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1970s Design Indulgence

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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 16 Jan 2022 at 7:17pm
Shall we add another reservoir capacitor in parallel to a bridge rectified power supply?

Shall we assume it has no ripple current, and only the first one has?

It would be magic if that was the case.

Perhaps we could isolate it with a switch? Then no charging current will flow and no energy would end up being stored in it. It would be bloody useless wouldn't it?

Suppose we operated the switch at say 1kHz?

Would it not be a case that the capacitor received intermittent (modulated) charge?

OK, we might put a resistor in series with the capacitor. Huh, don't have one suitable, so lets use a loudspeaker voice coil, say an 8 ohms one.

We'd better make the turnover frequency as low as it can go, say 10Hz? 2,000uF would do, so we can go for 2,200uF as it's the nearest available value.

So, what about the ripple appearing on the capacitor at 100Hz?

Modulated by the 1kHz switching frequency, it will still charge when the switch is on.

So, does the output capacitor have supply ripple current on it?
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Fatmangolf Quote  Post ReplyReply Direct Link To This Post Posted: 16 Jan 2022 at 7:56pm
I was puzzled by your previous post, on first read I thought are you sure? It is interesting how the power supply capacitor is both smoothing by turning ripple into heat, and providing a reservoir to release pulses of energy to met transient demand from the circuit at the same time. Long established and is it clever, stupid, or surprising this setup works... and more!

Even if you think you are over analysing Graham, this is really interesting. Thank you.

Jon

Open mind and ears whilst owning GSP Genera, Accession M, Accession MC, Elevator EXP, Solo ULDE, Proprius amps, Cusat50 cables, Lautus digital cable, Spatia cables and links, and a Majestic DAC.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 16 Jan 2022 at 8:49pm
Join the two terminals of a speaker drive unit together. What did you hear?

Answer: nothing.

Instead, place the terminals of a 1.5V battery across the terminals of the loudspeaker, but for only one second. What did you hear?

Answer: a click or pop.

What did you see?

Answer: the cone moved in one direction, and when the battery was disconnected, it returned to its original position.

If you put a switch in series with the battery, and repeat the above process - one second on, one second off, do you think the battery will eventually go flat?

Answer: yes (obviously)
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 16 Jan 2022 at 9:01pm
The lower transistor of this single supply amplifier connects both terminals of the speaker together, but you can hear half the signal frequency.

Must be a bloody miracle?

Not really. The output capacitor is an energy store, so as the inboard end is being pulled to 0V, the other end is negative in relation to 0V.

If it never receives charge, it will eventually be depleted of energy.

What if we take 1 amp from it? Will the quiescent current of 50mA put it back in time for the next half cycle?
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Post Options Post Options   Thanks (1) Thanks(1)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 16 Jan 2022 at 9:59pm
This image shows the rectified AC, and the lines that join the pulses represent the smoothing from the reservoir capacitor. The load uses up some energy, and it is replaced on the next half cycle.

ripple diagram


The formula is Q(charge) = CV = IT

We know T (0.01S for UK mains)
We know the maximum ripple voltage we can tolerate in Volts
We know the current (I) the load requires in Amps 

So the value of capacitance is

C = IT/V

The above cannot be denied (unless you're into DOUBLETHINK).

What it shows is that in this particular time window, the load discharges the capacitor, and it is replaced in the next time window.

So, say the load is 1A. Is the replenishment 1A?

Answer: No, it is approximately 6.2A (due to the timing of the rectifier being asymmetrical)

But if it were linear it would be 1A.

Pour 1 litre out of a jug, you need to put back 1 litre to make it the same as it was.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 16 Jan 2022 at 10:36pm
If V x I = W

but we don't know what V is, then, from Ohms law we can subsitute IR.

So IR x I = I^2 x R

So for 1 amp, 1 squared is 1, times 8 ohms is 8 watts.

So the voltage if I is 1A must be

V = W/I = 8/1 = 8 volts.

8V into 8 ohms is V/R = I = 8/8 = 1

W = rms current times rms volts, so peak volts is the square root of 2 times rms volts, and the same goes for current, such that peak watts is twice rms watts. And peak to peak is twice that, so

to drive the speaker positive requires 1.41428 amps, and to drive it negative requires 1.41428 amps.

The positive 1.41428 amps is from the power supply via the upper transistor.

The negative 1.41428 amps is from the output capacitor via the lower transistor.

The next cycle requires?

2.82856 amps via the upper transistor.

However, the entire cycle averages at 1.41428 amps.

The 2.82856 amps is for half the time window of the signal.

Twice current half the time.

But the rms current is 1A, and is always 1A for the 8W signal cycle, but draws 2A half the time.

If it were half, then it would not be 8 watts.



Edited by Graham Slee - 16 Jan 2022 at 10:37pm
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 16 Jan 2022 at 10:47pm
So if the upper emitter resistor is half the lower emitter resistor, and the current through it is twice the lower emitter resistor

then 2 x 0.5 = 1

and 1 x 1 = 1

So at minimum signal if that can be imagined, the positive and negative signals are equal, even if the upper quiescent voltage is half the lower one, because one is AC and the other DC.
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