Hello everybody,
so I was thinking about PWM and something doesn't add up. Here's what I mean:
Imagine an ideal 100kHz PWM signal with a 50% duty cycle, which is 0V when off and, say, 10V when on. If we feed a purely resistive load with that signal, the average power dissipated in that load will be, naturally, 50% of the power the same load would consume from a 10V DC source without the PWM. The average voltage will be 50% of the on voltage, and the RMS voltage (which i understand as the equivelent DC voltage to produce the given average power) will be 1/sqrt(2) of the on voltage, or 7.07V. So far so good.
What bugs me though is what would happen if we apply an ideal brick-wall LC filter at a frequency lower than the PWM frequency, say, 50kHz. An ideal LC filter should put out a true DC value, right? If we look at that PWM in frequency domain, it decomposes to a DC component (in our case 5V) a fundamental at 100kHz and odd harmonics at 300kHz, 500kHz.. and so on, each containing less power than the previous. Question is, after the ideal LC low-pass, the average voltage now IS the RMS voltage, because it is true DC and 1/2 the voltage generates 1/4 the power. If an ideal LC filter does not dissipate the power of the harmonics, but rather stores it to smooth out the voltage, where does that other 1/4 of the power (which is carried by the AC components) go to? If the PWM controller really puts out 1/4 and not 1/2 of the possible power, now that its output has been integrated by the LC filter, does this mean it will just consume less power on the input side?
Sorry if I'm overthinking this, I'm pretty sure I'm missing something really obvious here, but I'd really appreciate any thoughts you might have on this.
Have a nice weekend!
so I was thinking about PWM and something doesn't add up. Here's what I mean:
Imagine an ideal 100kHz PWM signal with a 50% duty cycle, which is 0V when off and, say, 10V when on. If we feed a purely resistive load with that signal, the average power dissipated in that load will be, naturally, 50% of the power the same load would consume from a 10V DC source without the PWM. The average voltage will be 50% of the on voltage, and the RMS voltage (which i understand as the equivelent DC voltage to produce the given average power) will be 1/sqrt(2) of the on voltage, or 7.07V. So far so good.
What bugs me though is what would happen if we apply an ideal brick-wall LC filter at a frequency lower than the PWM frequency, say, 50kHz. An ideal LC filter should put out a true DC value, right? If we look at that PWM in frequency domain, it decomposes to a DC component (in our case 5V) a fundamental at 100kHz and odd harmonics at 300kHz, 500kHz.. and so on, each containing less power than the previous. Question is, after the ideal LC low-pass, the average voltage now IS the RMS voltage, because it is true DC and 1/2 the voltage generates 1/4 the power. If an ideal LC filter does not dissipate the power of the harmonics, but rather stores it to smooth out the voltage, where does that other 1/4 of the power (which is carried by the AC components) go to? If the PWM controller really puts out 1/4 and not 1/2 of the possible power, now that its output has been integrated by the LC filter, does this mean it will just consume less power on the input side?
Sorry if I'm overthinking this, I'm pretty sure I'm missing something really obvious here, but I'd really appreciate any thoughts you might have on this.
Have a nice weekend!
Your PWM output is a 5V DC level, plus a 5V square wave. The filter stops the AC power from getting to the load, so the load only sees the 5V DC - so power is halved. Whether this AC is still generated by the source and just wasted as heat somewhere, or not taken from the DC supply when the filter is present will depend on the details of the amp design and the filter design.
Thank you for the reply. Just to clarify.. power is halved with regard to the unfiltered 50% PWM, so it is actually 1/4 of the non-modulated power right? As to the real-world design, i was thinking about a microcontroller (e.g. ATtiny) turning a P-channel MOSFET on and off (high side switching). So the high current path goes from the power supply, through the P-MOSFET, through an inductor into the load, with a capacitor between the load side of the inductor and ground.
Cheers!
Cheers!
Switching a 10V supply on and off via a switch (which you now describe) is not the same as switching an output from 10V to 0V via two switches (which your post 1 specified). In high efficiency design details matter.
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