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# Compensation of the fundamental frequency with a Capacitor in a Bridge rectifier

#### mopeters

##### Newbie level 6
Hi I want to compensate the fundamental frequency so that cos( φ1) = 0.8 in a Bridge rectifier.

I have already calculated it as I think it is correct and need confirmation if my calculations are correct. Thank you.

Your math may be correct as far as I know. It sounds as though you recognize something similar to power factor error, and wish to correct it or reduce it. Power factor error occurs when AC goes through an inductive load. Reactance causes the voltage waveform and Ampere waveform to shift out of alignment with each other. The circuit carries elevated Amperes all around. The conventional solution is to install the (correct value) parallel capacitor.

However the situation changes evidently with your power supply topology. Despite the incoming mains voltage waveform being a sine-wave, the Ampere waveform becomes less sine-like. The AC is switched On suddenly (which produces a different result than typical AC current lagging of sine voltage), and the AC is rectified, turning it into DC which responds in a different manner than reactively. So you may shift the Ampere waveform somewhat by adding the capacitor, although it's questionable how closely the result confirms your calculations.

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The exercise problem can't be calculated without knowing RL time constant and respective current waveform. Input current lag angle varies between 32.5 degree (L=0) and 50.7 degree (RL = 23.4 ms).

The exercise problem can't be calculated without knowing RL time constant and respective current waveform. Input current lag angle varies between 32.5 degree (L=0) and 50.7 degree (RL = 23.4 ms).
This is a normal AC/DC Converter.
L is large enough so that the current on the output side is a constant 40 A (so no ripple current)
Vrms = 230V is the effective value of a sine wave. V,grid(t) = 325*sin(wt)
Is this information sufficient? Normally you only have to calculate the reactive power (fundamental component) and then use this to calculate the capacitor ?
Only the reactive power is to be compensated.
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Your math may be correct as far as I know. It sounds as though you recognize something similar to power factor error, and wish to correct it or reduce it. Power factor error occurs when AC goes through an inductive load. Reactance causes the voltage waveform and Ampere waveform to shift out of alignment with each other. The circuit carries elevated Amperes all around. The conventional solution is to install the (correct value) parallel capacitor.

However the situation changes evidently with your power supply topology. Despite the incoming mains voltage waveform being a sine-wave, the Ampere waveform becomes less sine-like. The AC is switched On suddenly (which produces a different result than typical AC current lagging of sine voltage), and the AC is rectified, turning it into DC which responds in a different manner than reactively. So you may shift the Ampere waveform somewhat by adding the capacitor, although it's questionable how closely the result confirms your calculations.
The grid, i.e. the voltage Vrms,grid is a completely normal sine wave (from a European grid with 50 Hz). V(t)=325*sin(w*t). So Vrms = 230V (Root Mean Square).

So my calculations should be correct?

Below is a simulation which should clear things up. Values were adjusted to adapt to your parameters. (I put in the triac model to simplify operation. Results should be similar.) Notice the Ampere waveform at bottom right. The triac turns on at the proper time so it admits current. It's unhealthy for the triac if we try to pull current through it when it's supposed to be Off. In other words we can't expect to alter power factor.

An unexpected phenomenon occurs if we put a straight wire instead of the triac. The Ampere waveform is a square-ish wave instead of a sine. Another thing, the choke passing DC does not perform reactive behavior. Instead the arrangement of this output stage admits greater Amperes with each cycle. The level grows alarmingly. Only with the shortened duty cycle can we have a workable circuit.

By clicking the link below, you can navigate to Falstad's website, load the above schematic into his animated interactive simulator, and run it on your computer.

tinyurl.com/2bw9lpyy

Adjust values and rearrange components by right-clicking and select Edit.

L is large enough so that the current on the output side is a constant 40 A (so no ripple current)
Should have been mentioned, specifically because the previous post deals with finite rather than zero ripple. Nevertheless zero ripple is a sufficient specification.

Only the reactive power is to be compensated.
Yes and no. The exercise specifies compensation to cosphi 0.8 rather than 1, so real current matters for the calculation.
So my calculations should be correct?
It isn't.

Hi,

cos(phi) vs power factor:
* cos(phi) is phase shift, which means you get reactive power
* with power factor (PF) you don´t necessarily get reactive power.

And if I´m not mistaken, your circuit can not generate significant reactive power. Thus cos(phi) already is 1. But PF is below 1, since you have a non linear load.

This means: on the input side of Brian´s simulation: V(t) x (I(t) never gets negative. (Negative = reactive power; positive = true power)

So you surely can improve PF .. a little by adding a capacitor.
But this causes cos(phi) to drop below 1 ... while PF is shifted towards 1.

Klaus

Hi,

cos(phi) vs power factor:
* cos(phi) is phase shift, which means you get reactive power
* with power factor (PF) you don´t necessarily get reactive power.

And if I´m not mistaken, your circuit can not generate significant reactive power. Thus cos(phi) already is 1. But PF is below 1, since you have a non linear load.

This means: on the input side of Brian´s simulation: V(t) x (I(t) never gets negative. (Negative = reactive power; positive = true power)

So you surely can improve PF .. a little by adding a capacitor.
But this causes cos(phi) to drop below 1 ... while PF is shifted towards 1.

Klaus
My only requirement is that cos(phi) = 0.8
Everthing else is still to extensive for me at the moment.
And I should have achieved this goal with my capacitor after all.

I don't unterstand why my calculations should be wrong if the only goal is cos(phi) = 0.8

Hi,

Again: you calculate reactive power ... which I don´t think exists.

cos(phi) is only for sine shaped current, which you don´t have.

So - in my eyes - it is impossible to talk about cos(phi) at all.

Please do a simulation to verify my/your point.

Klaus

Hi,

Again: you calculate reactive power ... which I don´t think exists.

cos(phi) is only for sine shaped current, which you don´t have.

So - in my eyes - it is impossible to talk about cos(phi) at all.

Please do a simulation to verify my/your point.

Klaus
As a simplification, let's assume we have a sine shaped current. There must be a solution, because it comes from an old school sheet of mine.

Then my calculations should be right?

Hi,

How can one empty a glass of water if there is no glass of water?
How can one compensate for reactive power if there is no reactive power?

I´m not saying you are wrong ... mainly because I have a different view ...

Maybe what you did is what is expected in the sheet. Maybe. I don´t know.
I´m confused, thus I asked you for a simulation ... so we all can see what happens ... and if your solution and math is correct.

I´m not experienced with simulations ... maybe someone else can help.

Klaus

@KlausST Some of your conclusions are wrong, I think. According to Vout,avg spec, controlled rectifier is operating with 90 degree phase angle. Voltage and current are respectively involving both displacement and distortion power factor below unity. The exercise is abstracting from distortion and only calculating displacement power factor.

First step is to determine current waveform, magnitude and displacement angle of fundamental component. With the assumptions in post #4, output current is pure DC and input current bipolar square wave. 0-90° zero, 90-180° positive, 180-270° zero, 270-360° negative.

Hi,

I might be wrong, yes.

But reactive power is when V(t) x I(t) is negative. I.e. capacitive or inductive

In the given circuit .. I guess (!!) that V(t) x I(t) is positive all the time. This means all power is true power.

For sure, due to the phase angle, there is some kind of displacement. Displacement, because a part of the sine shape is missing. But never in a the meaning to create reactive power.

At least this is what my mind-analysis tells me.
To confirm whether I´m right or wrong ... a simulation would be helpful.

Klaus

I guess (!!) that V(t) x I(t) is positive all the time. This means all power is true power.
Product of instantaneous voltage and current is positive, but extracted fundamental current wave is 45° phase shifted sine and thus has cos(phi) of 0.71.

As a simplification, let's assume we have a sine shaped current. There must be a solution, because it comes from an old school sheet of mine.
Playing with my own simulation as I altered values of L & R (output stage) I discovered certain combinations that did result in sine-shaped current. You can observe the same by running simulations. I forget whether power factor changed much, or if the capacitor made much difference.

Playing with my own simulation as I altered values of L & R (output stage) I discovered certain combinations that did result in sine-shaped current. You can observe the same by running simulations. I forget whether power factor changed much, or if the capacitor made much difference.
I still don't know if my calculation is correct and I don't have the same circuit as I do in the simulation. My problem is that I have to dimension the capacitor so that cos (phi1) =0.8.

But I can't set the C anywhere in the simulation.

In your first post, you confirmed that my calculation is correct. Are you still of the same opinion? I am now a little unsure because many posters write otherwise, but without providing a solution or an explanation, unfortunately this is of no use to me.
So my calculation is correct?

The calculation goes like this:
Fundamental wave magnitude of 40 A bipolar square wave with 90 degree conduction angle according to fourier series table
40*4/pi*cos(45 degree) = 36.013 A
Irms = 25.465 A
Real and reactive component for 45 degree phase shift
Ir = Ix = 18.006 A
To achieve cosphi 0.8, Ix has to be reduced to Ir * tan(arccos(0.8)) = 18.006* 0.75 = 13.504 A
Required compensation current Ic = 4.502 A
Xc = 230/4.052 = 51.1 Ohm
C = 62.3 uF

The calculation goes like this:
Fundamental wave magnitude of 40 A bipolar square wave with 90 degree conduction angle according to fourier series table
40*4/pi*cos(45 degree) = 36.013 A
Irms = 25.465 A
Real and reactive component for 45 degree phase shift.
Ir = Ix = 18.006 A
To achieve cosphi 0.8, Ix has to be reduced to Ir * tan(arccos(0.8)) = 18.006* 0.75 = 13.504 A
Required compensation current Ic = 4.502 A
Xc = 230/4.052 = 51.1 Ohm
C = 62.3 uF
Thank you, I have finally understood the fundamental wave and how you arrive at the 25,465A.

But you have calculated:
"Real and reactive component for 45 degree phase shift"
Here we need a 90 degree phase shift, as the control angle is 90 degrees.
I don't quite understand that.

The center of the square wave is shifted 45 degree relative to sine voltage. 90 degree would mean pure reactive current which is obviously not the case.

The center of the square wave is shifted 45 degree relative to sine voltage. 90 degree would mean pure reactive current which is obviously not the case.
The rectangular mains current is shifted by 90 degrees in relation to the mains voltage. Because the thyristor only switches through at 90 degrees and then current flows. So it should be 90 degrees and not 45 degrees?