# How do you calculate the power factor of a power supply?

Is it possible to measure or calculate the power factor of a power supply without using expensive tools? If so how?

• – Adam Lawrence Nov 14 '14 at 18:07
• I'm not sure if this is what you're after, but my cheap mains power meter shows PF to 2 d.p. (not sure about it's accuracy) – tgun926 Nov 14 '14 at 18:55
• What kind of meter are you using? – Pete Nov 15 '14 at 19:46
• cf. my related question: "How do watt meters measure the power factor?" – Geremia Apr 14 '15 at 16:48

If you have a wattmeter you can take the ratio of real power to apparent power (VA). You might be able to use a current probe, voltage probe and multiply function on an oscilloscope to get true power.

A much cheaper option (accuracy might be open to question) would be to use one of the ubiquitous "Kill a Watt" consumer products (as, I think, @ tgun926 suggested), which actually have a PF indication. The specifications are somewhere between sketchy and non-existent. "0.2% Accuracy" (of what?), but the price is right at less than \$20 street price.

• Awesome I have one of those didn't realize it did power factor. – Pete Nov 15 '14 at 19:47
• How do these watt meters measure real and apparent power, though? (cf. my question here). – Geremia Apr 14 '15 at 16:45
• The Kill A Watt measures current and voltage samples and can calculate the in-phase product and product of magnitudes in firmware. Electromechanical wattmeters measure real power by using two coils (one for volts & one for amperes) so that the meter deflection is proportional to real power. – Spehro Pefhany Apr 14 '15 at 17:48

Power factor is defined as the ratio of real power to "apparent" power, where real power is the average of the result of multiplying the instantaneous voltage by the instantaneous current, and "apparent" power is the result of multiplying the RMS voltage by the RMS current.

With a purely resistive load, these two methods yield the same number, so the power factor is unity. With reactive loads, such as lightly-loaded motors, the current is out of phase with the voltage, causing the real power to drop relative to the apparent power and the ratio to be less than unity.

With nonlinear loads, such as power supplies, the current waveform is distorted, giving it a higher RMS value than that of a sinewave. This also causes the ratio to be reduced from unity.

So, to answer your question, one way to estimate the power factor of a power supply is to use a "true RMS" meter to measure its input current, and compare that result with the value that you get with a resistive load drawing the same amount of real power.

It's not so easy, actually.

In general, switching power supplies suffer from non-unity power factor due to harmonic distortion of the input AC current with respect to the input AC voltage. It's not a pure phase-shift situation like a purely inductive or purely capacitive load, although power supplies can and do have current lead and lag sometimes.

Linears also do "weird" things with the input current due to the action of the rectifiers and due to internal bulk capacitances. Again, not a pure phase-shift sort of situation.

Quoting Xitron's article on power factor:

There are actually many definitions for this. Generally, PF = W/VA is the one used, but PF = VA*cos(phase) is common, as is PF = 1/sqrt(1+THD 2 ). Aren’t these the same? Not in the real world!

The cos(phase) equation only yields the “correct” result when no distortion or other frequencies are present. The THD related equation only yields the “correct” result when there is no phase shift present. These are both simplifications of the W/VA definition. Did we know that? A lot of people don’ t. A common customer service call at Xitron is from a person with a rectification circuit. The question is that since the current is exactly in phase with the voltage (“I looked at it with a ‘scope and the current is right at the peak voltage so there’s no phase shift”) so the Power Factor “must” be 1! True, if you use the cos(phase) simplified definition, but not if you use the full W/VA definition.

PF should have polarity right? Correct, but not the one that many use. Many use the polarity of PF to indicate leading or lagging, this is not correct. The polarity of PF is the same as that of Watts (VA is a scalar, so has no polarity ). It is the polarity of VAR that follows the phase polarity. Why is this error so commonplace? Well, in the “old” days, instruments couldn’t display words like “lead” or “lag”, and they didn’t display VAR either. So, they indicated leading or lagging by means of a + or – symbol preceding the PF result (negative Watts are rather rare). This simple move to enable the user to see a leading or lagging indication has led to this popular misconception about the polarity of PF.

So, "good" power meters compute the power factor indirectly via THD, which is computed by performing FFT on the current waveform and doing some weighted computations on the harmonic content. If "expensive" means the price of an IEC61000-3-2 compliant power analyzer, or a fancy current probe + scope with FFT and a power analyzer module, I will say that a non-expensive tool solution isn't possible.

If you want to visualize the current and just get a feel for how bad the power factor may be, a wide bandwidth AC current probe hooked up to a scope can be quite revealing.

• I've got a Fluke 115 True RMS Multimeter. Expensive for me is anything more than a couple hundred bucks. – Pete Nov 14 '14 at 18:15
• You're out of luck then. – Adam Lawrence Nov 14 '14 at 18:16
• Thanks, random downvoter. Care to explain your -1? – Adam Lawrence Sep 30 '18 at 14:57
• @Adam is that $PF = \frac {1}{\sqrt {1 + THD^2}}$? If so you can use the MathJax $PF = \frac {1}{\sqrt {1 + THD^2}}$. "Watts" should be 'watts' when spelled out. (Their fault - not yours.) – Transistor Sep 30 '18 at 15:11
• Yeah just lazy today – Adam Lawrence Sep 30 '18 at 16:07

Add a capacitor in parallel and measure the currents coming from the supply. Remove it and measure again. Create the equivalent circuit equations and solve them.

• Is this how the Kill-a-Watt meter, for example, does it? – Geremia Apr 14 '15 at 16:47
• I don't think that it measures it this way. This would involve turning the power off, adding a cap, and turning it back on to measure again. – WalyKu Apr 15 '15 at 7:22