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How to get 1.5 volts out of a wall-wart power supply's 3 volt output...I am new to this group I believe anyways...Nice to meet you all!
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Daniel
May 28, 2013, 12:55:24 AM5/28/13
to
Hi, I have recently been trying to get 1.5 volts (or anything other than the power supply's rated 3 volts for that matter) out of a wall-wart 3 volt power supply. The wall-wart power supply is dc in output at 3 volts and even if I connect a 300 ohm resistor in series with the wall-wart power supply and the load (a 1.5 volt clock, old style analog clock)and when i measure the voltage with the voltmeter in place of the load (but remember the resistor is still in series) I keep getting 3 volts dc measured. I even put a 20 ohm resistor in parallel with the 300 ohm resister and still no change in measurement. I bought a older style wall-wart with a 1.5 volt setting and it measures at 3 volts with the voltmeter in series with the output leads. I can't get 1.5 volts (or anything other than 3 volts for that matter) no matter what I do and if with the older style wall-wart power supply I measure at the output leads with that power supply set to 12 volts, etc. the measured voltage is very high (19 volts for the 12 volt setting). Now, I know that without a load certain power sources don't measure the same as with a load but if I measure the voltage (I wish i didn't have to do that as the voltage measured was 3 instead of the 1.5 the clock is rated for, I might have done tiny damage to the clock for all I know), if I measure the voltage across where the battery would normally be it measures at 3 volts! SO I had to have been taxing the clock with it running at 3 volts instead of 1.5! Anyways - can someone shed some light on this? I am at the minimum above average in electronics skills and this problem I have been having has been driving me bonkers. The math shows that dropping 3 volts to 1.5 and allowing a current of .08 amps requires 18.75 ohms and I provided that like said above (300 ohms in parallel with 20 ohms equals 18.75 ohms) and it still measures at 3 volts dc! any help appreciated before I tear my hair out...Thanks, I think!
Phil Allison
May 28, 2013, 1:45:40 AM5/28/13
to
"Daniel"
>
Hi, I have recently been trying to get 1.5 volts (or anything other than the
power supply's rated 3 volts for that matter) out of a wall-wart 3 volt
power supply. The wall-wart power supply is dc in output at 3 volts and even
if I connect a 300 ohm resistor in series with the wall-wart power supply
and the load (a 1.5 volt clock, old style analog clock)
** Does this clock normally use a 1.5V, AA cell ?
Hi, I have recently been trying to get 1.5 volts (or anything other than the
power supply's rated 3 volts for that matter) out of a wall-wart 3 volt
power supply. The wall-wart power supply is dc in output at 3 volts and even
if I connect a 300 ohm resistor in series with the wall-wart power supply
and the load (a 1.5 volt clock, old style analog clock)
Does the cell last about 1 year - assuming Alkaline ?
The math shows that dropping 3 volts to 1.5 and allowing a current of .08
** How did you get the 0.08 amps figure ?
A 1.5V, AA cell would last about 1 day at that rate.
FYI:
Typical battery wall clocks consume about 150 micro amps - or 0.00015
amps. Two resistors of say 1000ohms wired in series ACROSS the 3V supply
will divide the voltage in half.
Add a 47 microfarad, 10 to 25volt, electro cap across the clock and you have
it done.
... Phil
vide...@ccountry.net
May 28, 2013, 2:54:56 AM5/28/13
to
Bottom posted.
On Monday, May 27, 2013 9:55:24 PM UTC-7, Daniel wrote:
> Hi, I have recently been trying to get 1.5 volts (or anything other than the power supply's rated 3 volts for that matter) out of a wall-wart 3 volt power supply. The wall-wart power supply is dc in output at 3 volts and even if I connect a 300 ohm resistor in series with the wall-wart power supply and the load (a 1.5 volt clock, old style analog clock)and when i measure the voltage with the voltmeter in place of the load (but remember the resistor is still in series) I keep getting 3 volts dc measured. I even put a 20 ohm resistor in parallel with the 300 ohm resister and still no change in measurement. I bought a older style wall-wart with a 1.5 volt setting and it measures at 3 volts with the voltmeter in series with the output leads. I can't get 1.5 volts (or anything other than 3 volts for that matter) no matter what I do and if with the older style wall-wart power supply I measure at the output leads with that power supply set to 12 volts, etc. the measured voltage is very high (19 volts for the 12 volt setting). Now, I know that without a load certain power sources don't measure the same as with a load but if I measure the voltage (I wish i didn't have to do that as the voltage measured was 3 instead of the 1.5 the clock is rated for, I might have done tiny damage to the clock for all I know), if I measure the voltage across where the battery would normally be it measures at 3 volts! SO I had to have been taxing the clock with it running at 3 volts instead of 1.5! Anyways - can someone shed some light on this? I am at the minimum above average in electronics skills and this problem I have been having has been driving me bonkers. The math shows that dropping 3 volts to 1.5 and allowing a current of .08 amps requires 18.75 ohms and I provided that like said above (300 ohms in parallel with 20 ohms equals 18.75 ohms) and it still measures at 3 volts dc! any help appreciated before I tear my hair out...Thanks, I think!
Hi Phil! Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5 volts. I accidentally lost track while typing and combined two projects, one powering a 1.5 volt AA mp3 player with a wall-wart power supply and of course the mentioned analog old style 1.5 volt alkaline wall clock. The mp3 took .08 amps and as you mentioned the all clock uses a lot less current than that. I don't follow your math for the solution but I am very happy with what you said so I can complete this project soon - thanks a million! Again, thanks a million Phil! By the way – I previously solved the mp3 issue by powering it with a wall-wart power supply that has a usb power source socket so I am happy with that. Thanks!
On Monday, May 27, 2013 9:55:24 PM UTC-7, Daniel wrote:
> Hi, I have recently been trying to get 1.5 volts (or anything other than the power supply's rated 3 volts for that matter) out of a wall-wart 3 volt power supply. The wall-wart power supply is dc in output at 3 volts and even if I connect a 300 ohm resistor in series with the wall-wart power supply and the load (a 1.5 volt clock, old style analog clock)and when i measure the voltage with the voltmeter in place of the load (but remember the resistor is still in series) I keep getting 3 volts dc measured. I even put a 20 ohm resistor in parallel with the 300 ohm resister and still no change in measurement. I bought a older style wall-wart with a 1.5 volt setting and it measures at 3 volts with the voltmeter in series with the output leads. I can't get 1.5 volts (or anything other than 3 volts for that matter) no matter what I do and if with the older style wall-wart power supply I measure at the output leads with that power supply set to 12 volts, etc. the measured voltage is very high (19 volts for the 12 volt setting). Now, I know that without a load certain power sources don't measure the same as with a load but if I measure the voltage (I wish i didn't have to do that as the voltage measured was 3 instead of the 1.5 the clock is rated for, I might have done tiny damage to the clock for all I know), if I measure the voltage across where the battery would normally be it measures at 3 volts! SO I had to have been taxing the clock with it running at 3 volts instead of 1.5! Anyways - can someone shed some light on this? I am at the minimum above average in electronics skills and this problem I have been having has been driving me bonkers. The math shows that dropping 3 volts to 1.5 and allowing a current of .08 amps requires 18.75 ohms and I provided that like said above (300 ohms in parallel with 20 ohms equals 18.75 ohms) and it still measures at 3 volts dc! any help appreciated before I tear my hair out...Thanks, I think!
Phil Allison
May 28, 2013, 3:32:54 AM5/28/13
to
<vide...@ccountry.net>
Daniel wrote:
Hi Phil! Yeah, the clock operates on one AA alkaline battery for about 1
year at 1.5 volts. I accidentally lost track while typing and combined two
projects, one powering a 1.5 volt AA mp3 player with a wall-wart power
supply and of course the mentioned analog old style 1.5 volt alkaline wall
clock. The mp3 took .08 amps and as you mentioned the all clock uses a lot
less current than that.
** Only about 530 times less.
Hi Phil! Yeah, the clock operates on one AA alkaline battery for about 1
year at 1.5 volts. I accidentally lost track while typing and combined two
projects, one powering a 1.5 volt AA mp3 player with a wall-wart power
supply and of course the mentioned analog old style 1.5 volt alkaline wall
clock. The mp3 took .08 amps and as you mentioned the all clock uses a lot
less current than that.
I don't follow your math for the solution
You need a 1.5V supply that can deliver 150 microamps average. The clock
actually takes short pulses of about 10 times that current every second,
hence the 47uF electro to handle the pulse current.
The two 1000 ohm resistors just split the voltage coming from the wart and
have low enough resistance so the average load current is not enough to make
any difference to this. You can connect the clock & electro across either
resistor, long as the polarity of both is right.
.... Phil
Phil Allison
May 28, 2013, 4:08:34 AM5/28/13
to
<vide...@ccountry.net>
Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
volts.
cells does.
PLUS if the AC power ever goes off for a while, the clock will stop and
thereafter show the wrong time.
Till you figure that fact out.
Wot a dumb idea.
... Phil
default
May 28, 2013, 7:51:36 AM5/28/13
to
(Depending on the measured voltage output of the wall-wart) Drop.6
volts per series diode under practically any load. Signal diodes are
good for ~100 milliamps (and smaller cheaper than rectifiers)
Or you can use the forward voltage drop of an led - something I use
when driving a clock from a processor running at 5 volts (cheap and
dirty way to check long time delays) - in that case the brightness of
the led changes with power draw - blinks once a second with a quartz
analog clock as the load. Red leds drop ~2 volts, green ~3, white and
blue ~3.4... load is limited to 20 milliamps with garden variety
leds, and quite a bit higher with high power leds.
Jamie
May 28, 2013, 8:48:16 AM5/28/13
to
Daniel wrote:
> Hi, I have recently been trying to get 1.5 volts (or anything other than the power supply's rated 3 volts for that matter) out of a wall-wart 3 volt power supply. The wall-wart power supply is dc in output at 3 volts and even if I connect a 300 ohm resistor in series with the wall-wart power supply and the load (a 1.5 volt clock, old style analog clock)and when i measure the voltage with the voltmeter in place of the load (but remember the resistor is still in series) I keep getting 3 volts dc measured. I even put a 20 ohm resistor in parallel with the 300 ohm resister and still no change in measurement. I bought a older style wall-wart with a 1.5 volt setting and it measures at 3 volts with the voltmeter in series with the output leads. I can't get 1.5 volts (or anything other than 3 volts for that matter) no matter what I do and if with the older style wall-wart power supply I measure at the output leads with that power supply set to 12 volts, etc. the measured voltage i
s very high (19 volts for the 12 volt setting). Now, I know that without a load certain power sources don't measure the same as with a load but if I measure the voltage (I wish i didn't have to do that as the voltage measured was 3 instead of the 1.5 the clock is rated for, I might have done tiny damage to the clock for all I know), if I measure the voltage across where the battery would normally be it measures at 3 volts! SO I had to have been taxing the clock with it running at 3 volts instead of 1.5! Anyways - can someone shed some light on this? I am at the minimum above average in electronics skills and this problem I have been having has been driving me bonkers. The math shows that dropping 3 volts to 1.5 and allowing a current of .08 amps requires 18.75 ohms and I provided that like said above (300 ohms in parallel with 20 ohms equals 18.75 ohms) and it still measures at 3 volts dc! any help appreciated before I tear my hair out...Thanks, I think!
The wallwart most likely isn't a regulated type and it most likely has
> Hi, I have recently been trying to get 1.5 volts (or anything other than the power supply's rated 3 volts for that matter) out of a wall-wart 3 volt power supply. The wall-wart power supply is dc in output at 3 volts and even if I connect a 300 ohm resistor in series with the wall-wart power supply and the load (a 1.5 volt clock, old style analog clock)and when i measure the voltage with the voltmeter in place of the load (but remember the resistor is still in series) I keep getting 3 volts dc measured. I even put a 20 ohm resistor in parallel with the 300 ohm resister and still no change in measurement. I bought a older style wall-wart with a 1.5 volt setting and it measures at 3 volts with the voltmeter in series with the output leads. I can't get 1.5 volts (or anything other than 3 volts for that matter) no matter what I do and if with the older style wall-wart power supply I measure at the output leads with that power supply set to 12 volts, etc. the measured voltage i
s very high (19 volts for the 12 volt setting). Now, I know that without a load certain power sources don't measure the same as with a load but if I measure the voltage (I wish i didn't have to do that as the voltage measured was 3 instead of the 1.5 the clock is rated for, I might have done tiny damage to the clock for all I know), if I measure the voltage across where the battery would normally be it measures at 3 volts! SO I had to have been taxing the clock with it running at 3 volts instead of 1.5! Anyways - can someone shed some light on this? I am at the minimum above average in electronics skills and this problem I have been having has been driving me bonkers. The math shows that dropping 3 volts to 1.5 and allowing a current of .08 amps requires 18.75 ohms and I provided that like said above (300 ohms in parallel with 20 ohms equals 18.75 ohms) and it still measures at 3 volts dc! any help appreciated before I tear my hair out...Thanks, I think!
a cap in there, which gives you a higher reading.. But putting a load
on the supply normally brings things down..
If you start with a 1.5 RMS (AC internally), that equates to ~2Volts
after rectified and filled wit out load. So lets try something different.
For what you're trying to do does not require anything very special.
Using 2 silicon diodes in series will drop your voltage ~ 1.2 volts
for every Si diode in series there is an ~650 mv drop.
use diodes that can handle the current rating..
Jamie
Tim Wescott
May 28, 2013, 12:45:28 PM5/28/13
to
current and actually having that current flow.
A 18.75 ohm resistor that has 80mA actually flowing through it will drop
1.5V. But connecting that resistor, waving your hands over it, and
saying "you may now flow 80mA" won't make current flow in the resistor.
What you're seeing is a voltage source, with a resistor connected, and
practically no current flowing.
All of Phil's practical comments are spot-on, and he hasn't even started
cursing and foaming at the mouth, so he's a useful USENET poster today.
An alternative would be to get a red LED from Radio Shack that's rated
for 1.4 or 1.5V, connect it up with a 150 ohm resistor in series to make
a light, and connect the clock in parallel with the LED. This will
provide some rather sleazy voltage regulation as well as a visual
indication that the clock is powered (and it'll be yet another
perpetually glowing LED adding to the light pollution in your house at
night -- we all need that).
--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?
Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Phil Allison
May 28, 2013, 7:08:40 PM5/28/13
to
"Tim Wescott"
> An alternative would be to get a red LED from Radio Shack that's rated
> for 1.4 or 1.5V,
Red LEDs have forward voltage fro 1.7 to 2.2 volts - depending on current
flow.
The 2 R and 1C solution I posted is the only one certain to work.
But " work" means increased running cost and far less reliable time
display.
... Phil
amdx
May 28, 2013, 7:54:25 PM5/28/13
to
> All of Phil's practical comments are spot-on, and he hasn't even started
> cursing and foaming at the mouth, so he's a useful USENET poster today.
>
His Usenet life could have been ruined!
Mikek :-)
Ian Field
Jun 1, 2013, 4:47:38 PM6/1/13
to
"Daniel" <vide...@ccountry.net> wrote in message
news:2debb754-eb4d-4f7c...@googlegroups.com...
> Hi, I have recently been trying to get 1.5 volts (or anything other than
> the power supply's rated 3 volts for that matter) out of a wall-wart 3
> volt power supply. The wall-wart power supply is dc in output at 3 volts
> and even if I connect a 300 ohm resistor in series with the wall-wart
> power supply and the load (a 1.5 volt clock, old style analog clock)and
> when i measure the voltage with the voltmeter in place of the load (but
> remember the resistor is still in series) I keep getting 3 volts dc
> measured. I even put a 20 ohm resistor in parallel with the 300 ohm
> resister and still no change in measurement. I bought a older style
> wall-wart with a 1.5 volt setting and it measures at 3 volts
What you need is an adjustable shunt regulator + dropper resistor - the
> the power supply's rated 3 volts for that matter) out of a wall-wart 3
> volt power supply. The wall-wart power supply is dc in output at 3 volts
> and even if I connect a 300 ohm resistor in series with the wall-wart
> power supply and the load (a 1.5 volt clock, old style analog clock)and
> when i measure the voltage with the voltmeter in place of the load (but
> remember the resistor is still in series) I keep getting 3 volts dc
> measured. I even put a 20 ohm resistor in parallel with the 300 ohm
> resister and still no change in measurement. I bought a older style
> wall-wart with a 1.5 volt setting and it measures at 3 volts
older TL431 only went down to 2.45V, but there are newer versions with 1.25V
reference:
http://www.nxp.com/documents/leaflet/75017148.pdf
As someone else mentioned - if the power goes out your clock will show the
wrong time, never tried it, but you might get away with backing it up with a
supercapacitor if the current draw is really that low.
David Eather
Jun 2, 2013, 5:19:53 PM6/2/13
to
transistor and 2 resistors if he wants to get fancy (and a cap for the
current peak when the clock 'ticks')
Ian Field
Jun 2, 2013, 5:37:03 PM6/2/13
to
"David Eather" <eat...@tpg.com.au> wrote in message
news:op.wx2ljfpswei6gd@phenom-pc...
fairly soft knee compared to a programmable zener of the TL431 general type.
Phil Allison
Jun 3, 2013, 8:54:29 AM6/3/13
to
"David Eather"
> He could just use 2 diodes in series for the shunt regulator
often they will not run on a NiCd or NiMH cell.
Three 1 amp diodes and a 330 ohm resistor might be the go.
... Phil
mrda...@gmail.com
Jun 12, 2013, 12:40:43 PM6/12/13
to
On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:
> <vide...@ccountry.net>
>
>
>
>
>
> Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
>
> volts.
>
>
>
>
>
> ** Changing to a wall wart will cost much more in electricity than using AA
>
> cells does.
It will???
> <vide...@ccountry.net>
>
>
>
>
>
> Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
>
> volts.
>
>
>
>
>
> ** Changing to a wall wart will cost much more in electricity than using AA
>
> cells does.
Here (California) we can get a 4-pack of AA batteries at the 99 Cent store ($1.08 after sales taxes). So, $0.27 per year.
Running 3 volts at 1000 ohms (I assume two 500 ohm resistors to provide a little extra juice to charge the capacitor) gives 0.009 watts (P = V^2/R).
1 yr x (365 d/yr) x (24 h/d) x ($0.15/kw-hr) x (1 kw/1000 w) x 0.009 W gives... $0.012/yr.
$0.27 >> $0.012
>
>
>
> PLUS if the AC power ever goes off for a while, the clock will stop and
>
> thereafter show the wrong time.
M
Jim Thompson
Jun 12, 2013, 12:46:07 PM6/12/13
to
On Wed, 12 Jun 2013 09:40:43 -0700 (PDT), mrda...@gmail.com wrote:
>On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:
>> <vide...@ccountry.net>
>>
>>
>>
>>
>>
>> Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
>>
>> volts.
>>
>>
>>
>>
>>
>> ** Changing to a wall wart will cost much more in electricity than using AA
>>
>> cells does.
>
>
>It will???
>
>Here (California) we can get a 4-pack of AA batteries at the 99 Cent store ($1.08 after sales taxes). So, $0.27 per year.
>
>Running 3 volts at 1000 ohms (I assume two 500 ohm resistors to provide a little extra juice to charge the capacitor) gives 0.009 watts (P = V^2/R).
>
>1 yr x (365 d/yr) x (24 h/d) x ($0.15/kw-hr) x (1 kw/1000 w) x 0.009 W gives... $0.012/yr.
>
>$0.27 >> $0.012
Plus Californica tax >:-}
>On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:
>> <vide...@ccountry.net>
>>
>>
>>
>>
>>
>> Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
>>
>> volts.
>>
>>
>>
>>
>>
>> ** Changing to a wall wart will cost much more in electricity than using AA
>>
>> cells does.
>
>
>It will???
>
>Here (California) we can get a 4-pack of AA batteries at the 99 Cent store ($1.08 after sales taxes). So, $0.27 per year.
>
>Running 3 volts at 1000 ohms (I assume two 500 ohm resistors to provide a little extra juice to charge the capacitor) gives 0.009 watts (P = V^2/R).
>
>1 yr x (365 d/yr) x (24 h/d) x ($0.15/kw-hr) x (1 kw/1000 w) x 0.009 W gives... $0.012/yr.
>
>$0.27 >> $0.012
>
>
>>
>>
>>
>> PLUS if the AC power ever goes off for a while, the clock will stop and
>>
>> thereafter show the wrong time.
>
>
>No argument there
>
>M
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85140 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
mrda...@gmail.com
Jun 12, 2013, 12:54:11 PM6/12/13
to
On Wednesday, June 12, 2013 9:46:07 AM UTC-7, Jim Thompson wrote:
> On Wed, 12 Jun 2013 09:40:43 -0700 (PDT), mrda...@gmail.com wrote:
>
>
>
> >On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:
>
> >> <vide...@ccountry.net>
>
> >>
>
> >>
>
> >>
>
> >>
>
> >>
>
> >> Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
>
> >>
>
> >> volts.
>
> >>
>
> >>
>
> >>
>
> >>
>
> >>
>
> >> ** Changing to a wall wart will cost much more in electricity than using AA
>
> >>
>
> >> cells does.
>
> >
>
> >
>
> >It will???
>
> >
>
> >Here (California) we can get a 4-pack of AA batteries at the 99 Cent store ($1.08 after sales taxes). So, $0.27 per year.
>
> >
>
> >Running 3 volts at 1000 ohms (I assume two 500 ohm resistors to provide a little extra juice to charge the capacitor) gives 0.009 watts (P = V^2/R).
>
> >
>
> >1 yr x (365 d/yr) x (24 h/d) x ($0.15/kw-hr) x (1 kw/1000 w) x 0.009 W gives... $0.012/yr.
>
> >
>
> >$0.27 >> $0.012
>
>
>
> Plus Californica tax >:-}
Well, sure, that adds $thousands to both sides of the equation... but you add them whether or not you use batteries XD
> On Wed, 12 Jun 2013 09:40:43 -0700 (PDT), mrda...@gmail.com wrote:
>
>
>
> >On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:
>
> >> <vide...@ccountry.net>
>
> >>
>
> >>
>
> >>
>
> >>
>
> >>
>
> >> Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
>
> >>
>
> >> volts.
>
> >>
>
> >>
>
> >>
>
> >>
>
> >>
>
> >> ** Changing to a wall wart will cost much more in electricity than using AA
>
> >>
>
> >> cells does.
>
> >
>
> >
>
> >It will???
>
> >
>
> >Here (California) we can get a 4-pack of AA batteries at the 99 Cent store ($1.08 after sales taxes). So, $0.27 per year.
>
> >
>
> >Running 3 volts at 1000 ohms (I assume two 500 ohm resistors to provide a little extra juice to charge the capacitor) gives 0.009 watts (P = V^2/R).
>
> >
>
> >1 yr x (365 d/yr) x (24 h/d) x ($0.15/kw-hr) x (1 kw/1000 w) x 0.009 W gives... $0.012/yr.
>
> >
>
> >$0.27 >> $0.012
>
>
>
> Plus Californica tax >:-}
One thing I do not know... how much power will a typical wall-wart consume, radiating heat, just from being plugged in? That, I do not know. When I get a spare moment I'll calculate the breakeven heat loss power cost ;)
M
Jim Thompson
Jun 12, 2013, 1:03:26 PM6/12/13
to
But I am toying with powering an outdoor clock via a wall-wart, since
the clock is made of flagstone and is a pain to take down.
mrda...@gmail.com
Jun 12, 2013, 1:06:14 PM6/12/13
to
On Wednesday, June 12, 2013 10:03:26 AM UTC-7, Jim Thompson wrote:
...
> >Well, sure, that adds $thousands to both sides of the equation... but you add them whether or not you use batteries XD
>
> >
>
> >One thing I do not know... how much power will a typical wall-wart consume, radiating heat, just from being plugged in? That, I do not know. When I get a spare moment I'll calculate the breakeven heat loss power cost ;)
>
> >
>
> >M
>
>
>
> For low power drain, a wall-wart is probably quite inefficient.
Yes... looks like if the wall wart wastes more than 0.2 watts, it will cost more than the battery... hmm...
> But I am toying with powering an outdoor clock via a wall-wart, since
>
> the clock is made of flagstone and is a pain to take down.
Yes! I also was struck with the genius of Phil's suggestion. Quite elegant, and needs few parts.
Michael
...
> >Well, sure, that adds $thousands to both sides of the equation... but you add them whether or not you use batteries XD
>
> >
>
> >One thing I do not know... how much power will a typical wall-wart consume, radiating heat, just from being plugged in? That, I do not know. When I get a spare moment I'll calculate the breakeven heat loss power cost ;)
>
> >
>
> >M
>
>
>
> For low power drain, a wall-wart is probably quite inefficient.
> But I am toying with powering an outdoor clock via a wall-wart, since
>
> the clock is made of flagstone and is a pain to take down.
Michael
Ian Field
Jun 12, 2013, 2:09:19 PM6/12/13
to
<mrda...@gmail.com> wrote in message
news:5e56fa73-9e09-4ce6...@googlegroups.com...
> On Wednesday, June 12, 2013 10:03:26 AM UTC-7, Jim Thompson wrote:
>
> ...
>
>> >Well, sure, that adds $thousands to both sides of the equation... but
>> >you add them whether or not you use batteries XD
>>
>> >
>>
>> >One thing I do not know... how much power will a typical wall-wart
>> >consume, radiating heat, just from being plugged in? That, I do not
>> >know. When I get a spare moment I'll calculate the breakeven heat loss
>> >power cost ;)
>>
>> >
>>
>> >M
>>
>>
>>
>> For low power drain, a wall-wart is probably quite inefficient.
>
>
> Yes... looks like if the wall wart wastes more than 0.2 watts, it will
> cost more than the battery... hmm...
You could always go for the "wattless dropper", its a little dodgy insofar
>
> ...
>
>> >Well, sure, that adds $thousands to both sides of the equation... but
>> >you add them whether or not you use batteries XD
>>
>> >
>>
>> >One thing I do not know... how much power will a typical wall-wart
>> >consume, radiating heat, just from being plugged in? That, I do not
>> >know. When I get a spare moment I'll calculate the breakeven heat loss
>> >power cost ;)
>>
>> >
>>
>> >M
>>
>>
>>
>> For low power drain, a wall-wart is probably quite inefficient.
>
>
> Yes... looks like if the wall wart wastes more than 0.2 watts, it will
> cost more than the battery... hmm...
as it puts the current waveform ahead of the voltage and cons the
electricity meter a bit.
There are a few gotchas to watch out for - especially for a very low power
use like a clock.
Around the late 70's; the UK TV maker TCE adopted the "wattless dropper" for
the 300mA series heater chain, this consisted of a 4.3uF capacitor in series
instead of a dropper resistor, for 50Hz applications the capacitance is
directly scalable for different current values - but don't forget to factor
in the reduced Xc for a 60Hz supply.
You need a significant resistor in series with the capacitor to absorb turn
on surge and any mains borne spikes, its basically a constant current supply
so you must use shunt regulation for intermittent loads (like an electronic
clock escapement).
The simplest configuration is the capacitor in series with the input to a
bridge rectifier - but then the DC rails swing +/- of the ground potential,
a better approach is to use the charge-pump style voltage doubler
configuration - at least then you have a ground referenced rail.
Also, don't forget a mains input bleed resistor to discharge the capacitor
when you pull the plug out the socket - needless to say, this resistor must
be rated to take continuous mains.
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