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1 second UPS

K

Ken Taylor

Jan 1, 1970
0
Chris Jones said:
I think if you open up the PC power supply and find the terminals of the
main input capacitor, these terminals could be brought to a shrouded
(finger proof) insulated connector on the back of the power supply. You
could then build a grounded box containing a bank of capacitors which would
be connected essentially in parallel with the internal capacitor of the PC
power supply. The larger capacitor could keep the supply running for a
second or more. In order to avoid very large surge currents when the PC is
first turned on, a high-power series resistor should be inserted so that
the external capacitor charges gently. A diode across the charging
resistor would allow the fairly large current to flow out of the capacitor
when powering the PC.

5Amp 400V
--------|<|-------
| |
.-----/\/\/------.
| 10kOhm |
| .-----------
| + | + |
----- ------------- \
----- ------------- /
| PC cap | external \ 100kOhm
| | cap / Bleed resistor
| | |
-----------------.-----------

There are numerous safety considerations when building something like this,
and if you don't understand the risks involved then it is better to learn
about the dangers before considering building it.

Chris

The 'main input capacitor' will be at several hundred volts (depending on
OP's location). This is therefore a very BAD idea, given the OP has little
idea of what's going on in these supplies. This is without even considering
how the input side of the power supply is going to react to the addition of
several extra amp's of surge current.

Ken
 
C

Chris Jones

Jan 1, 1970
0
Ken said:
The 'main input capacitor' will be at several hundred volts (depending on
OP's location). This is therefore a very BAD idea, given the OP has little
idea of what's going on in these supplies. This is without even
considering how the input side of the power supply is going to react to
the addition of several extra amp's of surge current.

Ken

Hence:
a) my warning not to build it until and only when the dangers are fully
appreciated since it is a potentially dangerous circuit, and

b) the series resistor which prevents any significant surge current above
what the power supply would normally produce.

Chris
 
R

Rich Grise

Jan 1, 1970
0
like

The 'main input capacitor' will be at several hundred volts (depending
on OP's location). This is therefore a very BAD idea, given the OP has
little idea of what's going on in these supplies. This is without even
considering how the input side of the power supply is going to react to
the addition of several extra amp's of surge current.

Ken

For the amount of money it would cost to get a qualified tech to do
a job like this safely (i.e., hack a Chinese PC power supply), you'd
be better off to just get a good UPS.

Good Luck!
Rich
 
P

petrus bitbyter

Jan 1, 1970
0
Chris Jones said:
No, the energy is 0.5 times the capacitance times the voltage squared.
or C*V*V/2

It is practical to power a computer for a second from a large capacitor,
but
that capacitor would be perhaps somewhat larger than a normal PC power
supply.

Chris

Aw, need some refresh too. Forgot to square the voltage. Nevertheless the
680uF capacitor cannot store 125Ws. Not at 150V that is.

petrus bitbyter
 
P

petrus bitbyter

Jan 1, 1970
0
petrus bitbyter said:
stefanv said:
Petrus,

To my knowledge a 250watts power rating are watts per hr. To make sure I
checked last months electric bill :) I used less than 400 kw/hr for the
month. This is 400,000 Watt/hr. At your claim for a computer supply at
250
watts /sec, 1 computer alone would run that in 1600 seconds or 27 minutes
per month. I think you are mixing up W/hr with w/sec. J are w/sec indeed.
I'm sure I won't get a long time, but 1/2 second is more than likely if I
get the rest of the circuit to run :)
A computer uses "only" 0.07w/s

StefanV

Stefan,

This is exactly why I advised to refresh your basics. Maybe the confusion
comes from the word "power". AFAIK it's used meaning both "force" and
"energy" which are two different things. If you have 110V on your outlet
and a fuse of lets say 16A then you have 110*16=1760W available. But
that's not what you pay for. (Unless you have a minimum contribution.) You
have to pay for the part of that 1760W * the time you use it. That's what
you find on your bill.

400kWh = 400,000Wh = 1440,000,000 Ws. Which directly shows why they use
kWh on the bills. (If they really put kw/hr on the bill they do definitly
wrong. Maybe you can even sue them for it. I would not be very astonished
if they do, as bills are often made by bookkeepers that may have some
knowledge of money but no knowledge at all of the products they sell.)

So if you need 250W for half a second, you need 125Ws. (What's that in
kWh? :) That's not much energy compared to the amount on your bill. Even a
small battery can produce it - but not within half a second. To store it
in a capacitor on the other hand requires quite a huge capacitor.
Calculate for your own, using the CV/2 rule.

To summerize the lecture :)

Energy = Force * Time -> [J] = [W] * (or [Ws] but not [W/s] !)

petrus bitbyter


Error: CV/2 should be CV^2/2

petrus bitbyter
 
M

Michael A. Terrell

Jan 1, 1970
0
Chris said:
I think if you open up the PC power supply and find the terminals of the
main input capacitor, these terminals could be brought to a shrouded
(finger proof) insulated connector on the back of the power supply. You
could then build a grounded box containing a bank of capacitors which would
be connected essentially in parallel with the internal capacitor of the PC
power supply. The larger capacitor could keep the supply running for a
second or more. In order to avoid very large surge currents when the PC is
first turned on, a high-power series resistor should be inserted so that
the external capacitor charges gently. A diode across the charging
resistor would allow the fairly large current to flow out of the capacitor
when powering the PC.

5Amp 400V
--------|<|-------
| |
.-----/\/\/------.
| 10kOhm |
| .-----------
| + | + |
----- ------------- \
----- ------------- /
| PC cap | external \ 100kOhm
| | cap / Bleed resistor
| | |
-----------------.-----------

There are numerous safety considerations when building something like this,
and if you don't understand the risks involved then it is better to learn
about the dangers before considering building it.

Chris


A dangerous suggestion, and useless against multiple short outages.
The power distribution system here attempts to reroute itself when there
is a failure, but sometimes it can't. in the process the power will go
off then back on every couple seconds till the controller decides that
there is an overload, or shorted line.
 
C

Chris Jones

Jan 1, 1970
0
Michael said:
A dangerous suggestion, and useless against multiple short outages.
The power distribution system here attempts to reroute itself when there
is a failure, but sometimes it can't. in the process the power will go
off then back on every couple seconds till the controller decides that
there is an overload, or shorted line.

Wiring inside switched-mode power supplies is certainly dangerous when
worked on by an incompetent, but the capacitors are not significantly more
dangerous than any other power wiring connected to the AC mains, except
that one must remember to wait before opening the cover. Either source of
power can be lethal and has about the same voltage. Rather than
discouraging someone from taking up an interest in electronics I would
encourage them to learn why something is dangerous so that they can become
able to work safely with that technology. If that doesn't suit you, then
pay someone else to do it all for you and sell it to you in a nice box.

If you read the OP's post, it sounds like the system he proposes is not
inherently safer than what I suggested, and I believe that my suggestion is
less complicated. If you believe that less high voltage wiring is safer
than more high voltage wiring then it is possible that my suggestion is
safer than the OP's suggestion. Either suggestion could be made as safe as
the original PC power supply by putting it in a grounded metal box, and
following the relevant standards for the construction of electrical
equipment. If there are particular things that you think need to be
pointed out from those standards, then I think that would be one
constructive way of helping to ensure that any equipment constructed as a
result of this thread would be safe.

On the other point, it sounds like the OP has an inverter that could power
his PC during a long train of repeated power outages, but that the delay to
switch over during the first outage is too long for the PC's power supply.
I think that provided the inverter starts up on the first outage and does
not switch back until the normal power has been restored and stable for a
certain period of time, the PC power supply may indeed benefit from a
larger capacitor.

Chris
 
S

Spajky

Jan 1, 1970
0
Chris Jones wrote:

§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§


doing stuff on output side of PSU voltages !!!
================================ mine idea: !


to computer-> 20pin conn.
|
| 2x 10A shottky_diode, few Ohm/few_watt Res.
|-------I>]-------------\/\/\/\---|
| |
->from--|--------[<I---------------------|
| PSU |
| | +
| -------------
| + -------------
----- | external cap
----- |
| (inside PSU cap |
| at output ) |
| |
-----------------.-----------------------

Caps inside ordinary PSU are around 2700µF @ output voltages & @ idle
keep PC on around max 1/4 of second before reseting or shuting it off
(in my case!). So for external caps you need 10x greater ones to be
sure for a second or two on +3,3V ; +5V ; +12V & +5Vsb outputs = 4x
same circuitry !

PSU-Out R (ohms) C
+12V around 6,8 22.000µF/16V or more
+5V around 1 22.000µF/10V or more
+3,3V around 1 22.000µF/6,3V or more

+5Vsb around 10 2.200µF/6,3V - this should
not be too much or too less (is to try!)
since when MoBo looses voltage on that last voltage, it switches PSU
OFF (even if on other caps is still enough power!).This last also does
not require so strong schottkey´s ..

Drawback: if this pack is added to a PSU, automatic shorting
protection of PSU IMHO is not working properly & if making a short,
there could something go wrong ...

this idea popped in my mind today canibalizing elements from a bunch
of dead PSU to throw away them later ... :) ; maybe will try it
myself tomorrow ..
 
S

Spajky

Jan 1, 1970
0
doing stuff on output side of PSU voltages !!!
to computer-> 20pin conn.
|
| 2x 10A shottky_diode, few Ohm/few_watt Res.
|-------I>]-------------\/\/\/\---|
| |
->from--|--------[<I---------------------|
| PSU |
| | +
| -------------
| + -------------
----- | external cap
----- |
| (inside PSU cap |
| at output ) |
| |
-----------------.-----------------------
PSU-Out R (ohms) C
+12V around 6,8 22.000µF/16V or more
+5V around 1 22.000µF/10V or more
+3,3V around 1 22.000µF/6,3V or more

+5Vsb around 10 2.200µF/6,3V - this should
not be too much or too less (is to try!)
this idea popped in my mind today canibalizing elements from a bunch
of dead PSU to throw away them later ... :) ; maybe will try it
myself tomorrow ..

I tried it & I must admitt, that is to toss it away ...
Well, it (kind a) works (helps a bit), but caps should be much much
greater! (much more than 10x bigger for a second!).
+5Vsb for 1 sec needs 10.000/6,3V (for 1A keeping it), so you could
imagine how HUGE should than other lines caps be. HUGE !!!
.... & that is expensive; much cheaper to buy an UPS .. :) Bye ..
 
C

Chris Jones

Jan 1, 1970
0
Spajky said:
Chris Jones wrote:

§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§
§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§
§§§§§§§§


doing stuff on output side of PSU voltages !!!
================================ mine idea: !


to computer-> 20pin conn.
|
| 2x 10A shottky_diode, few Ohm/few_watt Res.
|-------I>]-------------\/\/\/\---|
| |
->from--|--------[<I---------------------|
| PSU |
| | +
| -------------
| + -------------
----- | external cap
----- |
| (inside PSU cap |
| at output ) |
| |
-----------------.-----------------------

Caps inside ordinary PSU are around 2700µF @ output voltages & @ idle
keep PC on around max 1/4 of second before reseting or shuting it off
(in my case!). So for external caps you need 10x greater ones to be
sure for a second or two on +3,3V ; +5V ; +12V & +5Vsb outputs = 4x
same circuitry !

PSU-Out R (ohms) C
+12V around 6,8 22.000µF/16V or more
+5V around 1 22.000µF/10V or more
+3,3V around 1 22.000µF/6,3V or more

+5Vsb around 10 2.200µF/6,3V - this should
not be too much or too less (is to try!)
since when MoBo looses voltage on that last voltage, it switches PSU
OFF (even if on other caps is still enough power!).This last also does
not require so strong schottkey´s ..

Drawback: if this pack is added to a PSU, automatic shorting
protection of PSU IMHO is not working properly & if making a short,
there could something go wrong ...

this idea popped in my mind today canibalizing elements from a bunch
of dead PSU to throw away them later ... :) ; maybe will try it
myself tomorrow ..

Other drawbacks:
1): The PC power supplies that I have traced the schematics of had a "power
good" detector made from a comparator which would inform the motherboard as
soon as the +5V or +12V line went out of specification. If the voltage
dropped by the forward voltage of a Schottky (sp?) diode, then this would
trip the circuit and reset the CPU. You would be better off leaving out
all of the extra diodes you propose, if putting caps on the output, though
the current surge when charging could cause the PSU to be unhappy on
startup, at least there would be some hope of keeping the output in spec
under some conditions.

2): With my proposal, the regulating action of the switched mode power
supply means that the output voltages will stay in spec whilst a
significant fraction of the energy in the input capacitor of the PSU is
used up because the duty cycle of the PWM chip adjusts to compensate for
the decaying input voltage as the input capacitor is discharging, so adding
capacitance on the input side is fairly effective. On the other hand if
you add capacitance on the output side of the power supply, the output
voltage starts to drop as soon as the power cut occurs, there is no
regulating action during the power cut. This means that you have to add
capacitors with a much higher energy storage capacity if you add them on
the output side of the power supply. A smaller energy storage capacity
would be needed if adding capacitors on the input side.

3): You would have to determine the amount of charge to power each voltage
rail of the PC for a second, and the capacitors on the +5V line could not
help to support the load on the +12V line nor vice versa, so the charge in
some of your added capacitors will probably go unused, whereas if you add
capacitors on the input side of the power supply, the energy will
automatically go to whichever output needs it.

Still if you have an aversion to voltages above 100 Volts, and if you can
afford some of those nice Epcos hundred Farad capacitors, then your scheme
could work (minus the diodes of course).

Chris
 
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