Where does all the energy go using a full wave rectifier?

[Phrak]

Where does all the ripple energy go using a full wave rectifier?

Say I have some typical resistive load and a parallel smoothing capacitor feed by a full wave bridge from a nice 15 amp AC line. I have 169 peak voltage and a ripple voltage of 30 volts across an electrolitic 100 uF capacitor. The bridge consists of four 1N400x. Where does most of the energy go in bringing the capacitor back to near peak voltage?

 

[es1]

I haven't done the math but I would think most of the energy goes into the load because the diodes are in series with the load and Vf of 1N400x << 169Vp.

 

[uart]

Where does all the ripple energy go using a full wave rectifier?

Say I have some typical resistive load and a parallel smoothing capacitor feed by a full wave bridge from a nice 15 amp AC line. I have 169 peak voltage and a ripple voltage of 30 volts across an electrolitic 100 uF capacitor. The bridge consists of four 1N400x. Where does most of the energy go in bringing the capacitor back to near peak voltage?

Energy goes from the supply into the capacitor when it's charging and from the capacitor into the load when it's discharging.

 

[Averagesupernova]

I guess this seems a bit obvious to me so maybe I don't understand the question. Any energy that goes into a capacitor will go back out. A capacitor is a storage device. Of course there is some energy is lost when converted to heat by the losses in the capacitor.
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Edit: Didn't see UARTs post.

 

[Bassalisk]

Can we count in diode voltage drops? I think in full wave rectifiers you have 4 diodes. You do need energy to keep them running. Probably 90-95% energy is harvested on the ends of the capacitor.

 

[Phrak]

Hmm. I must have been obscure or misstated.

Every half cycle, there is a large dV/dt as the capacitor recharges. In the example presented, its about 15 kV/second. This shows up within the diodes and capacitor, traces and supply lines, as a relatively large current. I = C dV/dt.

Does most of the energy loss occur in the component leads, the copper foil, the supply lines, the silicon within the diodes, the aluminum foil within the electrolitic, or elsewhere?

P.S.

Or the diode bonding wires. But I don't think a 1N400X has any bonding wires, does it?

Or in green postmodernist, where is the energy harvested? Ya kill me, Bassalisk.

 

[uart]

The simple answer is that the energy don't have to be lost, as long is dv/dt is finite. This is NOT like the "DC source + switch + capacitor" problem, that perhaps you're comparing it to, where energy loss is unavoidable.

Just like any circuit of course there will be losses, but there is no unavoidable minimum. In other words, my answer is still the previous one I gave in #3 above.

 

[Skaperen]

A simplistic answer is that the energy "goes into the future". Eventually it ends up in the load, plus a very tiny amount in various other resistive aspects and/or magnetic fields.

 

[waht]

Where does most of the energy go in bringing the capacitor back to near peak voltage?

One thing to note is that all of the power that comes from the source is transferred to the load. Some is lost due heat but that's negligible.

Another thing to note is that there is a second time constant associated with a rectifier. It is the time where the actual power is transferred from a source to a load (RC) through the rectifier, it is not the RC time constant. But most often it's called the conduction angle.

This is due to the fact that diodes conduct when they are forward biased. It turns out when the stored voltage (peak) on the capacitor is large enough, for most of the time the voltage from the source will be less, and so, the diodes will be cut-off, or reverse biased. Only at the time when source voltage approaches peak again, will it be greater than voltage on the capacitor, and then conduction will occur.

So when the rectifier is conducting, the current coming from the source has two components, one goes to the load, and one goes to charge the capacitor C dV/dt which could be large but still comes from the source. When the rectifier is not conducting for most of the time, the capacitor takes over and supplies current to the load and discharges at RC.

 

[amaresh92]

but i think there is no energy loss expect heat.and capacitor is used only to just deliver that energy during the period when voltage across it is less than peak.so is there any other losses and whether my view is correct or not?