How to Calculate an AC Watt?

"Rcvr"

** But NOT both at the same time. ** Of RF energy - dickwad.

** Yep - they can.

SO ASK THEM !

...... Phil

"400 mA max." might mean something like 250-300 mA typical;

6V * 0.25 A = 1.5W

6V * 0.30 A = 1.8W

6V * 0.40 A = 2.4W

...so a 1-2 watt output power certainly doesn't sound unreasonable, does it?

Bob M.

It also says the voltage range is 6 to 30 volts. The 400mA may correspond to the 30 volt input, or about 12 watts, which won't be very efficient with only 2 watts coming out. The transistor will ge very hot. With a 6 volt input, the power might be 25 times less, or around

100mW.

-Bill

The minimum voltage and maximum amp specs doesn't correspond to the calculated wattage. Calculating AC power in real world is difficult and you need a specialty instrument to do it.

Vrms * Irms * power factor = power

To account for power factor, you have to instantaneously multiply the voltage and wattage at the same time in every part of the waveform and by integrating the absolute value of the product over a few cycles.

This can be done using a digital computer or a four quadrant analog multiplier and an analog integrator. The electric meter used on your house is a type of electromechanical multiplier (how fast the disk spins) and an integrator (counts how many times the disc has spun)

On 22 Jul 2006 08:48:52 -0700, in message , snipped-for-privacy@gmail.com scribed:

It's not difficult at all. It's very simple, and the only specialty instrument one needs is a bit of common sense. The design specs are for input voltage and current, and output power. Any power that comes in and doesn't go out is used up as heat in the generation of the output power, and counts against efficiency. E.g. 10V at 300mA in is 3W in. Output power at this point might be 1.5W radiated, with 1.5W being used to juice the transmitter/modulator. That's 50% efficiency. Simple.

Power factor has nothing to do with it. It's a DC power supply.

On 21 Jul 2006 13:35:46 -0700, in message , "Bill Bowden" scribed:

Actually, I would guess that the maximum input current corresponds to the minimum input voltage, since the input power is likely to be fairly consistent over the given range of input voltages. The transmitter might use 400mA at 6V (2.4W), or conversely 80mA at 30V (2.4W).

As the text claims, output power will be 1-2W, which leaves the range of input power from 100mW to 12W right out. I'll hazard that the actual range of output power is 1W at 6V (400mA, 2.4W in) and 2W at 30V (120mA, 3.6W in), which is just a guess (leaving wiggle room for more heat at the higher voltage). If one cared to purchase the kit, a mathematical circuit analysis could be undertaken.

X-No-Archive: Yes

I responded to what the OP asked in the title. "how to calculate an AC watt"

Output is output RMS AC voltage multipled by output RMS AC current into whatever load they used for specifying this - preferably some sort of antenna, since this device is supposed to be an FM transmitter. (and I suspect optimistically!) Keep in mind that radiated RF (radio frequency) power will be power dumped into the antenna multiplied by the efficiency of the antenna.

Another thing to keep in mind, in the USA: There is an FCC regulation that limits unlicensed wattage into the AM and FM broadcast bands at 100 mW, and I have heard that this is wattage into a final amplifier stage or into a "power oscillator", with further limits including and maybe not limited to antenna details including and maybe not limited to maximum overall length in longest dimension, maybe even if diagonal. I have also heard (but have yet to verify) an additional requirement lack of additional components to achieve resonant coupling or resonant tuning of an antenna that is non-resonant, including lack of coils in the antenna or in series with the base of the antenna.

I do invite anyone to correct me here with cites!

- Don Klipstein ( snipped-for-privacy@misty.com)

I don't care to wase through the rules but you can find them here:

So, you are saying the conduction angle is less as the supply voltage increases? In other words, the transistor is conducting most of the time at 6 volts, but only a few degrees during each cycle at 30 volts?

Well, you could assume a value for the emitter resistor R1 and a emitter voltage with a 6 volt input, and then try to figure out what happens at 30 volts. If we have say 6 ohms in the emitter with 2 volts on it, the current will be about 333mA continuous. Now you increase the input to 30 volts and the emitter voltage goes to 10 volts, and the peak current goes to 10/6 = 1.3 amps. So the duty cycle can only be 10% or maybe 36 degrees in a cycle to get the average current down to 120mA to meet your spec.

So, the question is, how do you work out that duty cycle?

Any ideas?

-Bill

On 22 Feb 2006 05:05:01 -0800, snipped-for-privacy@aol.com Gave us:

Look up the word VAR in a technical manual somewhere.

VAR: Volt Ampere Reactive

On Mon, 24 Jul 2006 00:57:09 GMT, "Ralph Mowery" Gave us:

Nice link. Thanks.

Why?

Doug

Take an AC volt, an AC amp, blend carefully and... voila ... an AC watt

No. That's a VA. Volt-Ampere aka 'apparent power'. Watts are quite different, as in *true power*. However..... since transformers are rated by VA and not watts, this is the one to use in this case.

Graham

On Mon, 02 Oct 2006 13:49:22 +0100, Eeyore Gave us:

No, asshole! Wrong again. You left out the R. It is Volt-Ampere REACTIVE, and it is quite real.

Also, quite different from the VA rating for a transformer.

On 2 Oct 2006 02:02:16 -0700, "The Dougster" Gave us:

Because that IS an "AC watt", which IS the question asked.