Capacitance multiplier question

philpoole

Member
2004-10-29 9:55 am
If using a capacitance multiplier in a PSU, like that described in http://sound.westhost.com/project15.htm , the amplifier (or load I suppose, might not be powering an amp) sees a multiplied capacitance.
What does the transformer and rectifier see? Does it see, and hence have to charge, just the physical capacitance present, or the multiplied capacitance? I am suspecting the physical capacitance, but I thought someone might actually understand this better than me, and be able to correct me if I'm wrong.

Cheers,
Phil
 
Hi,

The load appears to see a P.S. with very high capacitance in the
sense of low ripple for the current drawn. However in no way
can the supply provide a short circuit dump of current that is
the same as stored by a real high capacitance supply.

The transformer and rectifier see pretty much the first capacitor,
As the output is smoothed, ripple charge currents are small.

So mainly it recharges the first capacitor per half cycle.

:)/sreten.
 
Hi,
The output to the next stage is kept nearly constant.
The pass transistor changes it's "impedance" to draw, or not, sufficient current from the small smoothing cap that is now being multiplied.
I reckon the ripple on this small smoothing cap is worse than normal.
I see the transformer having to charge this smoothing, but it has to do no more difficult a job than in charging any undersized smoothing cap.

You can see from my unusual use of language that this is all off the top of my head.
 
I remember my Dad always raving about capacitance multipliers, so they've been well used and understood for decades :).
I know there's a paragraph or two in Horowitz and Hill, I'll have to dust it off and read up on it tonight.

Class A current demands aren't constant, but as a variation compared to quiescent current, surely there's less variation compared to class B?
I'm making this up off the top of my head, but I mean the current variation in a class A amp would be Iq+/- Irms, whereas for a class B, Iq would be zero, or close to it. So, as proportion to total current draw, the relative current variation is greater for class b (0-100%). This difference would be diminished if a larger output signal is being used.
I know what I mean, I'm just struggling to explain it!
 
philpoole said:
I remember my Dad always raving about capacitance multipliers, so they've been well used and understood for decades :).
I know there's a paragraph or two in Horowitz and Hill, I'll have to dust it off and read up on it tonight.

Class A current demands aren't constant, but as a variation compared to quiescent current, surely there's less variation compared to class B?
I'm making this up off the top of my head, but I mean the current variation in a class A amp would be Iq+/- Irms, whereas for a class B, Iq would be zero, or close to it. So, as proportion to total current draw, the relative current variation is greater for class b (0-100%). This difference would be diminished if a larger output signal is being used.
I know what I mean, I'm just struggling to explain it!
you know what you mean and I followed it.
The problem is that most do not appreciate that as the limit of ClassA output current is approached that the rail current drops to near zero.
The big advantage of ClassA is the avoidance of the half wave switch on/offs that ClassAB suffers
 

jnb

Member
2006-12-30 11:55 pm
Where I see that this discussion of current in class A amps culminates is that if you can eliminate ripple at quiescent conditions, then you've pretty much got the problem licked.

Another view of the capacitance multiplier is as a voltage regulator. If you look at the reference volage created by the dummy resistive load, it remains fairly constant with all but long term perturbations.
 
richie00boy said:
How can the rail current drop to near zero is there is a significant current in the load (assuming resistive load of course).

phil is on the right track.
Phil's right.
The rail current is Iq+-Iload.
If Iload approaches the ClassA limit i.e.Iload~=Iq (for single ended, or Iload~=2*Iq for push pull), then the limiting condition is rail current alternating between Iq+Iload (~=2Iq) and Iq-Iload (~=zero+voltage amp currents).

This amounts to rail current varying from near zero to twice quiescent current.

There are a few topologies that avoid this variable current but they are rarely adopted. A poster recently tried one for me and reported no audible difference to the normal topology.
 

philpoole

Member
2004-10-29 9:55 am
Hurray! I knew what I meant!
I'm still going to read this up tonight, but in the meantime ask dumb questions...
I see the transformer having to charge this smoothing, but it has to do no more difficult a job than in charging any undersized smoothing cap.
So, there is a benefit to using this topology as opposed to an equivalent physical capacitor? Should the peak charging currents demanded from the transformer be less if using a multiplier for the same effective capacitance?
 
Hi Phil,
I think you are right, the peak charging currents are no worse than for a very heavily loaded but, undersized smoothing capacitor.

If this is correct then the smoothing capacitor is suffering severe ripple current. A very good quality cap may be demanded for this duty. Ordinary smoothing caps are possibly not upto the job.
I wonder if this could be the reason that some builders have reported poor results from a multiplier?
 

philpoole

Member
2004-10-29 9:55 am
when the output AC current becomes significant that the ripple in the rail current becomes worse. To the point that at the highest ClassA output current the rail current can drop to near zero.
Ah, I see. Hopefully, that would be pretty rare, and hopefully resolved by increasing the capacitance (if the transformer is happy with the extra work).
I guess this is all about over engineering the power supply to avoid such situations.