Whats wrong with large filter caps for Gainclone?

[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




10000uF, 0.5 Ohm ESR




Pedja

[kropf]

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

[janneman]

Originally posted by DrewP


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

[millwood]

Originally posted by kropf
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.

[millwood]

Originally posted by janneman
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.

[Pedja]

Hi Jeremy,

Yes, you are probably right, the truth is somewhere in between. I tried to determine the boundaries of the problem. Actually, I just have prepared two graphs with 0.1 Ohms cap's ESR...

Pedja

[Pedja]

The waveforms if 100mA, 350mA and 1A is drawn by the speaker. Cap’s ESR is 0.1 Ohm.


1000uF




10000uF




Pedja

[apassgear]

Originally posted by DrewP



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.

[apassgear]

well adding to the previous post the voltage will be depleated more rapidly on the smaller cap, but that does not make much difference recharging the cap, but will sure make diference on music as you listen to it.

[janneman]

Originally posted by IanHarvey
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

Originally posted by IanHarvey


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