Half wave rectification circuits

Guys, I'm concerned about two rectification circuits.
Attached you will find a picture of the two circuits.
The first is a classic half-wave rectifier.
The second uses two anti parallel diodes.
My assumption was that each of the antiparallel diodes would act independently as half-wave rectifiers producing on its ones terminals (Va and Vb) a DC voltage that is equal in magnitude to the 1/√2 DC voltage of the the diode in the first circuit.
However, after simulations it turned it's not the case....
It rather seemed that the voltages were 1/2 of the half-rectifier DC voltage.

P.S. I would expect a 1/2 ratio if the diodes were parallel. But why does it happen now???
My intuition says that one of the diodes should be rectifying the first phase (+) and the other the second phase (-). Therefore the wave selects each time the appropriate path/diode depending on the polarity.

I would be glad to hear some ideas on this.

Thank you.

Re: Halfl wave rectification circuits

Va will be charged to the peak AC input voltage. Va will be -Vb. If it isn't then you are doing something wrong. A load resistor across each capacitor would be a good idea. Maybe you should post you simulation results.

Keith

Re: Halfl wave rectification circuits

This is what you should be seeing. Don't forget you will also get a diode volts drop, although with successive cycles that will diminish if you don't have a load.

Keith.

Re: Halfl wave rectification circuits

Thanks for the reply Keith.
I agree with you that Va=-Vb, this what I get also in my simulations.
The problem is: the DC energy rectified should be two times the DC energy
rectified in the simple half wave rectifier, because one diode rectifies the positive portion of
the wave and the antiparallel diode the negative one. However, the simulations give me the same DC energy for both circuits.
Isn't this wrong?

P.S. Have in mind that both circuits are fed with power source delivering the same energy.

Thank you.

I think you are misusing the work "energy". I assume you are really trying to compare a full wave rectifier with a half wave? If so, here is a comparison.

Keith.

Hi Keith,
No I'm not saying my circuit is a full wave rectifier.

I using the term energy because it's more clear to compare values between the two circuits. I think voltage would be more misleading...

I'll put my train of thought in an another way.
A sinusoidal wave has an amount of energy, half of it in the positive phase and half of it in the negative phase.
A half rectifier (circuit # 2 in my previous attachment) ideally "filters" one of the phases/portions.
Now, as a full wave rectifier has twice the amount of output DC energy than the half rectifier (because it exploits both phases)
the same should hold for my first circuit. Of course, we have two output voltages instead of one and they also have opposite polarity.
However, energy-wise we should have the same energy (DC) measured on Va and Vb with that of a full wave rectifier...
Am I missing something?

I think I understand your problem now. However, I still think you are misunderstanding "energy". In your original circuit the half wave rectifier has charged a capacitor. The full wave rectifier has charged two capacitors. So, the "energy" which is 1/CV^2 is double in the case of the full wave rectifier. Two capacitors instead of one. Double the energy. Same voltage

Keith.

Now things might get a little bit more complicated....
As I mentioned before I'm using a power source to make sure both circuits get the same energy.
This means that the voltage across the diodes will not be the same in both circuits.
The reason is that now (#1 circuit) we have two diode+capacitor branches in parallel, dropping the
overall impedance to half...
Having a power source means that the voltage in the antiparallel diodes circuit will drop as well
in order to maintain the same power...
We'll have V/sqrt(2) voltage now and therefore the stored energy in the capacitors will be the same again!!!!

The problem is power, time and energy are related. So if you start to restrict the power then you will also restrict the energy.

What are you actually trying to prove (or disprove)?

Keith.

You are right, from now on power.

The rectifying circuit is part of a detector.
My thought was to use the two antiparallel diodes circuit to increase the detected "power".
With this circuit I would need to have two electrodes measuring Va and Vb, whereas in the simple half wave rectifier I had only one.
Then I will have two signals instead of one. The total measured power should be twice compared to the simple half wave rectifier .... right? (AC source power is the same for both circuits)

P.S. Physical constraints restrict me from utilizing a proper full-wave rectifier.

Again, thanks a lot.

Two rectifiers should give you twice the power compared to one because you are rectifying for both phases which I guess was your original question. I think the difficulty in proving it with a simulation is to make sure that you set up a realistic simulation.

Keith.

Ok, Keith.
That was my implication as well, but the simulation results baffled me.
I'll check my simulation set up again then.

Thanks.

To give meaningful results related to power or energy efficiency, your circuit should have a source with an internal impedance and also a load. Otherwise, I don't see what you want to compare. A voltage source without impedance would deliver infinite power, ignoring all efficiency limitations. And a capacitor doesn't consume power at all.

If you are referring to the circuits I posted, those are for illustration purposes simplified.
As I mentioned above, the source in my simulations is a power source with a source impedance as well.

What I compare is the DC power measured (after you connect a resistor // to capacitor of course) in the two aforementioned circuits.
In the first circuit I have two voltages (Va and Vb=-Va) and in the second one (half rectifier) one voltage V.
So the question was: Should the DC power measured on circuit #1 be twice the DC power measured on circuit #2?

Remember:1) Both circuits are fed with same AC power
and 2) the source impedance is conjugately matched to the load impedance.

Thanks