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if your PWM frequency is within your DMM's measurement bandwitdh you can use the DMM in current measurement mode. Keep in mind that a PWM signal consists of of several sinusoidial waves by means of the fundamental frequency/signal and its harmonics. Have a look at fourier series of a square wave (PWM with 50 %) respectively a PWM signal.
So you have to ensure that the fundamental signal and a couple of harmonics (e.g. up to the 5th) is within your DMM measurement bandwidth. If your DMM is a TRMS multimeter you can easily calculate the current sourced by your MCU or your PWM driver.
My small fluke114 true RMS meter does "true RMS measurements" in AC mode only ... but it blocks DC.
This means it´s "true RMS without DC"
The datasheet says 45Hz ... 500Hz.
So neiter the HPF of 45Hz is good for you nor the 500Hz upper LPF frequency.
I doubt yo can get good informations from the FLUKE179.
Do a test:
Find a source with 0V / 5V square wave and 50% duty cycle.
* a true RMS inlcuding DC will show: 5V x 0.707 = 3.535V
* a true RMS excluding DC will show: 5V/2 = 2.5V
* an average measurement will show 2.5V (sadly the same value as above, it will differ with duty cycle)
Unless you have a multimeter with a BW of > 100kHz, you simply can't make the measurements you desire - at some point to do real power electronics - you need a 'scope ... and shunts, and CT's and 100x probes ....
why aren't you using your scope in combination with a low valued resistor to determine the current sourced by your PWM source, as suggested in post #10 ?
By investigating a waveform with a scope you should use a low Volt/Devision setting, which gives more insight by means of filling the screen in y-direction as much as possible (e.g. 500 mv/Div instead of 5 V/Div). Further, according to your recorded waveform it looks like your PWM never reaches the off-state (0 V). Also your PWM freuency is seems to be pretty low, is it your actual implemented freuquency?
decrease your V/Div as well t/Div setting, so your voltage resolution as well as your time resolution is increased (showing e.g. 2 or 3 periods). Nevertheless, I assume you are measuring the voltage at the output of your LED driver, which does not give any feedback about your sourced current.
Use a low valued shunt resistor and measure the corresponding voltage drop, which gives you the current by applying ohm's law.
You ask for current, show a scope picture with voltage and in the text you talk about power.
All three are different electrical measures. One measure can not replace the other ... (without circuit details)
I guess there is a misunderstanding of basics.....at least it is very confusing.