Full wave rectifier smoothing capacitor - Page 1
Forum section: electronics / beginners

author: raff5184
author_rank: Regular Contributor
timestamp: February 06, 2017, 11:36:22 pm
content: 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.pngThe 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.

author: Vtile
author_rank: Super Contributor
timestamp: February 07, 2017, 12:05:01 am
content: What is the waveform?

author: DG41WV
author_rank: Regular Contributor
timestamp: February 07, 2017, 12:53:42 am
content: 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.

author: danadak
author_rank: Super Contributor
timestamp: February 07, 2017, 12:14:54 pm
content: If the load is high z then cap will charge to a dc level. Ripple occurswhen current is drawn out of the cap. So in this case just parasiticleakage 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.phpRegards, Dana.

author: raff5184
author_rank: Regular Contributor
timestamp: February 07, 2017, 03:56:51 pm
content: Quote from: Vtile on February 07, 2017, 12:05:01 amA simple sinusoid

author: raff5184
author_rank: Regular Contributor
timestamp: February 07, 2017, 04:00:04 pm
content: Quote from: DG41WV on February 07, 2017, 12:53:42 amit goes into a capacitor in order to charge it, so I suppose that the R is almost zero (?)

author: raff5184
author_rank: Regular Contributor
timestamp: February 07, 2017, 04:07:46 pm
content: Quote from: danadak on February 07, 2017, 12:14:54 pmI 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_smoothSo if I got it right, just a "small" C_smooth should be fine. But how small?

author: Vtile
author_rank: Super Contributor
timestamp: February 07, 2017, 04:31:08 pm
content: So you have a load which is not constant, but do vary by time? How fast it the transition effect in that case?

author: Zero999
author_rank: Super Contributor
timestamp: February 07, 2017, 06:40:05 pm
content: Quote from: raff5184 on February 06, 2017, 11:36:22 pmWhat'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.

author: danadak
author_rank: Super Contributor
timestamp: February 07, 2017, 09:02:08 pm
content: If load is a transient then its a bit complicated, the interaction of source transformer/diode bridgeparameters and Load, C.But a crude approximationQ = C x VdQ/dT = Iload = C x dV/dTor C = ( Iload x dT ) / dVSo consider C w/o connection to power, step a current source into it for dT, andwith a spec in mind for dV calculate C. You should wind up with a C bigger thanexact 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.pdfEven more exact would be a spice sim taking into account transformer and bridge.Regards, Dana.

author: raff5184
author_rank: Regular Contributor
timestamp: February 07, 2017, 09:27:44 pm
content: Quote from: Hero999 on February 07, 2017, 06:40:05 pmThank 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)

author: Zero999
author_rank: Super Contributor
timestamp: February 08, 2017, 01:20:49 pm
content: 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.

author: Vtile
author_rank: Super Contributor
timestamp: February 08, 2017, 02:05:51 pm
content: 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 transientEDIT: 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.

author: Vtile
author_rank: Super Contributor
timestamp: February 08, 2017, 04:10:32 pm
content: Quote from: raff5184 on February 07, 2017, 04:00:04 pmI 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).

author: Zero999
author_rank: Super Contributor
timestamp: February 08, 2017, 07:21:46 pm
content: 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.