Discharging Capicators Automatically

J

Jonathan Mohn

Guest
I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6 seconds,
and the LED stays lit for the same amount of time. I believe that happens
because the 2200uF storage capacitor is discharging over this time period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more rapidly
when I turn the unit off? I think I could use a normally open push-button
switch to short the capacitor -- I would simply press it after turning the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor once
the power was switched off. Any ideas?

-Jonathan
 
Jonathan Mohn wrote:

I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6 seconds,
and the LED stays lit for the same amount of time. I believe that happens
because the 2200uF storage capacitor is discharging over this time period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more rapidly
when I turn the unit off? I think I could use a normally open push-button
switch to short the capacitor -- I would simply press it after turning the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor once
the power was switched off. Any ideas?
Click to expand...
Why? Why do you want to discharge the cap more rapidly?

--Jeff

--
Ho, ho, ho, hee, hee, hee
and a couple of ha, ha, has;
That's how we pass the day away,
in the merry old land of Oz.
 
you could use a switch/relay to short it how ever, the life of the
contacts would not really last..
the other way around is to use SCR as a trigure crow bar.
your switch should have a DPDT so that you can use the off possition
side to input voltage to the gate of the SCR which is tied accrossed the
cap..
that should kill the charge fast enough and will auto release as soon
as the charge gets down below the holding current of the SCR.
if you don't have this type of switch then have a ralay in there
using the NC side to bias the SCR.
P.S.
make sure you also use the same Relay on the other side to power the
supply to insure you don't have a timing issue between powering up etc..


Jonathan Mohn wrote:
I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6 seconds,
and the LED stays lit for the same amount of time. I believe that happens
because the 2200uF storage capacitor is discharging over this time period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more rapidly
when I turn the unit off? I think I could use a normally open push-button
switch to short the capacitor -- I would simply press it after turning the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor once
the power was switched off. Any ideas?

-Jonathan
Click to expand...
 
Subject: Discharging Capicators Automatically
From: "Jonathan Mohn" j.c.mohn@worldnet.att.net
Date: 2/8/2004 11:06 AM Central Standard Time
Message-id: <w0uVb.198673$6y6.3998779@bgtnsc05-news.ops.worldnet.att.net

I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6 seconds,
and the LED stays lit for the same amount of time. I believe that happens
because the 2200uF storage capacitor is discharging over this time period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more rapidly
when I turn the unit off? I think I could use a normally open push-button
switch to short the capacitor -- I would simply press it after turning the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor once
the power was switched off. Any ideas?

-Jonathan
Click to expand...
Hi, Jonathan. Try using a DPDT switch to turn on the line voltage for 1/2 of
the switch (normally open), and use the second half of the switch to provide a
discharge resistor (normally closed). For anything less than 30V, a 220 ohm, 3
Watt wirewound should work well for quickly discharging things. It would look
something like this (use fixed font or M$ Notepad to view):

DPDT SWITCH FOR LINE VOLTAGE IN/RESISTOR DISCHARGE OUT

.----------------.
| |
L1 o-----o + o---------o-----------o
o-----o--__ | | |
o- | | |
| | o
Line In | |DC Out \
| | \
L2 | | o \o
o---------------o | | |
| | |
| | .-.
| | | | R
| | | |
| | '-'
| | |
| - o-------------o--------o
| |
'----------------'



If you're a newbie, be very careful to keep the line voltage and the DC
separated. Use an ohmmeter to make sure you know what's going on with the
switch before you wire it up, and use a big switch (like the hardware store
kind -- don't use the minis available at Radio Shack) to make sure the wires
don't even get close to each other. If you're not familiar with wiring line
voltage, ask a friend who knows something to give you a hand. Make sure the
wires are really secured on the switch before you install it in the box. An
accident here could lead to electrical shock.

Good luck
Chris
 
"Jeffrey Turner" <jturner@localnet.com> wrote in message
news:102d5vmmjuodv50@corp.supernews.com...
Jonathan Mohn wrote:

I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6
seconds,
and the LED stays lit for the same amount of time. I believe that
happens
because the 2200uF storage capacitor is discharging over this time
period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more rapidly
when I turn the unit off? I think I could use a normally open
push-button
switch to short the capacitor -- I would simply press it after turning
the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor
once
the power was switched off. Any ideas?

Why? Why do you want to discharge the cap more rapidly?
Click to expand...

I don't know that I do. I built this power supply to use not only as a
tool, but as a project to help teach myself basic electronics. When I
noticed the effect of the discharging capacitor, I figured there must be a
method of handling it. I'm not sure whether I will actually build that
capability into my power supply, but one of the great things about projects
like this is that they invariably lead to new questions, and greater
learning.


--Jeff

--
Ho, ho, ho, hee, hee, hee
and a couple of ha, ha, has;
That's how we pass the day away,
in the merry old land of Oz.
Click to expand...
 
Thanks for the post and the schematic. I'll print it out but, because I am
a newbie (as you correctly guessed), I'll get a little more experience under
my built before I play around with this type of configuration. I'm a bit
paranoid about electrocuting myself!

-Jonathan


"CFoley1064" <cfoley1064@aol.com> wrote in message
news:20040208171053.13646.00001382@mb-m27.aol.com...
Subject: Discharging Capicators Automatically
From: "Jonathan Mohn" j.c.mohn@worldnet.att.net
Date: 2/8/2004 11:06 AM Central Standard Time
Message-id: <w0uVb.198673$6y6.3998779@bgtnsc05-news.ops.worldnet.att.net

I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6
seconds,
and the LED stays lit for the same amount of time. I believe that
happens
because the 2200uF storage capacitor is discharging over this time
period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more rapidly
when I turn the unit off? I think I could use a normally open
push-button
switch to short the capacitor -- I would simply press it after turning
the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor
once
the power was switched off. Any ideas?

-Jonathan

Hi, Jonathan. Try using a DPDT switch to turn on the line voltage for 1/2
of
the switch (normally open), and use the second half of the switch to
provide a
discharge resistor (normally closed). For anything less than 30V, a 220
ohm, 3
Watt wirewound should work well for quickly discharging things. It would
look
something like this (use fixed font or M$ Notepad to view):

DPDT SWITCH FOR LINE VOLTAGE IN/RESISTOR DISCHARGE OUT

.----------------.
| |
L1 o-----o + o---------o-----------o
o-----o--__ | | |
o- | | |
| | o
Line In | |DC Out \
| | \
L2 | | o \o
o---------------o | | |
| | |
| | .-.
| | | | R
| | | |
| | '-'
| | |
| - o-------------o--------o
| |
'----------------'



If you're a newbie, be very careful to keep the line voltage and the DC
separated. Use an ohmmeter to make sure you know what's going on with the
switch before you wire it up, and use a big switch (like the hardware
store
kind -- don't use the minis available at Radio Shack) to make sure the
wires
don't even get close to each other. If you're not familiar with wiring
line
voltage, ask a friend who knows something to give you a hand. Make sure
the
wires are really secured on the switch before you install it in the box.
An
accident here could lead to electrical shock.

Good luck
Chris
Click to expand...
 
Jonathan Mohn wrote:
"Jeffrey Turner" <jturner@localnet.com> wrote:
Jonathan Mohn wrote:


I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6
seconds,
and the LED stays lit for the same amount of time. I believe that
happens
because the 2200uF storage capacitor is discharging over this time
period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more rapidly
when I turn the unit off? I think I could use a normally open
push-button
switch to short the capacitor -- I would simply press it after turning
the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor
once
the power was switched off. Any ideas?

Why? Why do you want to discharge the cap more rapidly?

I don't know that I do. I built this power supply to use not only as a
tool, but as a project to help teach myself basic electronics. When I
noticed the effect of the discharging capacitor, I figured there must be a
method of handling it. I'm not sure whether I will actually build that
capability into my power supply, but one of the great things about projects
like this is that they invariably lead to new questions, and greater
learning.
Click to expand...
OK. What is the voltage across the capacitor when the supply is on?
Also, what is the voltage across the LED? If you replace R1 by a
zener diode which takes up the better part of that voltage difference
and a smaller resistor value to limit the current based on the
difference between the present voltage drop and that drop minus the
zener diode value, then the LED will go out much more quickly but the
capacitor will stay charged longer. If the problem is just the LED
then that might serve you. The charge on the capacitor itself
shouldn't be a problem.

--Jeff

--
Ho, ho, ho, hee, hee, hee
and a couple of ha, ha, has;
That's how we pass the day away,
in the merry old land of Oz.
 
"Jeffrey Turner" <jturner@localnet.com> wrote in message
news:102dlk0ok3c8i42@corp.supernews.com...
Jonathan Mohn wrote:
"Jeffrey Turner" <jturner@localnet.com> wrote:
Jonathan Mohn wrote:


I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6
seconds,
and the LED stays lit for the same amount of time. I believe that
happens
because the 2200uF storage capacitor is discharging over this time
period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more
rapidly
when I turn the unit off? I think I could use a normally open
push-button
switch to short the capacitor -- I would simply press it after turning
the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor
once
the power was switched off. Any ideas?

Why? Why do you want to discharge the cap more rapidly?

I don't know that I do. I built this power supply to use not only as a
tool, but as a project to help teach myself basic electronics. When I
noticed the effect of the discharging capacitor, I figured there must be
a
method of handling it. I'm not sure whether I will actually build that
capability into my power supply, but one of the great things about
projects
like this is that they invariably lead to new questions, and greater
learning.

OK. What is the voltage across the capacitor when the supply is on?
Also, what is the voltage across the LED? If you replace R1 by a
zener diode which takes up the better part of that voltage difference
and a smaller resistor value to limit the current based on the
difference between the present voltage drop and that drop minus the
zener diode value, then the LED will go out much more quickly but the
capacitor will stay charged longer. If the problem is just the LED
then that might serve you. The charge on the capacitor itself
shouldn't be a problem.

--Jeff
Click to expand...


Thanks for the post, Jeff.

The voltage across the capacitor is about 29 volts. Right now, R1 is
dropping about 27 V, and the LED is dropping about 2 V. I think I
understand your idea -- if I put a 25V zener in series with the LED and,
say, a 133 ohm resistor, then I should still get about 15 mA through the
LED. Then, when I turn the power off and the capacitor discharges, as soon
as the discharge voltage drops beneath around 25V, the LED should go out.

That's really clever. The timing on that suggestion is perfect, too,
because I've spent the last 30 minutes or so playing around with a zener
diode, testing its break-down voltage, current, voltage drops, etc. These
things are pretty useful little components. Thanks!

The one thing I like about the LED remaining illuminated is that it tells me
voltage is still being supplied to the voltage regulator and, hence,
anything I have hooked up to the output. I haven't, yet, installed the
in-line voltage meter shown in the schematic, so it is actually useful that
the LED stays lit until the capacitor is nearly discharged. I guess that it
is a pretty good reason to install the voltage meter, too. I had figured it
was just for the convenience of not having to use my DMM.

-Jonathan





--
Ho, ho, ho, hee, hee, hee
and a couple of ha, ha, has;
That's how we pass the day away,
in the merry old land of Oz.
Click to expand...
 
On Sun, 08 Feb 2004 19:38:12 -0500, Jeffrey Turner
<jturner@localnet.com> wrote:


OK. What is the voltage across the capacitor when the supply is on?
Also, what is the voltage across the LED? If you replace R1 by a
zener diode which takes up the better part of that voltage difference
and a smaller resistor value to limit the current based on the
difference between the present voltage drop and that drop minus the
zener diode value, then the LED will go out much more quickly but the
capacitor will stay charged longer. If the problem is just the LED
then that might serve you. The charge on the capacitor itself
shouldn't be a problem.
---
Click to expand...
You either have some sort of reading comprehension problem, or you don't
bother to read what posters want before you start spewing your useless
"advice".

For instance, if you'd have gone to the site the OP cited, you'd have
seen the schematic and the BOM for the power supply. With that
information in hand, you would have been able to determine the voltage
across the capacitor (well, maybe not _you_) and answered your own
question. You might also have found out from the OP's post that he
wants to discharge the cap quickly, which will automatically turn off
the LED quickly without having to use your hare-brained Zener scheme
which will, BTW, actually _exacerbate_ the discharge problem.

In short, don't be offering useless advice when you don't know what
you're talking about.

--
John Fields
 
On Sun, 08 Feb 2004 17:06:36 GMT, "Jonathan Mohn"
<j.c.mohn@worldnet.att.net> wrote:

I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6 seconds,
and the LED stays lit for the same amount of time. I believe that happens
because the 2200uF storage capacitor is discharging over this time period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more rapidly
when I turn the unit off? I think I could use a normally open push-button
switch to short the capacitor -- I would simply press it after turning the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor once
the power was switched off. Any ideas?
Click to expand...
---
Yes. First off, it is definitely _not_ a good idea to short the cap with
just a switch, since the only thing limiting the current through the
switch would be the ESR of the cap and the inductance and resistance of
the circuit, all of which would be very low. Because of that the
possibility exists that you could weld the switch contacts closed and
not know it, and then the next time you switched the supply on you'd
start blowing fuses, best case.

What I'd do would be to replace the ON-OFF switch with a DPDT switch
wired to short the cap in the OFF position, like this:


ON OFF
120AC>--[FUSE]--->\ <-- -->\ <--O----+------>TO EVERYTHING ON
\----------------\ | THE INPUT SIDE
O O | OF THE REGULATOR
| | |
| | [2200ľF]
[XFMR PRI] [100R] |
| | |
| | |
120ac>--------------+ +----------+------>DC GROUND

Notice that I placed the fuse _before_ the switch, which is the right
way to do it. Your schematic shows it after the switch, which will
leave the switch hot when the fuse blows. Not a good thing.

The 100 ohm resistor will be discharging the cap at about 1/2 amp if
it's charged up to 50V, but only for a short time, so something like a 1
watt resistor would be fine. I just tried it with a 100 ohm 1/2 watter
and a 3300ľF cap charged up to about 40V, and I could feel the resistor
warm up a little, so even that ought to be OK, but I'd use a 1 watter
anyway.

If you decide to try this, make sure you get a break-before-make switch.

Good luck, and BE CAREFUL.

--
John Fields
 
Thanks, John. Someone else had recommended a DPDT switch, too, but your
schematic makes it really clear how it would work (I wasn't 100% certain,
before). That is great, and is exactly what I'm looking for.

Thanks for the advice on the fuse. I wired as shown in the schematic. I'll
rewire it as you suggest, and I think that, while I'm at it, I'll drop in
the break-before-make DPDT switch, too.

Regards,

Jonathan




"John Fields" <jfields@austininstruments.com> wrote in message
news:g3ve20d89lh5m604onppl6mrds9dtu5tpd@4ax.com...
On Sun, 08 Feb 2004 17:06:36 GMT, "Jonathan Mohn"
j.c.mohn@worldnet.att.net> wrote:

I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6
seconds,
and the LED stays lit for the same amount of time. I believe that
happens
because the 2200uF storage capacitor is discharging over this time
period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more rapidly
when I turn the unit off? I think I could use a normally open
push-button
switch to short the capacitor -- I would simply press it after turning
the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor
once
the power was switched off. Any ideas?

---
Yes. First off, it is definitely _not_ a good idea to short the cap with
just a switch, since the only thing limiting the current through the
switch would be the ESR of the cap and the inductance and resistance of
the circuit, all of which would be very low. Because of that the
possibility exists that you could weld the switch contacts closed and
not know it, and then the next time you switched the supply on you'd
start blowing fuses, best case.

What I'd do would be to replace the ON-OFF switch with a DPDT switch
wired to short the cap in the OFF position, like this:


ON OFF
120AC>--[FUSE]--->\ <-- -->\ <--O----+------>TO EVERYTHING ON
\----------------\ | THE INPUT SIDE
O O | OF THE REGULATOR
| | |
| | [2200ľF]
[XFMR PRI] [100R] |
| | |
| | |
120ac>--------------+ +----------+------>DC GROUND

Notice that I placed the fuse _before_ the switch, which is the right
way to do it. Your schematic shows it after the switch, which will
leave the switch hot when the fuse blows. Not a good thing.

The 100 ohm resistor will be discharging the cap at about 1/2 amp if
it's charged up to 50V, but only for a short time, so something like a 1
watt resistor would be fine. I just tried it with a 100 ohm 1/2 watter
and a 3300ľF cap charged up to about 40V, and I could feel the resistor
warm up a little, so even that ought to be OK, but I'd use a 1 watter
anyway.

If you decide to try this, make sure you get a break-before-make switch.

Good luck, and BE CAREFUL.

--
John Fields
Click to expand...
 
Jonathan Mohn wrote:

"Jeffrey Turner" <jturner@localnet.com> wrote in message
news:102dlk0ok3c8i42@corp.supernews.com...

Jonathan Mohn wrote:

"Jeffrey Turner" <jturner@localnet.com> wrote:

Jonathan Mohn wrote:



I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6

seconds,

and the LED stays lit for the same amount of time. I believe that

happens

because the 2200uF storage capacitor is discharging over this time

period,

supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more

rapidly

when I turn the unit off? I think I could use a normally open

push-button

switch to short the capacitor -- I would simply press it after turning

the

unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor

once

the power was switched off. Any ideas?

Why? Why do you want to discharge the cap more rapidly?

I don't know that I do. I built this power supply to use not only as a
tool, but as a project to help teach myself basic electronics. When I
noticed the effect of the discharging capacitor, I figured there must be

a

method of handling it. I'm not sure whether I will actually build that
capability into my power supply, but one of the great things about

projects

like this is that they invariably lead to new questions, and greater
learning.

OK. What is the voltage across the capacitor when the supply is on?
Also, what is the voltage across the LED? If you replace R1 by a
zener diode which takes up the better part of that voltage difference
and a smaller resistor value to limit the current based on the
difference between the present voltage drop and that drop minus the
zener diode value, then the LED will go out much more quickly but the
capacitor will stay charged longer. If the problem is just the LED
then that might serve you. The charge on the capacitor itself
shouldn't be a problem.

--Jeff





Thanks for the post, Jeff.

The voltage across the capacitor is about 29 volts. Right now, R1 is
dropping about 27 V, and the LED is dropping about 2 V. I think I
understand your idea -- if I put a 25V zener in series with the LED and,
say, a 133 ohm resistor, then I should still get about 15 mA through the
LED. Then, when I turn the power off and the capacitor discharges, as soon
as the discharge voltage drops beneath around 25V, the LED should go out.

That's really clever. The timing on that suggestion is perfect, too,
because I've spent the last 30 minutes or so playing around with a zener
diode, testing its break-down voltage, current, voltage drops, etc. These
things are pretty useful little components. Thanks!

The one thing I like about the LED remaining illuminated is that it tells me
voltage is still being supplied to the voltage regulator and, hence,
anything I have hooked up to the output. I haven't, yet, installed the
in-line voltage meter shown in the schematic, so it is actually useful that
the LED stays lit until the capacitor is nearly discharged. I guess that it
is a pretty good reason to install the voltage meter, too. I had figured it
was just for the convenience of not having to use my DMM.

-Jonathan
Click to expand...
The capacitor will discharge quickly when you attach a load.

--Jeff

--
Ho, ho, ho, hee, hee, hee
and a couple of ha, ha, has;
That's how we pass the day away,
in the merry old land of Oz.
 
I looked for some break-before-make DPDT toggle switches at my local Radio
Shack today, but didn't see any specifically labeled as such. The store's
tech guy didn't seem to know what I was talking about. They had 3-position
DPDTs (on-off-on) and 2-position DPDTs (on-off). I picked up a 2-position
(part #275-666 if that means anything to you). Is there any way I can
determine whether it is break-before-make (I'm guessing that means, in this
application, that the 120V AC circuit is turned off before the capacitor
draining circuit is turned on). The RS tech told me that the switch should
be okay, but I'd rather get your opinion.

Also, since I got a couple of warnings about electrocution (which I do
appreciate, BTW), would y'all mind commenting on my plans for wiring this
switch? It has 6 pins on the back (see below), 1 and 4 are at the bottom
(off position), 2 and 5 are in the middle, 3 and 6 are at the top (on
position). When the switch is OFF, I get continuity between pins 1 and 2,
and also between pins 4 and 5. In the ON position, I get continuity between
pins 3 and 2, and also between pins 6 and 5. I was planning on wiring the
120V AC through 6 & 5, and the capacitor drain through pins 1 and 2, as
follows:


ON
6 3

5 2

4 1
OFF


Transformer:
*120 VAC (black wire) to Fuse to switch pin 6 (upper left pin, ON position)

*Transformer Lead-1 to pin 5 (middle left pin)

*Transformer Lead-2 to 120 VAC (white wire)


Capacitor Drain:
*Capacitor +lead to 100-ohm resistor to switch pin 1 (lower right pin, OFF
position)

*Switch pin 2 (middle right pin) to DC ground


I did a test run with this wiring on a prototyping board, but with 25VDC
instead of 120VAC, and it worked perfectly. Is there anything wrong with
this set-up?

Thanks for your time on this!!

-Jonathan





"John Fields" <jfields@austininstruments.com> wrote in message
news:g3ve20d89lh5m604onppl6mrds9dtu5tpd@4ax.com...
On Sun, 08 Feb 2004 17:06:36 GMT, "Jonathan Mohn"
j.c.mohn@worldnet.att.net> wrote:

I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6
seconds,
and the LED stays lit for the same amount of time. I believe that
happens
because the 2200uF storage capacitor is discharging over this time
period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more rapidly
when I turn the unit off? I think I could use a normally open
push-button
switch to short the capacitor -- I would simply press it after turning
the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor
once
the power was switched off. Any ideas?

---
Yes. First off, it is definitely _not_ a good idea to short the cap with
just a switch, since the only thing limiting the current through the
switch would be the ESR of the cap and the inductance and resistance of
the circuit, all of which would be very low. Because of that the
possibility exists that you could weld the switch contacts closed and
not know it, and then the next time you switched the supply on you'd
start blowing fuses, best case.

What I'd do would be to replace the ON-OFF switch with a DPDT switch
wired to short the cap in the OFF position, like this:


ON OFF
120AC>--[FUSE]--->\ <-- -->\ <--O----+------>TO EVERYTHING ON
\----------------\ | THE INPUT SIDE
O O | OF THE REGULATOR
| | |
| | [2200ľF]
[XFMR PRI] [100R] |
| | |
| | |
120ac>--------------+ +----------+------>DC GROUND

Notice that I placed the fuse _before_ the switch, which is the right
way to do it. Your schematic shows it after the switch, which will
leave the switch hot when the fuse blows. Not a good thing.

The 100 ohm resistor will be discharging the cap at about 1/2 amp if
it's charged up to 50V, but only for a short time, so something like a 1
watt resistor would be fine. I just tried it with a 100 ohm 1/2 watter
and a 3300ľF cap charged up to about 40V, and I could feel the resistor
warm up a little, so even that ought to be OK, but I'd use a 1 watter
anyway.

If you decide to try this, make sure you get a break-before-make switch.

Good luck, and BE CAREFUL.

--
John Fields
Click to expand...
 
"Jonathan Mohn" <j.c.mohn@worldnet.att.net> wrote in message
news:24VVb.450$hR.10041@bgtnsc05-news.ops.worldnet.att.net...
[questions/advice about power supply cap discharging elided]

Johnathan:

Although the switch idea is pretty good, one thing you could do which would
be a simpler/cheaper thing would be to bypass the 2200uF cap with a
resistor. If you used a 1k resistor, the cap would discharge very quickly
(with no other loads, the voltage across it will decrease by 1/2 every 1.5
seconds.) Since its already decreasing fairly quickly anyway, due to the
regulator, this would probably be adequate to turn off the LED almost
immediately. You could experiment with different resistor values to see
which one turned off the led quickly enough.

Note that at 30V, a 1k resistor will dissipate about 900mW (V^2/R), so it
could get a bit hot. A 2W resistor (or even one of those big 5W sandstone
resistors) might be better, since it wouldn't get as hot, due to the larger
surface area. That will make it last longer, I believe.

Regards,
Bob Monsen
 
On Mon, 09 Feb 2004 23:53:34 GMT, "Jonathan Mohn"
<j.c.mohn@worldnet.att.net> wrote:

I looked for some break-before-make DPDT toggle switches at my local Radio
Shack today, but didn't see any specifically labeled as such. The store's
tech guy didn't seem to know what I was talking about. They had 3-position
DPDTs (on-off-on) and 2-position DPDTs (on-off). I picked up a 2-position
(part #275-666 if that means anything to you). Is there any way I can
determine whether it is break-before-make (I'm guessing that means, in this
application, that the 120V AC circuit is turned off before the capacitor
draining circuit is turned on). The RS tech told me that the switch should
be okay, but I'd rather get your opinion.
Click to expand...
---
Break-before-make means that when you toggle the switch one contact is
broken before the other one is made.
---


Also, since I got a couple of warnings about electrocution (which I do
appreciate, BTW), would y'all mind commenting on my plans for wiring this
switch? It has 6 pins on the back (see below), 1 and 4 are at the bottom
(off position), 2 and 5 are in the middle, 3 and 6 are at the top (on
position). When the switch is OFF, I get continuity between pins 1 and 2,
and also between pins 4 and 5. In the ON position, I get continuity between
pins 3 and 2, and also between pins 6 and 5. I was planning on wiring the
120V AC through 6 & 5, and the capacitor drain through pins 1 and 2, as
follows:


ON
6 3

5 2

4 1
OFF


Transformer:
*120 VAC (black wire) to Fuse to switch pin 6 (upper left pin, ON position)

*Transformer Lead-1 to pin 5 (middle left pin)

*Transformer Lead-2 to 120 VAC (white wire)
Click to expand...
---
Yes. The black wire should go to mains hot and the white to mains
neutral.
---

Capacitor Drain:
*Capacitor +lead to 100-ohm resistor to switch pin 1 (lower right pin, OFF
position)

*Switch pin 2 (middle right pin) to DC ground
Click to expand...
---
No, it should be capacitor + lead to pin 1, and pin 2 to one end of the
resistor, with the other end of the resistor going to ground.
I did a test run with this wiring on a prototyping board, but with 25VDC
instead of 120VAC, and it worked perfectly. Is there anything wrong with
this set-up?
Click to expand...
---
Yes, the 100 ohm resistor must _not_ be in series with the capacitor.
Here's the schematic with your switch terminals shown.

BTW, the dashed line between the arms (the slashes) only indicates a
mechanical connection between both switches in the package. Do _NOT_
connect them electrically!!!


ON OFF
4 3
120AC>-[FUSE]-O-->\ <--o o-->\ <--O----+------>TO EVERYTHING ON
6 \----------------\ 1 | THE INPUT SIDE
O5 O2 | OF THE REGULATOR
| | |+
| | [2200ľF]
[XFMR PRI] [100R] |
| | |
| | |
120ac>--------------+ +----------+------>DC GROUND

--
John Fields
 
Thanks for all the posts. Gradually, I'm beginning to see things more
clearly! :)

Regards,

Jonathan




"Jonathan Mohn" <j.c.mohn@worldnet.att.net> wrote in message
news:24VVb.450$hR.10041@bgtnsc05-news.ops.worldnet.att.net...
I looked for some break-before-make DPDT toggle switches at my local Radio
Shack today, but didn't see any specifically labeled as such. The store's
tech guy didn't seem to know what I was talking about. They had
3-position
DPDTs (on-off-on) and 2-position DPDTs (on-off). I picked up a 2-position
(part #275-666 if that means anything to you). Is there any way I can
determine whether it is break-before-make (I'm guessing that means, in
this
application, that the 120V AC circuit is turned off before the capacitor
draining circuit is turned on). The RS tech told me that the switch
should
be okay, but I'd rather get your opinion.

Also, since I got a couple of warnings about electrocution (which I do
appreciate, BTW), would y'all mind commenting on my plans for wiring this
switch? It has 6 pins on the back (see below), 1 and 4 are at the bottom
(off position), 2 and 5 are in the middle, 3 and 6 are at the top (on
position). When the switch is OFF, I get continuity between pins 1 and 2,
and also between pins 4 and 5. In the ON position, I get continuity
between
pins 3 and 2, and also between pins 6 and 5. I was planning on wiring the
120V AC through 6 & 5, and the capacitor drain through pins 1 and 2, as
follows:


ON
6 3

5 2

4 1
OFF


Transformer:
*120 VAC (black wire) to Fuse to switch pin 6 (upper left pin, ON
position)

*Transformer Lead-1 to pin 5 (middle left pin)

*Transformer Lead-2 to 120 VAC (white wire)


Capacitor Drain:
*Capacitor +lead to 100-ohm resistor to switch pin 1 (lower right pin, OFF
position)

*Switch pin 2 (middle right pin) to DC ground


I did a test run with this wiring on a prototyping board, but with 25VDC
instead of 120VAC, and it worked perfectly. Is there anything wrong with
this set-up?

Thanks for your time on this!!

-Jonathan





"John Fields" <jfields@austininstruments.com> wrote in message
news:g3ve20d89lh5m604onppl6mrds9dtu5tpd@4ax.com...
On Sun, 08 Feb 2004 17:06:36 GMT, "Jonathan Mohn"
j.c.mohn@worldnet.att.net> wrote:

I recently built a variable regulated power supply (see
www.meridianelectronics.ca/circ/vps.htm). It works great.

When I turn the unit off, though, I still read a voltage for 5 or 6
seconds,
and the LED stays lit for the same amount of time. I believe that
happens
because the 2200uF storage capacitor is discharging over this time
period,
supplying current both to the voltage regulator and to the LED.

Is there a way to short the capacitor so that it discharges more
rapidly
when I turn the unit off? I think I could use a normally open
push-button
switch to short the capacitor -- I would simply press it after turning
the
unit off. I really don't like that idea, though. I would rather build
something into the circuit that would automatically drain the capacitor
once
the power was switched off. Any ideas?

---
Yes. First off, it is definitely _not_ a good idea to short the cap with
just a switch, since the only thing limiting the current through the
switch would be the ESR of the cap and the inductance and resistance of
the circuit, all of which would be very low. Because of that the
possibility exists that you could weld the switch contacts closed and
not know it, and then the next time you switched the supply on you'd
start blowing fuses, best case.

What I'd do would be to replace the ON-OFF switch with a DPDT switch
wired to short the cap in the OFF position, like this:


ON OFF
120AC>--[FUSE]--->\ <-- -->\ <--O----+------>TO EVERYTHING ON
\----------------\ | THE INPUT SIDE
O O | OF THE REGULATOR
| | |
| | [2200ľF]
[XFMR PRI] [100R] |
| | |
| | |
120ac>--------------+ +----------+------>DC GROUND

Notice that I placed the fuse _before_ the switch, which is the right
way to do it. Your schematic shows it after the switch, which will
leave the switch hot when the fuse blows. Not a good thing.

The 100 ohm resistor will be discharging the cap at about 1/2 amp if
it's charged up to 50V, but only for a short time, so something like a 1
watt resistor would be fine. I just tried it with a 100 ohm 1/2 watter
and a 3300ľF cap charged up to about 40V, and I could feel the resistor
warm up a little, so even that ought to be OK, but I'd use a 1 watter
anyway.

If you decide to try this, make sure you get a break-before-make switch.

Good luck, and BE CAREFUL.

--
John Fields
Click to expand...
 
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