and using a 4ohm dummy load on a single channel and we used a 100hz frequency generator as your input signal. Obviously, you may also have to take a consideration of your power supply and wiring as well to be close as possible.

 

So today I finally took the time to measure the voltage output of my RF amp while playing a steady tone into my Alpine sub.

It measured 63V bridged output. Using the equation and doing the math, that would be 992.25W? Not bad for a 750W rated amp.

Didn't have the oscilloscope hooked up, so I don't know how clean the signal was though.