Measuring input current of SMPS.. What am I doing wrong??? - Page 1
Forum section: electronics / beginners

author: Noxy
author_rank: Newbie
timestamp: March 04, 2020, 09:14:32 pm
content: Just asking this makes me feel like an idiot...Anyway, the other day I was trying to measure the efficency of different small smps at various loads.A DC load was taking care of everything on the output, while on the input a fluke 87V was measuring the mA drawn from the outlet.Problem is: The instant power calculated using the reading from the 87V was almost twice the one I could read from a digital power meter installed in the breaker box!In other words the 87V was trying to make me believe that a 40W smps at 90% load was drawing something like 80W (talk about efficiency...).I'm pretty much sure that both my DMM and the power meter in the breaker box works just fine so... What did I not take into account??

author: james_s
author_rank: Super Contributor
timestamp: March 04, 2020, 09:30:53 pm
content: Power factor. You are measuring VA, not watts.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 04, 2020, 09:45:47 pm
content: That sounds about right. You're measuring amperes without correlating it to volts, which means if you take |Vrms| * |Irms| you get apparent power. If the SMPS doesn't have power factor correction (PFC), it's probably drawing a power factor around 0.5, meaning your real power is about half the apparent power. Which will give an efficiency figure in line with expectations.To measure it properly, you need to correlate volts and amps -- just pick up a Kill-a-Watt or one of the clones, they're usually not too bad.Analytically -- you measure power by taking the instantaneous V and I, multiplying them, then averaging the result:P = avg(V*I)Typical devices, like the Kill-a-Watt, are doing this with a few sensors (typically a voltage divider, and a current shunt resistor), an analog-digital converter, and a microcontroller to do the math and show it on a display. It's pretty simple once you know what's going on.Apparent power is taking the RMS of each, i.e.S = sqrt(avg(V^2)) * sqrt(avg(I^2))RMS is where we take the square of a given variable -- since power through a resistor goes as V^2 / R or I^2*R -- and average that, then take the square root. The result is a DC voltage of equivalent (resistive-heating) value.When we multiply the RMS values, we get apparent power -- that is, how much power would be flowing into the load, if these DC equivalents were going into it. We've lost information, though -- AC can do loopy things that DC cannot, and that is where we have a disconnect between apparent and real power.This... isn't terribly convincing to describe in words, but you can plot two sine waves for example, their product, and the average of that product (its constant (DC) term). If the sine waves have equal frequency and phase, the product will beA B (1 + cos 2ωt) / 2the average of which is simply A B / 2; where A and B are thepeakvalues of the sine waves. (A simple calculation* shows the RMS of a sine wave is sqrt(2) times lower than the peak; two sqrt(2)'s together cancels the one-half, giving us the usual V * I = P.)*Albeit using calculus, although there's probably a good geometric explanation of the RMS operation as well. If nothing else, the RMS between two (perpendicular) points is suspiciously similar to the Pythagorean theorem: z = sqrt( (x^2 + y^2) / 2).If there is a phase shift between them however, the real power will be smaller, indeed the '1' term goes to zero at π/2 radians of phase shift. The real power goes to zero when the sines are out of phase, even though their amplitudes are still very much nonzero!Tim

author: Noxy
author_rank: Newbie
timestamp: March 04, 2020, 11:04:29 pm
content: Thank you for the explanation!