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I was learning about PWM circuits and something confused me:

If you have 1V going into a 1 ohm load for 1 hour we would get: 1V * 1A * 1 Hour = 1 Watt Hour

If you have 1V going into a 1 ohm load for 30 minutes we would get: 1V * 1A * .5 Hours = .5 Watt Hours

What if we had 1V at 50% duty cycle going into a 1 ohm load for 1 hour?

My intuition tells me it should be equal to running the 1V at 100% duty cycle for .5 hours and at 0% for .5 hours: 1V * 1A * .5 Hours + 0V * 0A * .5 Hours = .5 Watt Hours

However, another way to look at it would be: .5V * .5A * 1 Hour = .25 Watt Hours

Which is correct?

Evan
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    It is the "on" power * duty cycle. So 0.5 Wh. You are trying to calculate the average power, FYI. – user57037 Jul 19 '21 at 06:00
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    Since I currently see no comment in the answers regarding your "another way": You're halving current and voltage separately there ( (1V/2) * (1A/2) ), but you need to halve the power ( (1V*1A)/2 ) -- or the time, as in your intuition. PWM is a way to control the maximum power delivered over time. (You could also halve either current or voltage, but that's not that intuitive any more...) – orithena Jul 20 '21 at 13:48

3 Answers3

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The average voltage indeed is 0.5V in your example, however power is proportional to the square of voltage, so you cannot simply use the average of the voltage and the resistance (energy is the integral of power over the period).

The concept of RMS (root-mean-square) voltage takes care of this situation. The RMS voltage of the 50% PWM'd 1V is 0.707V.

So the 0.5Wh is correct.

The difference becomes ever more evident as the duty cycle drops. For example, a 100V pulse at 1% duty cycle has an average voltage of 1V, however the power into a 1 ohm load is 10kW during the on-time and thus the average power is 100W, not 1W.

Spehro Pefhany
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Your first guess is correct. PWM is Pulse Width so it does not do anything for the amplitude. Meaning it changes the time of ON versus TOTAL period time. So 50% DUTY would mean the full voltage is on 50% and it is of for other 50%. 1V at 50% duty cycle over 1 ohm load for 1 hour would indeed spend 0.5 Watthours of energy.

KASPAROLDENDORFF
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If one were to measure the voltage and current from a constant-voltage supply whose output is turned on and off at a 50% duty cycle, the current would be 0 half the time and the 1 amp half the time, but the voltage would be 1 volt all the time. An average of 0.5 amps at a constant 1 volt would thus be 0.5 watts.

If one were to measure the voltage and current from a constant-current supply whose output was sometimes shorted and sometimes had to flow through a 1 ohm load, then the current would be 1 amp all the time, but the voltage would be zero half the time and 1 volt half the time. A constant 1 amp and an average 0.5 volts would again be 0.5 watts.

The voltage and current experienced by the load in both of these scenarios would be identical to those under the original situation described, and there's no place other than the load for power to be dissipated, so the power dissipated in the load would have to be 0.5 watts.

supercat
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