Variable output voltage 0-5v with 9v battery?

Looking for a simple design to use a 9v battery to make a variable
0-5v output, using a potentiometer. Anyone able to hook me up on the
knowledge base? And possible Radioshack parts? Sort of a school
project...need help.
Thanks,
Aaron

 

Go buy a 1k potentiometer and a 9V connector from radio shack. Look at the
package the potentiometer comes with to figure out which of the pins are the
'wiper'. Connect the 9V battery connector terminals to the other pins (not
the wiper,) and turn the pot to the middle. Attach the battery. Now, measure
the voltage between the wiper pin and the - terminal of the 9V battery. That
voltage will vary as turn the screw on the pot. You'll need a multimeter to
set the voltage. If you don't have one, you can get one at radioshack.

When you connect something up to your new voltage source (meaning you
connect one side of whatever it is to the wiper, and the other to the -
terminal on the battery) you'll notice that the voltage will change from
what you set it to. This is normal, and just means that some of the current
is being diverted through your device (or 'load', as electronics guys like
to call it).

You can make more constant voltage outputs using chips called 'voltage
regulators', which will keep the voltage constant even when the load varies.
You can get a chip called an LM317 at radio shack that, along with a few
other parts, can do this. The circuit diagram is on the back of the package.
You can't get down to 0V with one of these, however, because of the way they
work. You can only get down to about 1.25V.

Regards,
Bob Monsen

 

Subject: Variable output voltage 0-5v with 9v battery?

Hi, Aaron. Saving the world with only Radio Shack components may be thought to
be a rather perverse pastime, like making life preservers with packing peanuts,
elmer's glue and gingham check, but for some reason, we have to keep doing it.

If you're willing to live with a "0V" output that's within 20 mV or so of GND,
and you're willing to live with 10 mA or so, this will do the trick.
Everything's available in blister packs at Radio Shack -- you won't even have
to go into their drawers ;-) View in fixed format or M$ Notepad:

0 to 5V Radio Shack P.S.
SW1
_/
.--o/ o--o-------------------o----------.
| | | |
| .-. | | .1uF
| | | .------|-----. ---
| 6.8K| | | | | ---
| '-' | | | |
| | | | | |
--- | | | | ===
- 9V | | |\| | GND
| | '----|-\ | +
| .-. | >----o---------------o
| | |<-----o---------|+/
| 10K | | | |/|
| '-' --- | 0 - 5V Out @ 10 mA max
| | --- 1/4 LM324|
| | .001uF| |
| | | o---------------------o
| | | | -
| | | |
=== === === ===
GND GND GND GND

If you can find a 7.5K resistor, your max output voltage will be closer to 5V
-- this will give you almost 6 as a max. Also, RS has a project box with a 9V
battery holder built in -- you'll be able to dress it up, too.

Remember that the LM324 doesn't like having floating inputs -- connect the
output of this op amp to the other 3 + inputs, and connect the - input to the
output on the other 3 amps like this:

.-----------.
| |
| |
| |\| |
'---|-\ |
| >----'
o---|+/
0-5V |/|

If you need more current, you can set this up with a transistor as a follower
for more current.

Good luck
Chris

 

I'm beginner in electronics but what the hey.

You can use a regulator but I think it's an overkill.

The other way is to drop 4V through a resistor. Let's say your circuit
requires 30mA.

So we have Vcc=9v I=30mA.

Now we can use OHMs law to figure out the resistor that can drop 4V.

V=4v I=30mA
V=IR
R=V/I=4/(30/1000)=133.33ohms

130ohms is good enough. Sometimes the resistance is good to be a little more
than the required so you might add another 10ohm resistor in series to make
it 140ohms.

Also, you have to consider the power of the resistor as well. Since we're
dropping 4V through the resistor and the current is 30mA=0.03A then using
the power equation, P=I*V=0.03*4=0.12Watts which is less than 1/4 so we can
use a resistor that is rated at 1/4Watts.

Check out the resistors at:
http://www.radioshack.com/search.asp?find=1/4w+resistor&SRC=1

You can put 4 resistors in series as follows: R=100+10+10+10=130ohm

So far we have the following:

R=130ohms
----------/\/\/\--------o
|
| Vcc=9V
---
-
|
|-----------------------o

Now we add the potentiometer which will vary the voltage between 0 and 5V.
That means it has to be able to drop 5volts at its max position. That means
the max ohm value you want is:

V=5v
I=30mA
V=IR
R=V/I=5/.030=166.7ohms

Find a pot with a max value close to that. Then connect it as in the
following diagram:

R=130ohms Pot=0ohms min, 167ohms max
--------/\/\/\-----(o)-----o Vout
|
| Vcc=9V
---
-
|
|--------------------------o Vgnd

Someone please correct my if I made a mistake. Like I said, this was a good
exercise for a beginner liky myself.

Say if a battery has rated at 600mAh(miliamp hours) and you need 30mA as
described above, you can use it for 20 hours non-stop. It will vary though
depending on the circuit you attach to the outputs Vout and Vgnd.

Or instead of using a resistor you can use a zener diode. Maybe someone with
more experience can help you out in the more efficient and correct ways to
accomplish this circuit.

Hope I could be of some help.

--Viktor

 

9v dc source
+>------/\/\/\/\-- 4.7k risistor
|
/
\<----------<+> Var out ~ 0..5 Vdc.
/
\ 5k pot.
|
->----------------/-----------<->

 

You can make more constant voltage outputs using chips called 'voltage

If you decide to do this method, you might be able to use a fixed 5V
regulator, then put in a voltage divider after (using the pot)t oadjust he
voltage within this range.

 

One reason not to do this is that the point of regulators it to keep voltage
constant with varying load. If you use a voltage divider, the voltage will
vary as the load varies, which usually isn't what you want.

Regards,
Bob Monsen