IIc+ Power Supply mod.
Jorge Chamorro Bieling
The input has a full wave rectifier that feeds a single 200v 100microF
electrolitic capacitor.
Do you think that replacing the full wave rectifier for a half wave
rectifier will do it ?
I would like to think that yes, but I really don't think so.
In fact I think that the only difference that this will make is that the
capacitor will take twice as much cycles to charge, but the voltage
across it will end up being the same as with the full wave rectifier and
the whole thing will explode or whatever...
In some switchers there are two capacitors, and the rectifier can be set
with a jumper to full wave bridge (220v) or full wave doubler (120v).
This power supply is not of this kind...
The main problem I see here is that the voltage I need across this
capacitor is half what I get from the mains plug. (well, that is quite
obvious)
Any (clever) ideas please ?
--
Jorge Chamorro Bieling
Luke Forat
your speculations are correct. Modding from 220V input to 110V is
possible with a wave doubler configuration (as you have noted, most non
"autoregulating" PSUs have that switch which does exactly this) but the
contrary is a major problem and yes you risk a cap explosion and a junction
breakdown - some nasty fireworks and smoke. The best and quickest thing you
can do is get a 220 to 110V autotransformer with at least a 1 Amp rating on
the input - or - replace the PSU with a 220V one.
Regards,
Luke
--
"Jorge Chamorro Bieling" <bie...@terra.es> ha scritto nel messaggio
news:1ha9anv.1anq4tc183e9sN%bie...@terra.es...
Jorge Chamorro Bieling
> Jorge,
>
> your speculations are correct. Modding from 220V input to 110V is
> possible with a wave doubler configuration (as you have noted, most non
> "autoregulating" PSUs have that switch which does exactly this) but the
> contrary is a major problem and yes you risk a cap explosion and a junction
> breakdown - some nasty fireworks and smoke. The best and quickest thing you
> can do is get a 220 to 110V autotransformer with at least a 1 Amp rating on
> the input - or - replace the PSU with a 220V one.
>
> Regards,
>
> Luke
Luke,
I think I'll try this BRUTE FORCE idea I've had:
What if I insert a 50w 110v zener in series with the bridge rectifier ?
The power supply is rated 36w, but that's maximum power.
This switcher may have 80-90% efficiency ?
Then the zener will have to dissipate just a bit more than the IIc+...
How much may be the average power consumption of the IIc+ ?
The how big is the radiator that I need ?
What am I missing here ?
:-)
Thanks.
--
Jorge Chamorro Bieling
Michael J. Mahon
Although that would work in principle, it would, as you note,
*double* the amount of heat dissipated in the IIc+, with most
of it concentrated in the location of the zener.
Further, one of the failure modes of an overheated zener is
to short, which would take out a lot of the power supply.
The transformer is the best idea, and is fairly small, inexpensive,
and *reliable*.
-michael
Music synthesis for 8-bit Apple II's!
Home page: http://members.aol.com/MJMahon/
"The wastebasket is our most important design
tool--and it is seriously underused."
Luke Forat
I don't think that's a good idea for several reasons - I'll try to
elucidate: Zeners need to be driven for reverse avalanche, you can't stick a
zener just "in series" with the bridge and by this I assume you meant on
it's Output (DC side) - it needs to be in parallel and "driven" by a
resistor in series. The output of a 240V full wave rectifier bridge across a
cap is not really 240V it is 240V * sqrt of 2 (340V) which is why the caps
are rated 350-400V on the output (on open load they will store up to the
peak). You would have to drop approx 200V across the resistor which "drives"
the zener which with a primary current of say .3 Amps being generously low
on the VA rating of the small PSU would mean about 60Watts of dissipated
heat (this output would then go to the main switching transistor circuit so
you would need to break the track(s) and rewire). You would need a large
enough Zener (say 5-10Watts as the current to keep a zener driven is in the
mA range) to keep the switched primary voltage "fluctuations" in tolerance
and avoid zener breakdown due to high voltage reverse peaks from the
switching primary (the primary transistor is protected for this, your zener
isn't and there is really no safe way you can with this wiring). In the
event of zener short the drive resistor would burn out. If your primary
current exceeds calculations your primary voltage will drop below 110V and
your switching circuit will "die" and/or eventually your resistor will burn
out as it is dissipating too much heat (W=I2R). You will need to naturally
change the ripple filter cap and place a 350V rated one.
So, if you manage with all the above and you turn on your PSU, you will
have a little heater in there that is just dissipating heat and when you
reach the max load on it, it will dissipate even more heat equivalent to a
100Watt light bulb which you don't see (drop resistor dissipation + primary
transistor dissipation + switching transformer dissipation).
Will it work? If wired and rated correctly it can. Is it a sensible way
to go? Not really because you risk a fire if not worse as the case is not
designed to dissipate that much heat - it will melt and/or the components
inside will die of overheating. So you would need to modify the case and
insert forced ventilation... but that is absurd on a "portable" switching
powersupply that lies on the floor and can be "chocked" so easily. It is one
of the main reasons why modern "dongle" switching laptop PSUs do not have
any external openings and must solely rely on internal dissipation - either
that or they can't be certified for that use. In other words it can be done
but it is not safe.
Trust me, get a 220 to 110V autotransformer, it is so much safer (and
it makes Tesla happy).
Luke
--
"Jorge Chamorro Bieling" <bie...@terra.es> wrote in message
news:1haau6w.1v5iyy1stdq4gN%bie...@terra.es...
Jorge Chamorro Bieling
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 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.
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...
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...
If I ever manage to get the hands on this zeners (the local supplier
said wow, 50W zeners ???) I'll let you know...
:-)
Regards,
--
Jorge Chamorro Bieling
bie...@terra.es
Yes, I know the transformer is the way to go, but I cannot resist the
idea of "tweaking" inside this power supply to make it go without using
an "ordinary" transformer ... :-)
How much heat does a 140*88/1000==12W "heater" generate ?
Do you think that this will require a massive heat sink ?
(please see my other post in this thread, I believe this dissipated
power figure is correct)
Thanks, Jorge.
Michael J. Mahon
Several square inches, with most of it in free air!
You will definitely cook the case in the area of the heat sink.
Think of putting a 50W light bulb in there...
Michael J. Mahon
> 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.