Calculating Power Output with PWM and 1ohm Load

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The discussion centers on calculating power output using PWM with a 1-ohm load and a 4V supply. The initial calculation suggests an average power output of 8W at a 50% duty cycle, but an alternative method yields only 4W. It is clarified that averaging both current and voltage requires using RMS values to obtain accurate results. RMS calculations apply broadly, not just to sine waves, allowing for correct power estimations regardless of waveform shape. The participants conclude with a better understanding of the importance of RMS in power calculations.
GlynnHeeswijk
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Hi
I have a question to do with using PWM
If for example I had a 1ohm load and a voltage supply of 4JC-1, I would get a power draw of 16JS-1, now if I had a duty cycle of say 50% I should get an average power output of 8JS-1. However if i thought of it as having an average current of 2CS-1 and a average voltage of 2JC-1 then the power output comes out to be 4JS-1, which is 4 times less and not half. If it should be the half what does the actual voltage and current average out to?

Thank you very much.
 
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(4V)^2/1 Ohm = 16W continuous = 4A * 4V

50% duty cycle

8W = 4A * 4V * 0.5

You can't average both the current and voltage, without taking an RMS average. RMS averages give you the right answer -- can you see why?
 
I can see that it does seem work. However I thought that RMS using route 2 was for sine waves? Am i missing something?

Thank you
 
GlynnHeeswijk said:
I can see that it does seem work. However I thought that RMS using route 2 was for sine waves? Am i missing something?

Thank you

No, it's used quite generally for power calculations. You can calculate the RMS value of an arbitrary function:

http://en.wikipedia.org/wiki/Root_mean_square

.
 
Be sure to keep in mind that .707 * Vpeak is the RMS equivalent for sine waves only. But real actual RMS doesn't care what the waveform is.
 
Thank you both very much for your help. I understand it a lot better now.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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