Correct implementation of R-C-R-C filtering in PSU

[Gopher]

Of the many examples of R-C-R-C filtered PSUs I've looked at, there are 2 variants:

(1) Bridge - C1 - R - C2 - R - C3

(2) Bridge - R - C1 - R - C2

Q1: Is there any advantage in introducing an R between the bridge and the first C, as in (2), or is it better to have a C directly after the bridge, as in (1), noting that (1) does have a higher total capacitance than (2)?

Q2: What are the suggestions for the optimum scaling of C1 and C2/C3? I've seen some examples of (1) where the value of C1 is half that of C2/C3 to reduce charging currents, but others where C1 is much higher than C2/C3 to reduce ripple. Does that logic still apply in (2) with a resistor before C1?

Q3: In most examples of the genre, the Rs are only found in the positive and negative supply lines. However, some examples have matching Rs before and between the capacitors in both the supply lines and the 0V lines. I always understood the 0V lines should be kept as low an impedance as possible, so why are the Rs there in those few cases?

I should add that I'm using a transformer with two separate secondary windings, not centre-tapped.

 

[Lo_Tse]

Of the many examples of R-C-R-C filtered PSUs I've looked at, there are 2 variants:

(1) Bridge - C1 - R - C2 - R - C3

(2) Bridge - R - C1 - R - C2

Q1: Is there any advantage in introducing an R between the bridge and the first C, as in (2), or is it better to have a C directly after the bridge, as in (1), noting that (1) does have a higher total capacitance than (2)?

Q2: What are the suggestions for the optimum scaling of C1 and C2/C3? I've seen some examples of (1) where the value of C1 is half that of C2/C3 to reduce charging currents, but does that logic still apply in (2) with a resistor before C1?

Q3: In most examples of the genre, the Rs are only found in the positive and negative supply lines. However, some examples have matching Rs before and between the capacitors in both the supply lines and the 0V lines. I always understood the 0V lines should be kept as low an impedance as possible, so why are the Rs there in those few cases?

I should add that I'm using a transformer with two separate secondary windings, not centre-tapped.

The case (1) that you show is a very common CRCRC (pi) filter and is well understood.

Case (2) looks curious - putting a resistor before C1 will drop the maximum voltage that C1 can be charged upto (how much depending on the value of the R), why? One can always choose a different transformer if the secondary voltage is too high? Wait a minute, is the R before C1 in series or in parallel?

Regards,

Lo_Tse

 

[Kindhornman]

Gopher,
I am wondering in the second example if that resistance is not just the secondary winding resistance that is being shown? Could this just be an example of someone trying to show all the parasitic resistance values in the circuit? Are they showing the resistance values or is it just a generic schematic?

Steven

 

[Gopher]

Steven, Lo-Tse

To answer your two responses together, the R between the bridge and first capacitor is a real resistor and I think the main argument for including it is to reduce charging currents into C1 for less radiated RFI.

 

[ashok]

Secondary resistance can be significant especially HV coils. Even in low voltage coils they might be significant like say 0.2 ohms or more which you see in series in some psu circuits.

 

[bobodioulasso]

PSUD2

 

[wintermute]

Hi Gopher, I used R's in the 0V lines on my regulated supply. I had not seen it done before, but it simmed well, and it made a difference when built as well. It pretty much halved the ripple magnitude for a given load.

I can't remember if it made a difference to the final voltage or not. Possibly the same ripple reduction would be obtained by only using resistors (of double the size) in the + - lines.

OK Just simmed that, overall ripple magnitude is identical with either 3r3 in both + - and returns, or double that at 6r6 in just the +- The actual simmed result is identical picture attached.

points monitored are after CRCRC wich is 4700u - 3r3 - 4700u - 3r3 - 1000u

one has 3r3 in both positive and zero return and the other has 6r6 in only the +ve

load was ~250mA after the reg circuit. measurements taken on input to reg circuit. Input voltage approx 18V

So if you are concerned about resistance in the return lines, the exact same ripple reduction should be possible by doubling the resistance in the +ve and -ve lines and leaving the return lines resistance free

Whether there are any differences that will show up in real life as opposed to a sim I don't know, as it wasn't something I tested with my scope when I was doing the real world tests.

Tony.

 

[Gopher]

Tony

It was your YARPS PSU I was thinking of actually, but I didn't want to name names.

Something about putting resistance in the 0V return lines just goes against everything I've read and seen previously so guess I'll stick with double the value in the supply lines instead.

Thanks for the input.

 

[wintermute]

Hi Gopher, no probs!! I was being unconventional in most respects with that PSU The idea of putting the resistors in the returns was that it would add further decoupling (just a thought and not based on any solid engineering). I'm not sure that it does and I would like to have tested it as I just simmed above when I was doing all of the testing, unfortunately I just didn't think about it!

My electronics knowledge is enough to be dangerous I wasn't sure whether the transition from simulation to real world was going to work, but it did, so I was happy

There may well be a very good reason not to put the resistors in the return lines. It is certainly something I've not seen in any other PS. Will cop the flack if it was a dumb idea!!

For me it just seemed the logical thing to do, especially since the YARPS uses a positive reg for both the positive and negative rails. It would have been rather more unsymetrical if I'd left out the resistors in the return lines and the point of using the two positive regs was to try and make the -ve and +ve lines as close to the same as possible

Of course whether it makes any difference at all I can't say, but it was an interesting excercise!! Just a shame I still haven't built (except a breadboard prototype) the crossover it is supposed to be powering

Tony.

 

[DF96]

Putting resistors in the 'ground' line could have the advantage of forcing the designer to think about correct PSU grounding, which many people get wrong. If you can pretend that every ground connection is a resistance (which it actually is!) then you can design good grounding. Then you don't need to actually include the ground resistors.

 

[Robert Kesh]

Always wondered, for a fixed total capacitance and resistance, what was the optimal way to distribute it. How many poles and how much in each.

 

[DF96]

I think "how much in each" is easy: divide the total up equally, at least for ripple suppression. A factor of two either way will not make much difference.

"How many poles" is a bit harder. I would guess that you need C and R in each to be big enough to provide some attenuation at the lowest frequency (usually twice line frequency): CR >> 1/( 2pi f ).

I am thinking out loud, so quite possibly talking rubbish! And, of course, ripple is not everything. There might be an advantage in making the first and last caps bigger than the others.

 

[Merlinb]

To answer your two responses together, the R between the bridge and first capacitor is a real resistor and I think the main argument for including it is to reduce charging currents into C1 for less radiated RFI.

This is quite right, and is one main reason why some people claim to like the 'sound' of valve rectifiers (which have high internal resistance). I have plotted the relationship between R and the ripple current here:
The Valve Wizard

Always wondered, for a fixed total capacitance and resistance, what was the optimal way to distribute it. How many poles and how much in each.

This interesting question was answered in:
Cathode Ray (1949), Smoothing Circuits: 1, Resistance Capacitance. Wireless World, October, pp389-93

See page 326 here:
(link to pirate site removed by moderators)

 

[Robert Kesh]

This interesting question was answered in:
Cathode Ray (1949), Smoothing Circuits: 1, Resistance Capacitance. Wireless World, October, pp389-93

See page 326 here:
(link to pirate site removed by moderators)

Thanks VW!

And I've plenty of bedtime reading now.

 

[KSTR]

While more stages will reduce supply ripple, the impedance at the output will rise significantly and therefore it won't be optimum for power amps -- the amp will pollute rails with its own load current especially when the amp leaves class-A region. Therefore, a two-stage RCRC filtering looks probably best in overall perfomance, with the second RC using a bigger cap (plus bypasses).

 

[DF96]

Page 326 of MJ3 puzzles me. His example of up to four 22uF and 2.5k total ends up opting for three sections; surely four would be better? Have I missed something, or has MJ?

Naively, ignoring intersection interactions and treating reactance as small, three lots of 22uF plus 833 gives 11.5 per section and 1527 in total. Four lots of 22uF plus 625 gives 8.6 per section and 5571 in total.

MJ says three 22uF plus 833 gives 997. This is taken from the table, yet his example does not match the table as he has slightly more capacitance.

Is the original Scroggie article available? He was always clear in what he said. I fear MJ may have garbled it a bit.

 

[Lo_Tse]

Steven, Lo-Tse

To answer your two responses together, the R between the bridge and first capacitor is a real resistor and I think the main argument for including it is to reduce charging currents into C1 for less radiated RFI.

I find this explanation a little hard to grasp. How much "less radiated RFI" can one expect? How can one correlate this resistance with this reduction in RFI? How much less RFI is acceptable?

Regards,

 

[DF96]

How many poles and how much in each.

Not having access to the original Scroggie article, I have done the maths myself. This is an estimate of the optimum number of poles. It could be off by one in some circumstances, but that probably doesn't matter too much.

Assume you have total R and C which gives you a ratio of 'a' when used as one pole; that is, ripple is reduced by 1/a. 'a' is approximately 2 pi f R C. If you split up the R and C N times you will get a ratio of approximately a/N^2 per pole, so the total will be (a/N^2)^N (assuming no interaction - approximately true). We want to maximise this.

Pretend for a while that N is not constrained to be an integer. Then to find a maximum (or minimum) we want the differential to be zero. In this particular case it is simpler to take logs; because 'ln' (natural logarithm) is a strictly monotonic function then maximising it is the same as maximising its argument.

Y = (a/N^2)^N
ln(Y) = N ( ln(a) - 2 ln(N) )
d/dN (ln(Y)) = ln(a) - 2 ( ln(N) + N/N ) = ln(a) - 2 ( ln(N) + 1 ) - using product rule
We require the differential to be zero so
ln(a) = 2 (ln(N) + 1 )
This gives
N = sqrt(a)/e
The result will need to be rounded to the nearest integer.

Putting this result back into where we started from, we find that the optimum ratio per pole is e^2 which about 7.4. This seems rather low, so it appears that most designers use a non-optimised smoothing scheme with fewer but bigger caps. This degrades ripple reduction, but may improve other things.

We can also invert the result to find what the best values for 'a' are: those which coincide with an integer N without rounding.
N=2 -> a=29.5
N=3 -> a=66.5
N=4 -> a=118.2
These appear to be about 2/3rds of the figures in MJ's table (after Scroggie) so maybe he used different approximations from me. (Or one of us has made a mistake).

Now you know what to say if anyone asks you what is the point of learning calculus!

 

[Merlinb]

I find this explanation a little hard to grasp. How much "less radiated RFI" can one expect? How can one correlate this resistance with this reduction in RFI?

Interesting question. My guess is you will substantially eliminate ringing across the PT when the circuit is critically damped (or more). I think this will happen when the total series resistance is equal to the reactance of the leakage inductance, at the resonant frequency (which is in turn dominated by the leakage inductance and reservoir capacitance)... Someone correct me.

 

[DF96]

Not reservoir capacitance, as that is disconnected by the rectifier. Stray capacitance, plus whatever snubber cap is used. In some cases it will not be necessary to eliminate ringing, but merely drop its frequency enough so that the wiring cannot radiate it particularly well. Given that the transformer is not intended to be good at high frequencies, the ringing circuit will already be quite lossy.