Rectified vs Unrectified AC and power delivered to load question (ANSWERED)

You will need one heck of a capacitor. According to Millman, Vdc = Vm - Idc/(4fC) Where Vdc is the average voltage, Vm is the peak from the rectifier, Idc is the DC load current, f is the power line frequency and C is the filter capacitor. In your case, if 11.4 is the RMS AC voltage, Vm = 16.1 V, Idc = 10 A. If we want Vdc to be about 15 V, C needs to be about 40,000 uF

Keith i am aware of this equation: C = I/(2FV){where C is the capacitance in farads, I is the current in amperes, F is the frequency of the wave and V is the voltage ripple] , which is essentially the same as your equation. I used 2FV though and you used 4FV. Why is that?

Thank you both of you for your help btw.

Because of full-wave rectification.

I am kind of confused now with the coefficient of f. You said that f is the frequency of the power line. If the frequency of the AC waveform is 50 hertz then shouldn't the frequency of the fully rectified waveform be 100 hertz? Why is it 200 as you suggest? Please explain.

First of all, these are approximations. f is set to be the power line frequency and they put in various factors depending on full-wave or halfwave and conduction angle assumptions.

Keith i think you got the equation wrong. Check the link out.

"Vripple = (Iload)/fc
Where: I is the DC load current in amps, ƒ is the frequency of the ripple or twice the input frequency in Hertz, and C is the capacitance in Farads.

The main advantages of a full-wave bridge rectifier is that it has a smaller AC ripple value for a given load and a smaller reservoir or smoothing capacitor than an equivalent half-wave rectifier. Therefore, the fundamental frequency of the ripple voltage is twice that of the AC supply frequency (100Hz) where for the half-wave rectifier it is exactly equal to the supply frequency (50Hz)."

Don't argue with me, argue with Millman! (This is from Electronics) Again, he is not using f as ripple frequency it is the power line frequency. He goes through two pages of derivations to prove it out. One factor of 2 comes from the powerline frequency, the other comes from dividing the ripple by 2 to get the average - as I stated "Vdc is the average voltage.

Sheesh, I couldn't have been more clear.

Thanx for the help Keith

Let's poke the bear. How can a passive device increase "energy" in this scenario? I say at best, an imaginary lossless device.

Simple. Every time the AC voltage is higher than the capacitor voltage, A lot of current flows and dumps a lot of charge into the capacitor. This quickly drives the voltage to the AC peak voltage. When the AC voltage is less than the capacitor voltage, the capacitor loses charge to the load and causes the voltage to "slowly" drop until the next time the AC voltage is greater than the capacitors voltage.

This drop is what the simplified equation calculates.

So you are saying that the ripple voltage increases power? How about "actual" measured power input to "actual" measured power output? Surely you're not proposing that this creates energy, the elusive perpetual motion machine of sorts.

Magician:
Yes, you will get higher average voltage, and consequently, more heat

Won't you also get lower peak the same RMS voltage resulting in exactly the same power transfer to the element, minus the losses in the capacitor from resistance? I'm really having trouble understanding where the "free" energy came from by adding a cap.

There is no power increase. You are simply using the energy storage of the cpacitor and the ability for the transformer/rectifier/diode to provide high peak currents to boost the average DC voltage available.

In the example above, lets say that the diodes conduct 20% of the time, or about 2 ms. The ripple voltage is about 2 V. So the capacitor loses 2 V in the 80% period and must be charged up during the 2ms time period. Since C = q/v, for a 40,000 uF capacitor to increase by 2 V, you need .08 coulombs. SInce an amp is one columb per second, you need to dump .08 coulombs in 2 ms, or 40 A of peak current during the conduction time.

So, no magic involved, assuimg all the components can handle the peak currents, the capacitor is kept charged at a DC voltage higher than the RMS value of the AC voltage.

I was wrong above. There is a power increase, and the power going out is less than the power going in - because of losses. Hoever, when you drive the load with the ac voltage and no rectifier/capacitor, it spends relatively little time at the peak voltage where the current (and instantaneous power!) is highest. By using the energy storage of the capacitor you are stretching out the AC peak voltage to supply a higher than RMS voltage at the load. While the AC voltage going in is sinusoidal, the current definitely is not! It draws a lot of current at the AC peaks, and little to none the rest of the time. This funky current shape is what allows the load to draw more power than during the AC RMS case.

If you don't believe me, let SPICE show you.

You can't really understand what is going on unless you look at the waveforms. You can do this theoretically with some software or on a scope. In practice it is very complex and very non linear. You can work it out by approximation using v=it/c .This is true for small changes in v. Eventually you will reach a stable waveform with increases in volts during charge and decreases in volts during discharge when the driving voltage is above or below the stored voltage, plus the diode drop. There isn't a lot of point in deep analysis as the components especially the capacitance will vary.

I'd like to see a test setup using something like this - LMG500 - Power Analyzer - ZES ZIMMER and some real results posted. Theoretical vs Actual would tell the real story.

You still don't believe me?

By the way, where did this actual vs theoretical come from? I thought this was a theoretical discussion?

I was going to simulate it in ADS, but lost interest.

I read this forum as General Electronics not Theoretical Electronics. In theory, a human can run 300 miles per hour. Step on a hot coal and for an instant in measured time, that foot is moving about 300 mph. Now, is that 300 mph actual or theoretical?

Well, you were the one that said:

I say at best, an imaginary lossless device.

MisterResistor:
I read this forum as General Electronics not Theoretical Electronics. In theory, a human can run 300 miles per hour. Step on a hot coal and for an instant in measured time, that foot is moving about 300 mph. Now, is that 300 mph actual or theoretical?

There really isn't a distinction between theory and practice. There is what we measure and what we predict. Measurements are subject to error and theory is subject to limitation. In engineering we extend and refine our theoretical models as far as we need to. This is where perfect models come in at the simplest level. Like an ideal voltage source or current source or resistor. Then the models get extended to better represent what happens. We can model things very accurately, but most components are not that repeatable, their parameters vary. In that becomes a waste of time to consider models that are more accurate than the components.

At the physics level there are many regions that are not well understood. The electron is a good example. But that takes us into the quantum world where uncertainty is an occupational hazard and things are very strange. We still manage to produce devices that have well enough defined behaviour to make reliable systems.