Whats wrong with large filter caps for Gainclone?

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So far, nobody seems to have mentioned the following:

The gainclone and most other amps have a split power supply. Every cycle of the mains the power supply capacitors are charged: current flows into the + terminal of the +ve supply rail capacitor, out of its - terminal, into the + terminal of the -ve supply cap and finally out of its - terminal back to the transformer.

This is a nasty, spiky current pulse whose magnitude is directly proportional to the capacitance involved. If there are problems in the layout or grounding, this current will find its way into the output signal - via finite lead resistances, or inductive coupling. If it does, smaller PSU capacitances are one way of solving the problem...

There is another, slightly more subtle, issue to check here. When the PSU capacitances are mismatched, there is a balance of charging current which flows from the junction of the two capacitors back to the centre tap of the transformer. This current is injected directly into the grounding system; its magnitude is proportional to the absolute difference in capacitances, which in turn increases directly as the capacitors get bigger.

Looking at some of the layouts posted here (with PSU caps coupled directly to signal grounds), I wonder if this is what's happening in some cases...

Cheers
IH
 
IanHarvey said:
This is a nasty, spiky current pulse whose magnitude is directly proportional to the capacitance involved. If there are problems in the layout or grounding, this current will find its way into the output signal - via finite lead resistances, or inductive coupling. If it does, smaller PSU capacitances are one way of solving the problem...
The currents aren't that nasty if you have reasonable large caps.

The main obviuos disadvantage ís lower output power and I suspect higher distortion due you the high ripple voltage.

Yes, I agree with Jocko here :yes:

But the best you can do is to test! If you like it, it's good....for you.
 
Every cycle of the mains the power supply capacitors are charged: current flows into the + terminal of the +ve supply rail capacitor, out of its - terminal, into the + terminal of the -ve supply cap and finally out of its - terminal back to the transformer.

Umm, DC current flow through a capacitor? As far as I'm aware, caps don't pass DC, that's what they do, not pass DC.

I think the whole point here is to have a small reservoir that can be very quickly topped up by the transformer as opposed to a large reservoir that can be emptied to a greater extent. It seems that dropping headroom on HUGE signals due to a smaller reservoir is preferable to having a big slow supply of DC to the amp.

That being said, i'll wait till my own version with 1000u Nichicon Muse per rail at the outboard PSU and 100u Black Gate local at the chip is done before I nail my colours to the flagpole.

drew
 
IanHarvey said:
The gainclone and most other amps have a split power supply. Every cycle of the mains the power supply capacitors are charged: current flows into the + terminal of the +ve supply rail capacitor, out of its - terminal, into the + terminal of the -ve supply cap and finally out of its - terminal back to the transformer.IH


the same analysis applies to full wave rectification as well.

IanHarvey said:
This is a nasty, spiky current pulse whose magnitude is directly proportional to the capacitance involved. IH


I think there are two things to be concerned: one is the spiky current going through the diode (and slightly smaller ripple current going in the cap), and another is the fluctuation in PS voltage.

Smaller caps cause larger voltage fluctuations; and larger caps cause larger spiky current.

It is probably hard to argue which one is more harmful.
 
peranders said:

The currents aren't that nasty if you have reasonable large caps.

These are the charging currents (through transformer and rectifier) I'm talking about.

The time-averaged current in the charging pulses must exactly equal the time-averaged discharge current (i.e. delivered to the load).

The capacitors are charging whenever the diodes are conducting i.e. when the transformer output voltage is greater than the capacitor voltage.

A larger capacitance has a smaller ripple voltage, meaning that the voltage drops by less during the discharge phase. This means that the diodes are conducting for a smaller part of the cycle (it helps to draw a graph here).

In turn, this means that the peak current is higher (the same average current in a shorter time), and so will have greater effect. The shorter duration of the current pulses also implies higher frequencies (and therefore more effective inductive coupling to other parts of the circuit).

Cheers
IH
 
DrewP said:


Umm, DC current flow through a capacitor? As far as I'm aware, caps don't pass DC, that's what they do, not pass DC.

They "don't pass DC" by virtue of increasing the voltage across their terminals at a rate proportional to the current. Think of it as recharging a battery - eventually the voltage increases to the point where the charger will give up.

Does that help?

Cheers
IH
 
Noise generated by the caps, part 2

The FFTs in post #55 are done for the cases of usage of ideal caps, i.e. without the ESR. Raising the capacitance and ESR, one time things are not differ much, but at one point waveform starts to look this way and this is, I suppose, the reason to expect some HF/RF content in it.

An externally hosted image should be here but it was not working when we last tested it.



Spectrums in certain cases look like this.

1000uF, 0.5 Ohm ESR

An externally hosted image should be here but it was not working when we last tested it.



10000uF, 0.5 Ohm ESR

An externally hosted image should be here but it was not working when we last tested it.



1000uF, 1 Ohm ESR

An externally hosted image should be here but it was not working when we last tested it.



And it starts to rise with 10000uF, 1 Ohm ESR

An externally hosted image should be here but it was not working when we last tested it.



It can’t be seen from these graphs, but what is happening all this time at higher frequencies and lower levels? Yeah, there are some differences...

Pedja
 
To make a correct conclusions about the previous graphs, note that they show the situations when 100mA is drawn by the speaker. Now, 350mA is drawn and this matches to 1W at 8 Ohms.


1000uF, 0.5 Ohm ESR

An externally hosted image should be here but it was not working when we last tested it.



10000uF, 0.5 Ohm ESR

An externally hosted image should be here but it was not working when we last tested it.



Pedja
 
capacitor ESR

Thanks for the sims.

I looked some ESR values in the Digikey catalog. For the few Panasonic electrolytic capacitors I checked, typical ESRs for 1000uF and 10000uF were less than 0.1 ohm. From the trend I see in your plots, this will reduce the harmonic content at low frequencies, but I expect that it will shift the harmonic content to higher frequency, where it could be more problematic from a PSRR perspective.

I'd be interested in seeing a log-log scale and higher frequency contributions. I'll be checking this myself, for sure, but it will be a while before I post any plots.

Jeremy
 
DrewP said:


Umm, DC current flow through a capacitor? As far as I'm aware, caps don't pass DC, that's what they do, not pass DC.

I think the whole point here is to have a small reservoir that can be very quickly topped up by the transformer as opposed to a large reservoir that can be emptied to a greater extent. It seems that dropping headroom on HUGE signals due to a smaller reservoir is preferable to having a big slow supply of DC to the amp.

That being said, i'll wait till my own version with 1000u Nichicon Muse per rail at the outboard PSU and 100u Black Gate local at the chip is done before I nail my colours to the flagpole.

drew

Who said it was DC? The whole point of a capacitor being charged, my friend, is that current flows into it. Discharge; current flows out of it. You may call it "topping up" or "emptying", but it is current flowing.

And a small cap can indeed by charged by the current quickly, but equally quickly discharged by the load. If your reasoning was correct, a cap of .1uF as supply reservoir would be ideal. Not so; if you would have looked at the diagrams in this thread you would see that the charging is almost always quicker than the discharge anyway. But possible Nichicon Muse will change the laws of nature?

Jan Didden
 
Re: capacitor ESR

kropf said:
From the trend I see in your plots, this will reduce the harmonic content at low frequencies,
Jeremy


it depends critically what you view is the interference source.

Pedja's sims are on output voltage. so if you are concerned about circuitry picking up rail voltage fluctuations, you want to use large caps.

another source of interference is the rectifiers. there, they emit more and more high-frequency interference as the cap gest bigger and bigger.

I will try to simulate what will happen if one uses snubbing caps on the diodes and see what will happen there.

Personally, I see the current-driven interference as a bigger concern but I don't have concrete evidence of that, yet.
 
janneman said:
And a small cap can indeed by charged by the current quickly, but equally quickly discharged by the load.

absolutely correct.

the charging is almost always quicker than the discharge anyway. Jan Didden

that is probably because the "output impendence" of the diodes + transformer is far smaller than the "input impendence" of the load + cap ESR. so the time constant on the charging side is far smaller than that of the discharging side.
 
DrewP said:



I think the whole point here is to have a small reservoir that can be very quickly topped up by the transformer as opposed to a large reservoir that can be emptied to a greater extent. It seems that dropping headroom on HUGE signals due to a smaller reservoir is preferable to having a big slow supply of DC to the amp.

That being said, i'll wait till my own version with 1000u Nichicon Muse per rail at the outboard PSU and 100u Black Gate local at the chip is done before I nail my colours to the flagpole.


The amount of current drawn from the amp is the same, having small or big caps. So the time for recharge should also be the same for either caps.
 
IanHarvey said:
So far, nobody seems to have mentioned the following:

The gainclone and most other amps have a split power supply. Every cycle of the mains the power supply capacitors are charged: current flows into the + terminal of the +ve supply rail capacitor, out of its - terminal, into the + terminal of the -ve supply cap and finally out of its - terminal back to the transformer.

This is a nasty, spiky current pulse whose magnitude is directly proportional to the capacitance involved. If there are problems in the layout or grounding, this current will find its way into the output signal - via finite lead resistances, or inductive coupling. If it does, smaller PSU capacitances are one way of solving the problem...

There is another, slightly more subtle, issue to check here. When the PSU capacitances are mismatched, there is a balance of charging current which flows from the junction of the two capacitors back to the centre tap of the transformer. This current is injected directly into the grounding system; its magnitude is proportional to the absolute difference in capacitances, which in turn increases directly as the capacitors get bigger.

Looking at some of the layouts posted here (with PSU caps coupled directly to signal grounds), I wonder if this is what's happening in some cases...

Cheers
IH


IanHarvey said:


These are the charging currents (through transformer and rectifier) I'm talking about.

The time-averaged current in the charging pulses must exactly equal the time-averaged discharge current (i.e. delivered to the load).

The capacitors are charging whenever the diodes are conducting i.e. when the transformer output voltage is greater than the capacitor voltage.

A larger capacitance has a smaller ripple voltage, meaning that the voltage drops by less during the discharge phase. This means that the diodes are conducting for a smaller part of the cycle (it helps to draw a graph here).

In turn, this means that the peak current is higher (the same average current in a shorter time), and so will have greater effect. The shorter duration of the current pulses also implies higher frequencies (and therefore more effective inductive coupling to other parts of the circuit).

Cheers
IH

Ian,

I have been trying to say this for ages it seems. Probably I never took the time to explain it well enough. You said it well. Thanks.

Jan Didden
 
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