Re: Help: Need DC Voltage Reducer


John Popelish wrote:
Greg wrote:
I have a DC circuit powered by a 9v battery and I need to power a 1.5v
timer from the same battery. Is there a simple way to reduce the 9v to
1.5v? I have seen the voltage divider circuit in the Forrest Mims book
from Radio Shack: Two resistors in series are connected to the two
terminals of a battery. V+ out is taken from between the two resistors,
V- from the neg. battery contact. V out is determined by the values of
the resistors. I have no problem doing the math, but wouldn't that
circuit short out the battery?

You might use a small 3 terminal voltage regulator, like the LM317LZ,
with the output voltage set with a couple resistors.

http://cache.national.com/ds/LM/LM317L.pdf

If you can live with 1.25 volts out, instead of 1.5, you need only a
minimum load resistor. The data sheet shows a 240 ohm resistor from
output to the ADJ pin, to handle the absolute worst case for minimum
load current, but it will probably work fine with a 560 to 1000 ohm
resistor there (for lower battery load). The data sheet says that the
typical minimum load current is about 1.5 mA with a 3 to 15 volt
input. That would be 1.25/.0015 = 833 ohms, especially true if this
thing will never see its full rated temperature.

Hi, John. If the OP wants to go to Radio shack, he'll have to settle
for an LM317T:

http://www.national.com/ds.cgi/LM/LM117.pdf

Radio Shack page:
http://tinyurl.com/ze5q9
Catalog #: 276-1778
$2.29 USD

The LM317 IC is a 3-pin linear voltage regulator available at Radio
Shack, that will maintain a steady output voltage for a varying load.
Not only that, but it will only use 3.5mA extra, instead of the 10mA we
were talking about for your voltage divider. Not quite as good as the
LM317L, but still OK for battery use.

If the OP gets one of these, he can follow the pinout on the package
after reading the data sheets, and put this fairly simple circuit
together:

|
| _____
| IN | |OUT
| .-------o-----|LM317|-o-----o-----.
| | +| |_____| | +| |
| | --- | .-. --- |
| +| --- ADJ| | | --- |
| --- 10uF| |360| |10uF| |
| - | | '-' | |
|9V | | | | | |
| | | o----' | / \
| | | | | ( T )
| | | | | \_/
| | | .-. | |
| | | 68| | | |
| | | | | | |
| | | '-' | |
| | | | | |
| '-------o--------o----------o-----'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

He'll need a 360 ohm resistor, a 68 ohm resistor, and two 10uF caps
(all of which he should be able to get at Radio Shack). And as John
says, if your timer can live with 1.23V, you can forget the 68 ohm
resistor, and just connect the ADJ pin and the other side of the 360
ohm resistor to GND. But whether the load is an LCD clock or a quartz
clock, you really should include the 360 ohm resistor to GND -- either
load is pulsed current, and you should always make sure there's at
least 3.5mA going through the LM317.

The great thing about this circuit is, he will be able to get a steady
1.5V output over the entire current range of his transistor battery
(from zero to 100mA). Also, he only "wastes" 3.5mA. This would mean
that, for a 1mA load, he'll only need 4.5mA from his battery rather
than 90mA.

Such a deal.

Thanks for the spot, John.

Cheers
Chris

.

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