Re: 5V switching IC

On Mon, 28 Nov 2005 16:55:53 +0000, Mike Young wrote:

> "Don Foreman" <dforeman@xxxxxxxxxxxxxxxxxxxx> wrote in message
> news:974ho1pnforfpol4vkpjs89b0fvc7c9e1r@xxxxxxxxxx
>> On Sat, 26 Nov 2005 11:24:18 GMT, "Roger Hamlett"
>> <rogerspamignored@xxxxxxxxxxxxxxxxxxx> wrote:
>>
>>
>>>I'd suggest going discrete!.
>>>If you look at:
>>>http://www.romanblack.com/smps.htm
>>>
>>>This circuit costs the least of any design that I know of, with reasonable
>>>efficiency, and the parts cost will be less than the IC solution,
>>>especially once you have added the discrete parts to the latter. :-)
>>
>> That is a neat circuit!
>
> Can you help me understand this better? Even with Roman's explanations, I
> don't have enough of a basic grasp to figure out what's doing what when. I
> would like to drop 48VDC unregulated to 12VDC and 5 or 3.3 VDC supplies. It
> seems straightforward enough, but the waveforms in SPICE are not very
> encouraging. The duty cycle on the main chopper can be quite short depending
> on load, and this has strange effects on the oscillator. I can continue
> changing the inductance and capacitance values randomly until it looks good,
> but that would like to understand the relationships better.
>

There is a feedback loop, through Q2's emitter to the base of the PNP pass
transistor (Q1). That is the source of the instability that causes it to
oscillate. When Q1 is on, current is poured into the smoothing cap through
the inductor, causing the output voltage to increase. Once it increases
enough to shut off Q2, Q1 gets shut off too. The current through the
inductor stays on for a bit, and thus the voltage continues to increase as
its magnetic field collapses, but eventually stops, allowing the load to
run on the smoothing cap. Thus, the output voltage eventually falls. When
the output voltage finally goes Vbe below the fixed 5.6V threshold, Q2
starts up again, which again starts up Q1, restarting the cycle.

The thing that makes it oscillate is that the big inductor creates a
delay between turning off Q1, and when the output voltage actually
responds to the cutoff. The voltage continues to go up even after Q1 turns
off, and then droops once the inductor's magnetic field is gone. Same is
true of turn-on; turning it on results in current through the inductor,
which gradually builds up to the point where it overcomes the load and
starts charging the capacitor. So, the frequency will depend on the size
of the inductor (how far the charge lags the voltage changes), the size of
the smoothing cap (how long it takes for charge to change the voltage),
and the current draw of the load (how quickly the inductor will overcome
the load).

The cap C2 gives the turnon-turnoff a tiny bit of speed up, in that when
Q1 turns on, it'll yank Q2 on just a bit more with positive feedback, thus
making it return the favor to Q1. When Q1 turns off, it'll yank Q2 off
just a bit more, speeding up the Q1 turnoff. This sharpens up the edges a
bit, leading to a bit better efficiency.

Sadly, for certain loads, I've found that the circuit will simply fail to
oscillate, reverting to being an inaccurate linear regulator without
current protection or temperature compensation. If you are trying to drop
48V to 3.3V with any kind of current, that is clearly a bad thing.
Dissipation will go from Iload*3.3V/E to Iload*48...

---
Regards,
Bob Monsen

A great discovery solves a great problem, but there is a grain of discovery
in the solution of any problem. Your problem may be modest, but if it
challenges your curiosity and brings into play your inventive faculties,
and if you solve it by your own means, you may experience the tension and
enjoy the triumph of discovery.
- George Polya
.

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