Cap-coupled bridge rectifier

[Matador]

If I have a circuit like this:

Does anyone know of a quick figure of merit for deciding how much DC current can be pulled from the negative rail, as a function of C1, C2, C3 size?

One way might be to figure out the equivalent impedance of C1 and C2 at mains frequency, to see what the AC voltage drop is across them. If I have a 3-terminal regulator after the bottom bridge, and I need 500mA of DC current, and I can tolerate no more than 5VAC drop across C1 and C2 (before my regulator goes outside of its minimum dropout), then C1 and C2 cannot be more than 10 ohms at 60Hz, which means the minimum size for C1 and C2 would be about 330uF (which is about 8 ohms at 60Hz). Would probably do 680uF to be safe.

(I get this isn't an ideal way to get a bipolar supply, however in my situation I'd like to have a completely floating secondary supply, that can be either positive or negative wrt. the first rectifier, which is why I'm capacitor coupling the second bridge so it's completely floating wrt. the first bridge). I'd also like to stick with a single secondary, since the current requirements of the second supply are only a few hundred mA at low voltages).

 

[Bo Deadly]

This is a job for LTSpice if there ever was one.

Or drag yourself out of the past and just stack two SMPS.

 

[JohnRoberts]

the quick easy way is to melt some solder.

JR

 

[abbey road d enfer]

Does anyone know of a quick figure of merit for deciding how much DC current can be pulled from the negative rail, as a function of C1, C2, C3 size?

No. there are two many parameters involved. particularly what voltage the xfmr delivers, how much it sags on peaks (dependant on DC resistance and leakage inductance), how much ripple is tolerable.

(I get this isn't an ideal way to get a bipolar supply, however in my situation I'd like to have a completely floating secondary supply, that can be either positive or negative wrt. the first rectifier, which is why I'm capacitor coupling the second bridge so it's completely floating wrt. the first bridge). I'd also like to stick with a single secondary, since the current requirements of the second supply are only a few hundred mA at low voltages).

This arrangement is quite common for delivering 48V phantom power from a 24V xfmr. I order to give you an answer you would need to give us more info, particularly what current is supposed to be drawn. "a few hundred mA at low voltages" is not enough info.
As Bo mentioned, it's typically a job for a simulator. But a simulation needs to be verified by experimentation, as JR mentioned.

 

[Matador]

In my case, it is an existing 9VAC secondary, used to create a 3.3V supply via a small LDO regulator. The regulator needs to supply about 200mA to a small microcontroller and a handful of other circuits.

I had PCB's made, and picked 1000uF for the coupling caps. The circuit works as expected, and delivers the current as expected (the regulator doesn't drop out at 200mA load current).

It did make me think however how small I could have made these caps and still had the circuit work properly. In spice, the smaller the cap, the more droop there is on the rectified voltage at a given load current, but that is all I could easily glean. I can always shotgun different values but that's always unsatisfying.

 

[gyraf]

(sorry for the derailing, but for anything micro I'd go with an isolated buck converter)

 

[abbey road d enfer]

I can always shotgun different values but that's always unsatisfying.

Why? Using a .step command, which varies one or several parameters at each new simulation is perfectly legit. It's not shotgun, it's evaluation.

 

[JohnRoberts]

It did make me think however how small I could have made these caps and still had the circuit work properly. In spice, the smaller the cap, the more droop there is on the rectified voltage at a given load current, but that is all I could easily glean. I can always shotgun different values but that's always unsatisfying.

The size of the cap will have a roughly linear relationship with current draw, likewise linear with the charging frequency.

For an extreme example of how small the cap could be made years ago I made a cap doubler (tripler) for a phantom supply inside a product that was already using a DC to DC convertor. Instead of recharging the boost cap at typical mains frequency of 60Hz, I used the 100kHz+ frequency of the switcher. I was able to get adequate current (for 1 mic) using tiny 0.1uF SMD caps in the boost circuit.

JR

 

[Bo Deadly]

In my case, it is an existing 9VAC secondary, used to create a 3.3V supply via a small LDO regulator. The regulator needs to supply about 200mA to a small microcontroller and a handful of other circuits.

That would be totally solved with an $8 DC converter the size of your pinky nail. There are probably 5 different manufacturers that make that part.

 

[MisterCMRR]

I think it's easier to think about power supplies in the time domain, rather than frequency and impedances. To a good first approximation, the behavior of these circuits is governed by the basic relationship CE = It, where C is capacitance in Farads, E is voltage in Volts, I is current in Amps, and t is time in seconds. What you're after is the value of C that allows the capacitors themselves to charge or discharge no more than a small percentage of the peak AC voltage (basically, setting the output ripple voltage). Rearranging the equation, C = It/E or, more properly here, C = I x delta t (one half cycle at mains frequency) / delta E (the peak-to-peak ripple). Of course, diode drops aren't included, but are a minor factor for voltages over about 10 VAC. Given the broad tolerances of electrolytic caps, doing this simple calculation and choosing the standard value on the high side is generally a very good starting place.

 

[Matador]

So delta t would be 8ms for 60Hz mains?

Using that equation, I would need 800uF for every 100mA if I wanted peak-to-peak ripple to be 1V or less?

 

[moamps]

No one has mentioned this so far but I would suggest that capacitors C1 and C2 be designed for large ripple currents, in your case for ripple current at least equal to the rated current of the load.

 

[sahib]

So delta t would be 8ms for 60Hz mains?

Using that equation, I would need 800uF for every 100mA if I wanted peak-to-peak ripple to be 1V or less?

That is perfectly realistic. I think you needed 200mA clean. With charging current you are looking into minimum 300mA. Even as a rule of thumb I would use minimum 2200uF for that.

Moamps has a point but generally a run of the mill 2200uF 16V cap with a diameter of about 16mm will have a ripple current >1A. However, better to check the data sheet of course.

Edit: Below is Suntan specs.

 

[CJ]

do those caps need to be bi-polar?

 

[abbey road d enfer]

do those caps need to be bi-polar?

No. In normal operation they are correctly oriented on the OP's schemo.
However if the upper rail is shorted, the caps receive a variable dose of AC.

 

[merlin]

To quote

If I have a circuit like this:

Does anyone know of a quick figure of merit for deciding how much DC current can be pulled from the negative rail, as a function of C1, C2, C3 size?

To quote myself:
the reactance of C1 and C2 at mains frequency must be small enough that large AC voltages cannot appear across them, since electrolytic capacitors cannot tolerate reverse voltages larger than about one volt. An easy rule of thumb is to make them larger than Vpp/(2 π f Rload), or [Load current]/314.