Hi all.
What is the correct method for calculating the efficiency in a basic bridge rectifier circuit. Basic circuit: AC source -> bridge rectifier -> bulk cap and resistor load. The power in the resistor is easy. The power delivered from the AC supply is where I have a problem. You can't use the RMS voltage X RMS current , due to the very non linear current waveform. Even VICos(theta) does not even come close to the correct answer. I'll attach a LTSpice sim. That gives the correct answer , but without that I'd be stuck :0)
Using the simulation the power in load (R1) is 39.89W , and the power supplied by V1 is 47.9W. That leaves approx 7W dissipated in the bridge and cap ESR etc. This is totally realistic and what I would expect. However , if you measure the source voltage and current (separately) and calculate the power then you get 21.06 x 3.2 = 67.4W , which is obviously wrong. You would need a phase angle of about 45deg to get VI x Cos(angle) to work , and the phase angle is nowhere near that.
Any pointers to where I'm going wrong would be great :0)
Cheers
Rob

 

How do you get the power from V1? I see how to measure the power in the load.

 

Hi Ron,
Two images of V1 power .
E
Duplicated image due to my finger trouble.

Update:
Added keying sequence.

1.Place cursor on V1 symbol
2. Press and hold ALT key, cursor changes to Thermometer symbol
3. Left click mouse to plot power
4. Place cursor on plot window V(n002.... header title
5.Press and hold Control Key
6. Left click mouse

 

To get the RMS (real) power for a non-sinusoidal waveform such as the high peak current spikes generated by a diode-capacitor supply, you need to multiply the incremental voltage times the incremental current at each point in time.
In practice this can be done with an A/D converter that samples both the current and the voltage at the same time, with many samples per cycle.
You then multiply each sample of voltage by the same time-sample of current, sum them together for one period of the waveform, and then take the average of the multiplied sample values.

Ltspice does that for the calculated voltages and current samples in the simulation to get the Power.
You can also do that graphically from the voltage and current waveforms.

 

Hi Eric, How do you get the waveform window? I have never seen that before. Thanks Ron.

 

Hi Ron,
1.Place cursor on V1 symbol
2. Press and hold ALT key, cursor changes to Thermometer symbol
3. Left click mouse to plot power
4. Place cursor on plot window V(n002.... header title
5.Press and hold Control Key
6. Left click mouse

The Bold text section.
E

 

Hi Eric, I have not updated in a while. I get a different window. Ron

 

Hi Ron.
I am using LTSpice XVII(x86)

IIRC it has always worked on previous versions of LTS.
E
BTW: Check that you are pressing Left mouse key, I get your result when I press the Right mouse key.

 

Thanks, now I have my Right and my other Right figured out. lol

I am getting 6.25 watts total in the diodes.
AND
0.49 watts in the ESR in the power source.
0.336 in the ESR of C1.
That is close to the missing 7 watts.

 

Thanks, now I have my Right and my other Right figured out. lol

I am getting 6.25 watts total in the diodes.
AND
0.49 watts in the ESR in the power source.
0.336 in the ESR of C1.
That is close to the missing 7 watts.

Missing 7W was not the issue , it's the 67.4W you get from Vrms x Irms :0)
You can do it as crutschow mentioned ( or via sim) as I have done , but there must be a better / easier way to at least get close to the correct answer.

 

Hi,
This asc circuit shows the individual Wattages
E
40.137W+6.25W+.315W= 46.702 Watts

 

Crutschow has it right, integrate V x I. In the case of your circuit the positive and negative half cycles are symmetrical and I worked the math in 200 us increments for half a cycle and came up with 48.1 W. LTSpice reports 46.7 W. A 3% difference no doubt due to me using such coarse increments. I got 133.9 watts for 3.6 ms and the answer is 133.9W x (3.6ms/10ms) = 48.2 W.

 

So basically , at the end of the day , without running a sim or doing complicated maths, there is no way to reasonably accurately calculate the input power.

 

So basically , at the end of the day , without running a sim or doing complicated maths, there is no way to reasonably accurately calculate the input power.

Yes.
Sometimes there are no simple options.