ilimzn,
I`m sorry to say, but I find your delivery in this thread a little bit circumstantial.
I`m sorry to say, but I find your delivery in this thread a little bit circumstantial.
The ripple voltage is caused by the capacitance doing it's work badly due to being insufficient in relation to the load current. The formula above speaks for itself. There´s no need for backbreaking, time-consuming investigations.
circumstantial. Thats when you take thing out of context and treat it as a whole.....Why do you constantly seek conflict....why not raise questions for all to get wiser....Or maybe you already know all..??
I for one would surely like to see how the current paths are in a real amplifier..Where and how does the currents flow. Class a and maybe most interesting Class B..
I for one would surely like to see how the current paths are in a real amplifier..Where and how does the currents flow. Class a and maybe most interesting Class B..
I think he is trying to say that if you have enough uF, you don't have to worry about ripple.
What Ilimzn is talking about, is mitigating the gross mistakes made by typical uninformed
showpiece approaches to amplifier PSUs.
It would pay, as MiiB suggestrs, to listen to experience and results rather than interject from
what is just another "circumstantial" POV, built only on elementary theory.
What Ilimzn is talking about, is mitigating the gross mistakes made by typical uninformed
showpiece approaches to amplifier PSUs.
It would pay, as MiiB suggestrs, to listen to experience and results rather than interject from
what is just another "circumstantial" POV, built only on elementary theory.
Hmmm... I do agree to a point (doing it's work badly due to finite capacitance, to paraphrase your post) but the formula above does not show directly what happens to Toff when the capacitance rises, all else being equal. In theory, with infinite capacitance, the ripple voltage all but dissapears, at the expense of ripple current increase in most simple PSUs, where the transformer is to be considered 'ideal', in effect the problem is shifted to the rectifier and transformer. It is certainly possible to design a PSU around that concept, and in fact it's done (mostly for tubes) when you can get series inductance without having to pay for it with too much series resistance and provoking problems with rectifier recovery. For most regular applications it's difficult to afford 'sufficient' capacitance so a middle way has to be found, and this was the point of most of my posts.
Eh... although I see your argument, there is a flaw in your logic regarding this post. A split supply amp will maintain DC level on it's output powered from discharging caps as long as the caps are sufficiently charged. A single supply amp with a capacitor in series with the load and feedback pick-off point at the load will maintain DC level on it's output as long as the power and output caps are sufficiently charged. Without the rectifier working, under DC conditions the principle is really virtually the same. In both cases the caps are attempting to simulate voltage sources.
However, with the rectifier in the circuit, the situation becomes very different. Only in the case of the cap in series with the output can one state that the load current passes through the cap (and since there is no rectifier to provide independent charging, the amp will again only maintain some semblance of DC as long as the cap is sufficiently charged). For all other cases/caps, some portion of the load current plus some portion of the rectifier current passes through the cap, the sum acting on a complex and non-linear impedance, which certainly does make a whole lot of difference to the cap (how much the influence can be seen on the output of the amp is a whole different story and depends on the amp).
What Mooly is trying to say is, that saying the cap is in series with the load under these circumstances, is not correct and not only from a semantic standpoint, it's simply not correct from the standpoint of electrical network analysis. We are analyzing a complete amp/PSU, in a steady state, not just while some part of the output current waveform is indeed only passing through a cap.
If two components A and B are in parallel and this combo is in series with component C, you cannot even say A is in series with C and B is in series with C, let alone analyze it as such. Simply, (and in parralel with B) is in series with C and there's no way around it. In case A is a capacitor and B is a transformer with rectifier in series, you cannot even disregard the influence the rectifier HAD, even though at the given moment it may be virtually disconnected from the cap. You MIGHT make an educated guess how much this influence would be based on the specs of A and B and knowing that the amplifier uses (G)NFB and is well designed, and say, that the influence is minor (on a log scale, mind you, and in the frequency band of interest), then build it and verify this by measurement - this would be called experience. But, when analyzing a general case, no, you simply cannot ignore a component in the network. This problem, BTW, arises routinely in amplifier simulation - stick a battery as the power source and away we go. Believe me, it pays off to simulate a simple power supply with realistic components (including parasitics), even just as a 'thought experiment'. Sometimes I wish I could see the reactions to the results
Last edited:
Hi,
the signal path doesn't look like a pipeline.
Since the power supply rails in their role as AC grounds are connected together through the power supply capacitor, the power supply capacitor will signal wise be in series with the load regardless of circuit topology. (As smoothing capacitor it is in parallel with the load).
the signal path doesn't look like a pipeline.
Since the power supply rails in their role as AC grounds are connected together through the power supply capacitor, the power supply capacitor will signal wise be in series with the load regardless of circuit topology. (As smoothing capacitor it is in parallel with the load).
A single supply amp with a series connected cap and load can indeed maintain at its output a DC output voltage. What it can not do is maintain a constant DC voltage across the load. They are two very different things.
Believe me, it pays off to simulate a simple power supply with realistic components (including parasitics), even just as a 'thought experiment'. Sometimes I wish I could see the reactions to the results
Yes! I couldn't agree more and to wander off topic a little I wish more would do this, particularly when they ask about grounding and "my amplifier hums/buzzes etc Why ??" . Once you start to consider all conductors and print as resistances and start to think "what if" that or this current changes and "will it affect the output in any way" then all starts to become much clearer. It becomes so much easier to design and build amplifiers that have no channel interaction and have no noise and hum once you start thinking in these terms. Think and analyse the amp at DC. Ask what happens if the ground conductor (and all others too) have significant resistance (put something ridiculous in like 10 ohms) and imagine say 1 amp flows down the conductor and then design and wire such that the volt drops along that conductor have no impact on performance by getting the important reference points and critical nodes correctly connected.
The post I wrote was specifically aimed at the case with the PSU removed and initially charged capacitors, which you used previously as a demo. If the NFB pick-off point is after the output cap (and the rest of the NFB is DC conencted) it will actually attempt to retain DC at the output, as long as the output cap is sufficiently charged. In this case it will behave virtually the same as a dual supply (from caps) amp, at some point the load will discharge the cap below a minimum necessary to maintain proper operation. Granted, the mechanism of maintaining DC may fail in different manners between the dual supply and single supply + output cap case, depending on the actual size of caps and load. Point being, it was not the best case to demonstrate your point (which i got a long time ago
).
If anyone would care to draw a simple amp (feedback connected any way you wish) with AC coupling to the load (i.e. load and cap in series, connected any way round you wish, and then power that circuit from whatever PSU you wish rather than just reservoir cap/s in isolation as in my example (which gives you much more freedom) and show that it can maintain a DC voltage across the load then I would be very interested to see it.
The DC coupled amp as per my 741 example can do this even when powered via only the discharging rail capacitors... just the components as shown.
That shows by experiment all I have been trying to say... that the PSU caps are "out of the equation" and can not be considered in the same way as the series load/cap combination or that the PSU caps are now "just the output cap moved to the other side of the load". It is not
ilimzn,
However, the output voltage is solely determined by the capacitor, whose voltage can only be changed by changing its charge. The bigger the capacitor, the more charge it takes to change its voltage. The discharge through the load occurs with a time constant of RC seconds.
Ipd = T1 / T2 x Idc
where
Ipd = peak diode current (much higher than the load current)
T1 = power supply frequency (the ripple frequency output of a full-wave rectifier is twice as high as the input AC signal frequency)
T2 = diode conduction time
Idc = load current
Of course, the best choice is the time-invariant battery supply.
This is a relevant point. Well, a larger capacitor shortens rectifier conduction period and increases the peak diode current, thereby the ripple voltage.
However, the output voltage is solely determined by the capacitor, whose voltage can only be changed by changing its charge. The bigger the capacitor, the more charge it takes to change its voltage. The discharge through the load occurs with a time constant of RC seconds.
Ipd = T1 / T2 x Idc
where
Ipd = peak diode current (much higher than the load current)
T1 = power supply frequency (the ripple frequency output of a full-wave rectifier is twice as high as the input AC signal frequency)
T2 = diode conduction time
Idc = load current
Of course, the best choice is the time-invariant battery supply.
Member
Joined 2009
Paid Member
Mooly, the picture you need to draw is not related to the dc, just consider the ac flowing through the load. This ac current must flow in a loop. Can you draw the loop without including a capacitor or output transformer in this loop ? There's only one way to do this, that's with a bridge amplifier where both sides of the bridge are SE and fed with CCS. All other topologies you can draw and trace the ac current through the load will include a capacitor or an output transformer. Whether or not the capacitor has an influence on the sound likely depends on implementation details.
OK so far
No. This is ripple current. The change in current due to charging and/or due to discharging.
What a mix up. Half is right (or close enough) but a confusing explanation of the mechanism that is at work.
The load current determines the rate of discharge. The mains voltage period without recharge determines the time that the discharge lasts for.
The ripple voltage on the smoothing caps depends on the load current and on the charging frequency.
You must simplify it in your head before you explain it to others less knowledgeable than you.
Adrew,
I`m not saying that it´s a nice explanation, but I see no reason to revise it. I don`t willingly regard ripple in terms of current. Changes in the input voltage can provide current according to I = V / R and a low-pass filter function. What I meant is that a larger capacitance is preferable, despite the fact that replacing a larger charge loss means higher diode peak current under the forward biased period.
I`m not saying that it´s a nice explanation, but I see no reason to revise it. I don`t willingly regard ripple in terms of current. Changes in the input voltage can provide current according to I = V / R and a low-pass filter function. What I meant is that a larger capacitance is preferable, despite the fact that replacing a larger charge loss means higher diode peak current under the forward biased period.
Last edited:
I can't give Wuyit the benefit of saying approximation after your suggested correction.
He has defined Ipk wrongly.
I think he has given Iavg over the short charging period.
Ipk is even larger.
Either we have a language problem and he can't explain his ideas in english, or he does not understand what he is trying to explain (rather badly).
He has defined Ipk wrongly.
I think he has given Iavg over the short charging period.
Ipk is even larger.
Either we have a language problem and he can't explain his ideas in english, or he does not understand what he is trying to explain (rather badly).
His (corrected) formula would be OK for a rectangular charging pulse, which is a rough approximation to reality. As you say, it actually gives the average over the charging period.
But there is not a larger charge loss with a bigger cap. The same charge loss (due to the load), and hence a smaller voltage loss. The higher peak charging current is due to the shorter charging period, because the higher cap voltage means the rectifier turns on later.WuYit said:
I don't see any problem although you should be aware that there are drawbacks to DC coupled design. The advantage is that you can amplify signals down to zero hz. While this is not necessary or desirable for amplifying audio signals it can be done.
Here's a starting point for you to research it. I'm sure there are many designs for them on the internet and in textbooks.
Direct coupled amplifier - Wikipedia, the free encyclopedia
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.