Full wave rectifier smoothing capacitor

[raff5184]

Hi I am designing a full wave rectifier, a simple 4 diodes, like this: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-3/rectifier-circuits/#03264.png

The problem is that the load is just a capacitor, so I don't have a resistor. The incoming wave is a 700kHz signal. Is there a formula that I can use to find the smoothing capacitor? I found one but it uses the load resistor R.

[Vtile]

What is the waveform?

[DG41WV]

the R is in the formula because the smoothing capacitor depends on the load. so the capacitor value would depend on where the rectified signal is going.

[danadak]

If the load is high z then cap will charge to a dc level. Ripple occurs
when current is drawn out of the cap. So in this case just parasitic
leakage will affect size of C needed, and can be quite low in value.

Its all about the load.     

http://www.radio-electronics.com/info/circuits/diode-rectifier/rectifier-filtering-smoothing-capacitor-circuits.php


Regards, Dana.

[raff5184]

What is the waveform?

A simple sinusoid

[raff5184]

the R is in the formula because the smoothing capacitor depends on the load. so the capacitor value would depend on where the rectified signal is going.

it goes into a capacitor in order to charge it, so I suppose that the R is almost zero (?)

[raff5184]

If the load is high z then cap will charge to a dc level. Ripple occurs
when current is drawn out of the cap. So in this case just parasitic
leakage will affect size of C needed, and can be quite low in value.

Its all about the load.     

http://www.radio-electronics.com/info/circuits/diode-rectifier/rectifier-filtering-smoothing-capacitor-circuits.php


Regards, Dana.

I had seen that site before, but since the DC signal is given in input to a capacitor the I_load is not constant (it's a transient), hence I don't know what value of I_load I should consider to calculate C_smooth


So if I got it right, just a "small" C_smooth should be fine. But how small?

[Vtile]

So you have a load which is not constant, but do vary by time? How fast it the transition effect in that case?

[Zero999]

Hi I am designing a full wave rectifier, a simple 4 diodes, like this: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-3/rectifier-circuits/#03264.png

The problem is that the load is just a capacitor, so I don't have a resistor. The incoming wave is a 700kHz signal. Is there a formula that I can use to find the smoothing capacitor? I found one but it uses the load resistor R.

What's the voltage and current?

And you need fast diodes at that frequency. The ordinary 1N4001 isn't going to cut it.

You could also consider using a transformer with a centre tap to reduce the losses in the diodes.

[danadak]

If load is a transient then its a bit complicated, the interaction of source transformer/diode bridge
parameters and Load, C.

But a crude approximation

Q = C x V
dQ/dT = Iload = C x dV/dT

or C = ( Iload x dT ) / dV

So consider C w/o connection to power, step a current source into it for dT, and
with a spec in mind for dV calculate C. You should wind up with a C bigger than
exact design given we ignored transformer/bridge. But it will do for most applications.

A more exact calculation can be had here - http://www.nhu.edu.tw/~chun/BE-Ch11-Capacitor%20Charging%20&%20Discharging.pdf

Even more exact would be a spice sim taking into account transformer and bridge.


Regards, Dana.

[raff5184]

What's the voltage and current?

And you need fast diodes at that frequency. The ordinary 1N4001 isn't going to cut it.

You could also consider using a transformer with a centre tap to reduce the losses in the diodes.

Thank you. Indeed the problem are not the diodes or the voltage, but the current (that I don't know, don't know how to calculate, is not defined) because my load is not a resistor, but a pure capacitor which is charged by the rectified voltage, which creates a transient before becoming an open circuit (current=0)

[Zero999]

If there is no load, other than the capacitor, then there will be theoretically no ripple, as the capacitor will just charge up. In practise there will be ripple, due to the parasitic capacitance of the diodes. Using fast diodes with a low capacitance will minimise any ripple.

[Vtile]

There is also parasitic inductances and parasitic resistances in wires/lines. (attachment) Propably something you can just forget.

The question is what one will do with a such device? If you do even need to measure it you will have some sort of load (current) in it, a really small but still a load. I think we are just not getting what you are really after.

You can model and calculate it ie. with Laplace methods or use spice modelling softwares (LTspice is free) or..

.. Or are you after the switch on transient

EDIT: PS. If you have some formula for a ripple, but it uses a resistor (which you do not have), just put some huge value for it for calculation, lets say 100 000 000 000 = 100Gigaohms. You get that much leakage current through surface contaminations anyway.

[Vtile]

the R is in the formula because the smoothing capacitor depends on the load. so the capacitor value would depend on where the rectified signal is going.

it goes into a capacitor in order to charge it, so I suppose that the R is almost zero (?)

I manage to skip this question before. 

You got the answer on the PS section of my previous post.

If 100Gohm is too much for ie. calculators field, use as high figure it can handle, but over 100 Megaohms (=100 000 000)

Typical DMM would make 10 000 000 ohms Rload (aprox)..

If there is also a resistor before capacitor like in the attached picture, use some small figure (or go with the capacitor datasheet to look up the equivalent series resistor value the manufacturer gives or doesn't give).

[Zero999]

Attached is a simulation of ripple on a bridge rectifier circuit with no load. Unfortunately I don't have any crappy 1N4001 diodes in my model library, otherwise I'd use them, as the ripple would be worse.

The RMS input voltage is 10V. The ripple is only 200mV peak to peak. This might not seem bad but the power supply rejection of lots of circuits is poor at 700kHz. A much larger capacitor than 1nF should be used. It was just a demonstration that there can be ripple on a rectifier circuit, even with no load.