How to get 1.5 volts out of a wall-wart power supply's 3 vol

Daniel · May 28, 2013

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

"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 ?

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
amps requires ...

** 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

Phil Allison · May 28, 2013

<videoman@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.

I don't follow your math for the solution

** It's really simple.

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

<videoman@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.

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

May 28, 2013

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!

default · May 28, 2013

On Mon, 27 May 2013 21:55:24 -0700 (PDT), Daniel
<videoman@ccountry.net> 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!
Put three or four silicon rectifiers in series with the plus lead?

(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

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
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

On Mon, 27 May 2013 21:55:24 -0700, 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!

Your error comes when you misinterpret the difference between allowing a
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

"Tim Wescott"

An alternative would be to get a red LED from Radio Shack that's rated
for 1.4 or 1.5V,

** Huh ???

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

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.

Yep, Daniel dodged a bullet.
His Usenet life could have been ruined!

Mikek

Ian Field · Jun 1, 2013

"Daniel" <videoman@ccountry.net> wrote in message
news:2debb754-eb4d-4f7c-bd80-bc5024b38b87@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
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

On Sun, 02 Jun 2013 06:47:38 +1000, Ian Field
<gangprobing.alien@ntlworld.com> wrote:

"Daniel" <videoman@ccountry.net> wrote in message
news:2debb754-eb4d-4f7c-bd80-bc5024b38b87@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
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.

He could just use 2 diodes in series for the shunt regulator or a
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

"David Eather" <eather@tpg.com.au> wrote in message
news

p.wx2ljfpswei6gd@phenom-pc...

On Sun, 02 Jun 2013 06:47:38 +1000, Ian Field
gangprobing.alien@ntlworld.com> wrote:

"Daniel" <videoman@ccountry.net> wrote in message
news:2debb754-eb4d-4f7c-bd80-bc5024b38b87@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
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.

He could just use 2 diodes in series for the shunt regulator or a
transistor and 2 resistors if he wants to get fancy (and a cap for the
current peak when the clock 'ticks')

A common trick used with the MK484 (ZN414) AM radio, but diodes have a
fairly soft knee compared to a programmable zener of the TL431 general type.

Phil Allison · Jun 3, 2013

"David Eather"

He could just use 2 diodes in series for the shunt regulator

** The voltage is too low at 1.1V to 1.3V to reliably operate a 1.5V clock -
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

Jim Thompson · Jun 12, 2013

On Wed, 12 Jun 2013 09:40:43 -0700 (PDT), mrdarrett@gmail.com wrote:

On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:
videoman@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 >:-}

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

Jim Thompson · Jun 12, 2013

On Wed, 12 Jun 2013 09:54:11 -0700 (PDT), mrdarrett@gmail.com wrote:

On Wednesday, June 12, 2013 9:46:07 AM UTC-7, Jim Thompson wrote:
On Wed, 12 Jun 2013 09:40:43 -0700 (PDT), mrdarrett@gmail.com wrote:

On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:

videoman@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

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

Ian Field · Jun 12, 2013

<mrdarrett@gmail.com> wrote in message
news:5e56fa73-9e09-4ce6-bc2f-7b6ba1dd1ec8@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
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.

Jun 12, 2013

On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:

videoman@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

Jun 12, 2013

On Wednesday, June 12, 2013 9:46:07 AM UTC-7, Jim Thompson wrote:

On Wed, 12 Jun 2013 09:40:43 -0700 (PDT), mrdarrett@gmail.com wrote:

On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:

videoman@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

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

Jun 12, 2013

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

How to get 1.5 volts out of a wall-wart power supply's 3 vol

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