# Power factor of simple transformer + rectifier?

Discussion in 'General Electronics Chat' started by daviddeakin, May 3, 2014.

1. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
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27
I am simulating a simple transformer, bridge rectifier, reservoir capacitor, plus a resistive load, and I want to find the power factor.

Is it a simple matter of measuring the average power in the resistive load, and dividing it by the average input power to the whole circuit?

The reason I ask is that until now I have been measuring the RMS value of the ripple current and multiplying it by the transformer secondary voltage to get the average secondary power (i.e. complex power), but sometimes this gives a figure that is way higher than the average input power to the whole circuit

2. ### MrChips Moderator

Oct 2, 2009
12,849
3,526
You are mixing apples and fruitcake.

Power factor is the cosine of the phase shift of AC power into a reactive load.

The power factor of a simple transformer, bridge rectifier, reservoir capacitor, plus a resistive load should be 1.00.

3. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
207
27
That can't be right, considering the load is part resistor, part reservoir capacitor (i.e. reactive).

4. ### MrChips Moderator

Oct 2, 2009
12,849
3,526
Consider the power factor of a sine-wave AC signal into an RC series circuit.

For a bridge rectifier circuit, R is very small and C is relatively large. The voltage across the capacitor is practically DC. There is very little AC component.

5. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
207
27
The load resistor is in parallel with C, not in series with it. The transformer is producing a sinewave voltage, but the current which flows from the transformer isn't remotely sinewave-like, so the power factor must be less than 1...

Feb 17, 2009
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7. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,018
1,121
So for bridge bridge rectifier + capacitor and load we have.

This situation looks like this

As you can see the source voltage is sill a sine wave but the source current no longer looks like a sine wave. And this is why PF is not equal to 1.

Vin_rms = 7.07V and Iin_rms = 161.3mA
So the APPARENT POWER = S = 7.07V *161.3mA = 1.14VA
And the REAL POWER is transfer by first harmonic RMS source current.

I_V1_rms = 104.21mA * 0.707 = 73.5mA

So the REAL POWER = P = 7.07 * 73.5mA = 0.52W

And finally

PF = 0.52W/1.14VA = 0.45

After some more thoughts about this I think that the we can find REAL POWER in ltspice directly.

All we need is to plot V_In * I_in = V(N002,N003)*-I(V1)
and move the mouse to the label of the trace, hold down the control key and left mouse click.

So the P = 728.47mW

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Last edited: May 3, 2014
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8. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
207
27
Isn't the real power simply the DC voltage * DC load current? In you example that would be about 10V * 0.1A = 1W ?

9. ### MrChips Moderator

Oct 2, 2009
12,849
3,526
Thanks. I was waiting for someone to prove me wrong so that I don't have to look it up.
But aren't we calculating efficiency rather than power factor?

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,018
1,121
If you ignore diode power dissipation then yes.
But for my example P_load = 571mW

But PF is indeed a circuit power efficiency. Because we as a consumers pay only for real power (0.7W) but power station need must deliver to us 1.14AV. Also don't forget that this non-sinusoidal currents (high frequency components) creates additional losses in the power cord. And this is why powerhouse is not happy with this situation. And force as to use a power factor correction circuit in new equipment.

11. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
207
27
But surely your DC output voltage was close to 10V? That means the real power would have to be close to 1W (because the load is 100 ohms)? 571mW sounds too low...

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,018
1,121
No, run the simulation file that I attached in post 7 (LTspice) and you will see that Vout_DC = 7.5V
Also I picked Vin = 10 peak so Vout_max = 10V - 1.4V = 8.6V

13. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
207
27
OK that makes sense. I still don't understand why the total average input power to the transformer primary seems to be less than the complex power measured in the secondary (i.e RMS ripple current * secondary voltage)

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,018
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I don't understand, can you give some example ?

15. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
207
27
OK here's the example I'm looking at:

Input voltage: 240V rms
Secondary voltage (off load): 225V rms
Ripple current: 0.956A rms
Average DC voltage: 276V

Real power: 276$^{2}$ / 633 = 120.3W
Complex power: 225 * 0.956 = 215.1W
Power factor:120.3/215.1 = 0.56 (I think?)
Average source power (from simulation): 131.6W <-- so what happened to the 215.1W I calculated?

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16. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,018
1,121
The apparent power consumed by your circuit is equal to:
S = V_G1_rms * I_G1_rms
The real power (average) is close to DC load power. But in reality this power is larger than 120W because we need to add bridge and transformer losses. And this is why your simulation shows 130W instead of 120W.
The rest of the power is a reactive/distorted power. And this power don't do any useful work.
And apparent power provide insight into the amount of current drawn by circuit. And this help us choose the proper wire/fuse size. Notice how I_G1_peak is larger than I_load.

17. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
207
27
Aha, so my calculations were more or less correct. I didn't appreciate that average power IS the real power. I thought it was apparent power, d'oh!