Maker Pro
Maker Pro

How to get 1.5 volts out of a wall-wart power supply's 3 voltoutput...I am new to this group I belie

D

Daniel

Jan 1, 1970
0
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 evenif 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 itmeasures 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 tohave 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!
 
P

Phil Allison

Jan 1, 1970
0
"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
 
Bottom posted.

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 voltpower 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 measurethe 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 inmeasurement. 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, Ithink!

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 supplyand 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!
 
P

Phil Allison

Jan 1, 1970
0
<[email protected]>
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
 
P

Phil Allison

Jan 1, 1970
0
<[email protected]>


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
 
J

Jamie

Jan 1, 1970
0
Daniel said:
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
 
P

Phil Allison

Jan 1, 1970
0
"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
 
A

amdx

Jan 1, 1970
0
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 :)
 
P

Phil Allison

Jan 1, 1970
0
"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
 
<[email protected]>





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
 
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
 
On Wednesday, June 12, 2013 10:03:26 AM UTC-7, Jim Thompson wrote:

....
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
 
Top