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Re: IIc+ Power Supply mod.



Jorge Chamorro Bieling wrote:
Luke,

I understand what you say in the previous message. I believe that's the
typical way you'd connect a zener to get a regulated voltage source, but
this is not the case. I'm only interested in the big voltage drop a
zener has to offer, and I think that I only have to take care to stay
within the zener's specs... I'll try to explain why I still think that
this will work.

The "BRUTE FORCE" plan has evolved and now it is to put *2* zeners in
series with the bridge's output, so as to halve the power that each one
dissipates.
The voltage drop across each one is going to be the zener voltage if I
manage to keep the current through them within specs. Hopefully with
some aid from the input capacitor (acting as a low pass filter).

The input filter capacitor is not a low-pass filter on the
input side!  In fact, it is responsible for the line current
being quite pulse-like.  Although the *average* current is
determined by the power being delivered to the load, the *peak*
current may be many times that.  This is easily understood if
you consider that current only flows when the line voltage (minus
whatever zener drop you install) is greater than the voltage
on the input capacitor, which is not much less than the peak
input voltage.

Put another way, a capacitor in *series* with a device behaves
like a high-pass filter from the point of view of current flowing
through the device.

The math I've come to goes likes this:

As you say, the target voltage across the input capacitor is
120v*sqr(2)~==169v. I'm going to be more optimistic about this as the
label on the power supply says 136v Max, so lets say that the target
voltage was 130v*sqr(2)~==183v. This will help to reduce both the
current and the needed voltage drop across the zeners.

When plugged to 220v mains, we've got 220v*sqr(2)~==311v

So we'll need a voltage drop of 311v-183v==128v.

So as for voltage, two zeners of ~70v would suffice. That would give us 311v-70v-70v= 171v across the capacitor's leads.

You calculation is correct.  But also consider that the line
voltage is not constant.  Variations of +/-5% are quite common,
and since your voltage "adjustment" is purely subtractive, this
will translate into larger relative variations at the input filter.

Not a killer, but something to think about.  Put another way, the
line regulation of the supply will be degraded by almost a factor
of two.

Now we need to check the currents.

The specs of the apple IIc+ say it's power consumption is 10w continous
15w peak (maximum) (I suppose that's with the disk drive operating and a
ram card installed ?). So the max *averaged* dc current going out from
the input capacitor is 15w/171v==88mA.

What about the peak currents ?

1.- I'm going to believe that the input capacitor acting as a low pass
filter and the overrated 50w zener will be enough to cope with the
current peaks of the switching transistor.
2.- There's going to be a big peak at power up when the capacitor is
discharged. There's an NTC in every switcher to help to deal with that.
Again, I will hope that this and the overrated 50w zener is enough...

No, the input capacitor *will not* average the zener current--quite the
opposite (see above).  Only an *inductive* input filter has a current
averaging effect.

Looking at the datasheet for a BZY93 75v, **20w** zener, the test zener
current is 0.2A, that makes for 15w of dissipated power (75v*0.2A==15w).
So I guess a 50w zener must be able to withstand at least double that.
Given the above math the average current is going to be 88mA but the
zeners are going to be rated for 400mA...

And that *may* be enough to handle the 120Hz recurrent peak current
pulses, assuming that the diodes are conducting at least 20% of the
cycle when the supply is loaded.

If I ever manage to get the hands on this zeners (the local supplier
said wow, 50W zeners ???) I'll let you know...

Good luck.

-michael

Music synthesis for 8-bit Apple II's!
Home page:  http://members.aol.com/MJMahon/

"The wastebasket is our most important design
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