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power consumed by SMPS

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alphacat

New Member
Hey.
In the last thread, i was suggested to measure the true power consumed by an SMPS in the following way.
Turning the existing SMPS:


Into this:


And measure the current through the sense resistor (R1) and the voltage across C1 or C2 (there shouldn't be a significant difference between their voltages).

My question is, why connecting the sense resistor between C1 & C2, and not after C2?
After all, The current through R1 (at the current configuration), is Iin + I_C2, and I dont care about I_C2, but only about Iin.

I dont care about I_C2 because C2 doesnt exist in the original SMPS, and beacuse the current through C2 creates reactive power, and i care only about true power.

I'd like to receive your opinion on it.

Thank you in advance.
 

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dknguyen

Well-Known Member
Most Helpful Member
THere's no difference in theory, but in practice there is. THe voltage at C1+ is going to have more ripple than the voltage at C2+. So the resistor is probably just there to ignore some ripple, but not all of it. The resistor also helps with filtering (like an RC filter)...just by a very very small amount because the value is so low. Seems more like a miniscule detail of preference to me than something enormous.
 
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alphacat

New Member
Hey,
I picked a low resistor's value since I wanted the resistor to have the smaller effect as possible on the real current that flows in the original circuit.

The quesion is, can the current I_C2 can be ignored?
Iin is about 150mA.
 
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Mr RB

Well-Known Member
I'm sure this was answered in the last thread... Your half phase rect is going to produce a LOT of ripple voltage (10uF is just silly), so make both capacitors large then check the ripple on a scope is less than 1% or so. So now you have 2 DC voltages one on each cap, and a DC current through R1 to measure the current. Once you have measured DC current by R1 and DC voltage at C2 you have the input power that is going in to your buck regulator.
 
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