# 120V AC to 1.5V DC

Discussion in 'Electronic Basics' started by Jonathan Mohn, Jan 16, 2004.

1. ### Jonathan MohnGuest

Just for fun, I'm trying to modify one of those Taylor temperature probes
(used for cooking) so that I can power it from a wall outlet, rather than
from a 1.5V AAA battery.

I have a wall plug transformer that converts 120V AC to 9V DC (100mA). I
also have a potentiometer, which I figured I could use to step the voltage
down to 1.5V. So, I've done just that. I've connected the poitive end of
the temp prob unit to the 1.5V end of the potentiometer. The thing works,
but I have some problems.

Each time I push a button to set the timer, the temperature alarm, etc., the
display will briefly flash off, and then on. Also, the alarms/beeps aren't
as loud as they are when I have the battery connected. It seems that I'm
not getting enough current.

I hooked up a multimeter to test that theory, and I think that it is
correct. When a battery is connected and I push one of the buttons, I get a
current of 5 - 10 milliamps. When I've got it plugged into the wall and I
push a button, I get a current of less than 1 milliamp.

Can anyone help me out on this? I imagine that the potentiometer is the
wrong component to step down the voltage, but I don't know why. What should

-Jonathan

2. ### TimGuest

Hi Jonathan,

That was a good try, but the wrong components.

The simplest way to get about 1.5 volts (1.8 volts last time I checked) was
to use three 1N4001 diodes in series with a 220 ohm resistor connected to
the 9vdc power supply.

V+ - resistor - Vout - diode - diode - diode - GND

Vout of 1.8 volts between resistor and first diode. With 220 ohm resistor,
you should get about 30ma maximum current before Vout starts to sag. That
should supply your probe just fine.

If you need any more current than that, reduce the resistor value a little,
but you will need to use a 1/2 watt resistor rather than 1/4 watt. You might
want to use a 1/2 watt resistor anyway. I don't like being that close to a
power limit with a resistor. They do get warm.

Hope that helps.

Tim

3. ### John PopelishGuest

Connect a couple silicon diodes (cathode end toward negative) between
the negative end of the pot and the negative supply output. Feed the
positive side of the load from the wiper. Use a 500 to 1000 ohm pot.
This should stiffen the voltage quite a bit. Of get an LM317 and
build an actual regulator (see the data sheet). They run about a buck
ar Radio Shack. You can use the same pot to set its output voltage.

http://cache.national.com/ds/LM/LM117.pdf

4. ### Jonathan MohnGuest

Thanks for the posts.

As you've probably already guessed, I'm a real novice at this. I thought
that diodes were used to convert AC to DC. How does the diode help in my
application?

-J

5. ### Robert C MonsenGuest

If you put a diode in series with a resistor, and connect a positive voltage
source to ground through it, the diode will drop something close to 0.6V.
Its often used in circuits to provide a fixed 0.6V voltage drop. Thus, using
three in series produces 1.8V of voltage drop, or two in series 1.2V. Try
it...

Regards,
Bob Monsen

6. ### Guest

| Just for fun, I'm trying to modify one of those Taylor temperature probes
| (used for cooking) so that I can power it from a wall outlet, rather than
| from a 1.5V AAA battery.
|
| I have a wall plug transformer that converts 120V AC to 9V DC (100mA). I
| also have a potentiometer, which I figured I could use to step the voltage
| down to 1.5V. So, I've done just that. I've connected the poitive end of
| the temp prob unit to the 1.5V end of the potentiometer. The thing works,
| but I have some problems.
|
| Each time I push a button to set the timer, the temperature alarm, etc., the
| display will briefly flash off, and then on. Also, the alarms/beeps aren't
| as loud as they are when I have the battery connected. It seems that I'm
| not getting enough current.
|
| I hooked up a multimeter to test that theory, and I think that it is
| correct. When a battery is connected and I push one of the buttons, I get a
| current of 5 - 10 milliamps. When I've got it plugged into the wall and I
| push a button, I get a current of less than 1 milliamp.
|
| Can anyone help me out on this? I imagine that the potentiometer is the
| wrong component to step down the voltage, but I don't know why. What should

When the current is changing on the load, its resistance is changing. When
that is in series with a fixed resistance, the voltage drop over the load
changes as the ratio of resistance changes. A voltage regulator backed up
with an overcurrent resistor would be what you need.

7. ### Guest

| As you've probably already guessed, I'm a real novice at this. I thought
| that diodes were used to convert AC to DC. How does the diode help in my
| application?

A diode has a non-linear transfer curve. As voltage varies, current either
varies more by many times, or not at all. The effect over certain voltage
ranges is that the diode is a nearly constant voltage drop. A zener diode
is specially decide to take advantage of this for voltage and current levels
a power supply needs.

8. ### TimGuest

Each diode will produce about .6 volt drop when connected anode to a
positive current supply and cathode to negative or ground.

I would certainly not use this on a design requiring an exact voltage
because of diode barrier voltage changes with current and temperature. But
to replace a 1.5v AA battery? Sure! Bear in mind a AA battery output voltage
varies considerably with current draw and age.

Tim

9. ### John PopelishGuest

While resistors drop a voltage in proportion to the current through
them, diodes drop a voltage that is (roughly) proportional to the log
of the current through them, so they produce a much more stable drop
over a wide range of current than resistors do.

10. ### Jonathan MohnGuest

Thanks, everyone, for educating me. It makes sense to me, now.

This is the first time I've visited this ng, and I already love it.

-J

11. ### Jonathan MohnGuest

I ended up buying the LM317T, as John suggested, and built a voltage
regulator circuit. These things are pretty handy. If I read the specs
right, I can draw up to 1.5A and still maintain the voltage I've set with
the pot. Now I just need to find some other 1.5V applications I can tie
into this thing!

-Jonathan

12. ### Tim DicusGuest

You can get 1.5 amps for the LM317 only if you are using a DC source capable
of supplying it. The transformer and rectifier must be up to the task.

If you are still using the 9V 100ma "wall wart" (plug-in transformer) you
mentioned originally, you are limited to 100ma from the LM317, else the wall
wart will fail.

The LM317 is the proper tool if you require an exact voltage.

Tim

13. ### Jonathan MohnGuest

Thanks for pointing that out, Tim. You are right -- I am using the 9V 100ma
"wall wart" (I like that) I mentioned earlier.

I tried hooking the 1N4001 diodes up in series with a 220 ohm resister, as
you suggested, and it worked great! That is a really clever way of doing
things. The diodes I purchased had a 0.75V drop each, so I only needed two
to give me 1.5V.

Perhaps you could clear something up for me, though. I'm not sure I
understand why the 220 ohm resister is required. Is it simply to limit the
current?

-Jonathan

14. ### Tim DicusGuest

Very good! It determines the current the circuit will consume as long as you
do not exceed 30ma from Vout. That diode circuit will always take about 30ma
with a 9 volt supply, as long as the load is less than 30ma. Even if you
exceed 30ma (say you decide to ground the resistor instead of connecting it
to the diodes), it will only draw 40ma maximum on a 9 volt supply.

That is another advantage of using the LM317 besides the accurate
regulation. It only draws the current of the load plus a little for its own
use. But 30ma is not bad as long as it is powered by a wall wart (AC power).
If you want to power things with a battery or other limited current sources
like solar cells, voltage regulators are the only way to go.

If you decide to use that LM317, I would recommend obtaining a 12v to 15v
transformer (2amp minimum), a bridge rectifier (2amp minimum), a 2000uf 35v
capacitor, a 1uf tantalum cap, a heat sink, and a fuse and holder. That
would complete your power supply so you could use it for its rated current.
They are a great starting project. I have found mine to be really handy. I
added a meter so I can see the output voltage.

Tim

15. ### Jonathan MohnGuest

Thanks for the suggestions, Tim. I've printed your message out and will
spend some time doing research to make sure I understand everything. I've
seen some schemtatics for bridge rectifiers while perusing the internet, but
I haven't, yet, taken time to understand how they work (or how to use them,
for that matter).

I've also pulled down a schematic for a voltage regulator circuit that I can
build, using transformers, diodes, capacitors, etc. I don't intend to build
it, but I would like to understand how these voltage regulators work. It
sure is convenient to be able to buy an IC like the LM317, though!

-Jonathan

16. ### Watson A.Name - Watt Sun, Dark RemoverGuest

and/or calibration may be off. It would be safer to use two diodes,
or two diodes and a schottky diode like the 1N5817. I would try for
a LM317L with two 470 ohm resistors between output and adjust
terminals, and a 47 ohm resistor from adjust terminal to ground. That
should give a stable 1.5V supply.

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17. ### Tim DicusGuest

No worries now. Jonathan found with a voltmeter that three diodes in fact
was too much, resorted to two, and all is apparently well. At that current,
the 1N4001 barrier voltage is .75 volt. I should have checked.

The LM317 is still my favorite for the odd voltage supply that is not
covered by the 78xx series regulators.

Thanks,

Tim

18. ### Jonathan MohnGuest

At the risk of wearing out my welcome, I have one more simple question that

I've assembled the 1.25V - 25V adjustable regulator depicted on the first
page of the LM317 spec sheet -- http://cache.national.com/ds/LM/LM117.pdf --
only I used two 1 uF tantalum capacitors, one for the input side and one for
the output side, and a 10K cermet potentiometer. If I understand how the
capacitors are being used in this circuit, it should not matter that I used
1 uF instead of the recommended 0.1 uF on the input side, no?

My main question is on the pin-out of the pot. This is one of those blue,
rectangular pots with a screw on one side for moving the wiper. At first I
connected pin 1 to ground, and the middle pin to the adj pin of the LM317
and to the 240 ohm resistor, as shown in the depiction. But I'm having
problems with pin 3. At first, I hooked it up to the Vout side of the
LM317. It seemed to work just fine, but I grew worried about doing it that
way. So then, I hooked up pin 3 to the positive side of my 9V power source,
and, again, it worked fine. Finally, just out of curiousity, I disconnected
pin 3 altoghether, leaving pin 1 hooked up to ground, and the middle pin
hooked up as described above. To my amazement, everything still worked
fine! What gives?

I finally settled on hooking pin 3 up to the positive side of my 9V power
source, but can anyone explain what the appropriate pin-out is and why?

Thanks a bunch.

-Jonathan

19. ### John PopelishGuest

The three pins on the pot are the two ends of a resistive element, and
one sliding contact. Since this schematic calls for a two terminal
variable resistor, you should use one end terminal and the sliding
contact, only. I question the resistance or the pot. I think a 5000
ohm pot would be better (you will not use half of the 10k pot's
rotation to reach full output), and if your input voltage is less than
27 volts, a 2000 ohm pot might be even better (for a 1.2 to ~12 volt
adjustment). Did you solve the formula for the required value of R2?

The capacitors should be fine.

20. ### Jonathan MohnGuest

I did solve for the required value of R2. I wasn't certain what Iadj should
be in the formula, so I first assumed that it is 0, and I found R2 should be
48 ohms to achieve an output voltage of 1.5V. The way the formula is
written, any value of Iadj that is greater than 0 will result in an R2 of
less than 48 ohms. So 48 ohms seems to be the maximum required (again, for
1.5V).

I used a 10K ohm pot because it was all that they had available at the Radio
Shack in my neighborhood. Eventually I'll update the circuit with a 5K.
Will I improve on its performance by going with a 5K pot?

I'll stick with 5K or up because I like this regulator so much that I want
to be able to use it for other apps, so will want the full range of