Power factor of a SMPS

Thread Starter

florian_1034

Joined Apr 2, 2022
14
Hello everyone,
I have a problem with understanding the power factor of a switch mode power supply. More specifically, a phone charger.

So it all started with me trying to measure the efficiency of my phone charger. I just measured the voltage, then the current after loading the charger at nominal amperage. I have multiplied the voltages and currents (in and out) and calculate the efficiency with Eff=Pout/Pin. The results where 45% efficiency. I start thinking that I'm doing something wrong. I know that by multiplying voltage x current it gave me apparent power (AC side), but in my idea, a phone charger should not consume reactive power and I write later why.

I will try to explain what I know about power factor and I hope that somebody will correct me where I'm wrong.
I know that usually the energy in an electric circuit it flows from the source to the load. Especially in DC circuits, but in some cases this may not be always true. I understand that a capacitor or a coil can absorb energy from the AC line while the sine is ramping up and then pushing that energy back in the lines when the sine is descending (or I guess it can push energy back anytime some kind of component from the circuit can generate a bigger voltage then the source). The same thing apply for the negative part of the sine.
In a pure DC restive circuit, measuring the current between the load and the source can give us a clear indication where the energy is heading, in an AC circuit on the other hand, the current and voltage is continuously changing direction. In order to establish where the energy is heading in an instant moment, we need to take the voltage in consideration so that any multiplication of the current and voltage that gives a negative value, will conclude that the load is pushing energy in to the source.
In my mind is clear that by having a sinusoidal current and voltage that are out of phase, there will be periods where their product will be negative. Integrating that products for a period of time and we get the reactive energy.
My conclusion is that in order to have reactive power, somewhere in time there has to be a negative V*I and the load must have some kind of battery (capacitor, inductor, chemical battery or other things) that can store the energy and released it later in the source.

Taking the clear case of a capacitor connected to AC, it is clear that when the sine goes up the capacitor charges, and when the sine goes down the capacitor release his energy back. A motor is like an inductor so is easy to understand, or an ac transformer.

I do also know that a bad power factor is also considered when a load is not consuming energy in a sinusoidal fashion. A SMPS may consume energy only when the sine reaches 90% of is amplitude, it get charged and after that is consume zero till the next 90% is reached in the negative alternance. That is why you want to use an active PFC, which act like a DC to DC converter to charge the SMPS main cap even at lower voltages.

Now, here is what I do not understand. Given the fact that we have a classic SMPS with a full bridge rectifier on the input that feed the main capacitor. How can the energy flow back to the source if the diode should prevent any reverse flow?

Is either they are leaking current when the SMPS MOSFET is switching the transformer and some high voltage spikes appear or, the diodes is not closing perfectly or fast enough and some energy can escape back.

I have only a one channel oscilloscope home and I can plot only the current of my phone charger. But still I can presume where the voltage is based on the huge spike current gets when the cap begin to charge.
I CAN SEE NO REVERSE FLOW FOR THAT CURRENT.
I found some kind of explanation on the internet about some kind of harmonics but, I can't imagine how harmonics works in this case, need more explanation here.
I attach here some images with what I measured and my presumption where the voltage should be. The yellow plot is the voltage measured across a 10 ohm shunt resistor mounted on a 230V 50Hz supply connected to a phone power adapter of 5V 2amps running at 2 amps. The orange plot is the voltage that I presume it looks like.

So, how can 15% of the energy stored in the phone charger get back in the grid? (assuming a pf of 0.85)
 

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Ian0

Joined Aug 7, 2020
5,556
There are two different versions of Power Factor. The first is the usual "inductive load" power factor where the current waveform is phase-shifted in relation to the voltage waveform.
The second applies to non-sinusoidal current waveforms and the power factor in this case is the ratio of the true power to the product of the rms current and rms voltage.
 

Thread Starter

florian_1034

Joined Apr 2, 2022
14
Thank you guys for the material but still is not answering my question. How can that harmonics bypass the diodes of a SMPS?
Danko link explained pretty well how to calculate harmonics but they don't give any hint of how can they increase the current flow in a conductor. How does they add up to the reactive power?

Can it be that the power lines acts like a reactance and when the current increase and decrease rapidly they spike voltages that carries more current in all direction?
 

Ian0

Joined Aug 7, 2020
5,556
Thank you guys for the material but still is not answering my question. How can that harmonics bypass the diodes of a SMPS?
Danko link explained pretty well how to calculate harmonics but they don't give any hint of how can they increase the current flow in a conductor. How does they add up to the reactive power?

Can it be that the power lines acts like a reactance and when the current increase and decrease rapidly they spike voltages that carries more current in all direction?
There is no reactance or phase shift involved in the second definition of power factor (post #2)

In your sin2.jpg the average current is quite low, but the rms current, because it involves average the square of the current, is much higher.
For instance, if a steady DC current of 1A is replaced by pulse currents lasting 1ms in every 10ms (10%), then the pulses have to be 10A.
The rms value of the steady DC current is still 1A, but the pulse currents have a mean square value of 10% of 10^2, so the rms value is 3.16A.
 

Thread Starter

florian_1034

Joined Apr 2, 2022
14
There is no reactance or phase shift involved in the second definition of power factor (post #2)

In your sin2.jpg the average current is quite low, but the rms current, because it involves average the square of the current, is much higher.
For instance, if a steady DC current of 1A is replaced by pulse currents lasting 1ms in every 10ms (10%), then the pulses have to be 10A.
The rms value of the steady DC current is still 1A, but the pulse currents have a mean square value of 10% of 10^2, so the rms value is 3.16A.
So what you are saying is that the RMS current of a distorted sinusoidal waveform is no longer equal to the equivalent DC current value. Even if my current meter is true RMS.
And by that means that there is no reactive power coming back from the charger. Is just my equipment being deceived by that distortions.
Did I get it right?
 

Ian0

Joined Aug 7, 2020
5,556
So what you are saying is that the RMS current of a distorted sinusoidal waveform is no longer equal to the equivalent DC current value. Even if my current meter is true RMS.
And by that means that there is no reactive power coming back from the charger. Is just my equipment being deceived by that distortions.
Did I get it right?
Yes.
Not only that, the rms value of an UNdistorted sinewave is not equal to the average DC value when it is rectified. One is √2 and the other is 2/π.
You need to be careful with that when charging batteries. It is the average DC current which gives the amount of charge that goes into the battery, but the rms value which gives the amount of heat generated in the charger. A true RMS meter is not always such a great asset if you want to know how much you have charged a battery!
 

Thread Starter

florian_1034

Joined Apr 2, 2022
14
Yes.
Not only that, the rms value of an UNdistorted sinewave is not equal to the average DC value when it is rectified. One is √2 and the other is 2/π.
You need to be careful with that when charging batteries. It is the average DC current which gives the amount of charge that goes into the battery, but the rms value which gives the amount of heat generated in the charger. A true RMS meter is not always such a great asset if you want to know how much you have charged a battery!
Thank you very much for your explanations. I will run some simulations to better convince myself now.
Is there a way for me to calculate the active power for the charger? I have expensive watt-meters at work but I think they are doing the Fourier transform of the signal to get first harmonic and then calculate the current.

Putting a capacitor across the current shunt can get me closer to the real current value?
 

Ian0

Joined Aug 7, 2020
5,556
If you sample voltage and current, multiply V by I for each sample, then average the result, you get true power.
(Or you can do it in analogue with an analogue multiplier and a simple RC filter on the output)
 

Ian0

Joined Aug 7, 2020
5,556
Putting a capacitor across the current shunt can get me closer to the real current value?
You need a series R as well (say about 1k to 10k, if you meter has 10M impedance), because the resistance of the shunt would be so low that you would need an enormous capacitor.
That would give you average current, but it would not give you true power.
 

Thread Starter

florian_1034

Joined Apr 2, 2022
14
You need a series R as well (say about 1k to 10k, if you meter has 10M impedance), because the resistance of the shunt would be so low that you would need an enormous capacitor.
That would give you average current, but it would not give you true power.
Ok, I will try and then verify with the expensive watt meter.
Thank you once again.
 
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