Greg said:
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?
Thanks,
Greg
Hi, Greg. Congratulations on getting the Forrest Mims book -- it's
probably the best total newbie start to learning electronics. You can
easily do what you're suggesting, and it won't short out the battery
(view in fixed font or cut&paste to M$ Notepad):
|
| .--------. .
| | |
| | .-. |
| | 750| | |I
| | | | |
| | '-' V
|9V | |
| +| o-----------.
| --- | |
| - .-. |
| | 150| | | _ |
| | | | |I1 / \ |I2
| | '-' | ( T ) |
| | | V \_/ V
| | | |
| | | |
| '--------o-----------'
|
(created by AACircuit v1.28.5 beta 02/06/05
www.tech-chat.de)
This will provide you with a steady 1.5V at the timer -- as long as it
doesn't use very much current.
Look at the circuit above. I = I1 + I2. If I2 is near zero, then
there's about 900 ohms load across your battery, and I = 10mA (Ohms Law
-- 9V / 900 ohms = 0.01 amps = 10mA). If you've got one of those small
LCD watch battery clocks, they generally use about 1mA of current.
That won't affect the voltage divider too much -- a 1mA timer load will
bring the divider voltage down to 1.35V or so. You should still be OK
here.
From a practical standpoint, I'd assume the current draw from the
little timer isn't quite steady -- it probably requires more current in
very short pulses for the timer crystal and the display IC. This is
probably going to interfere with the timer operation. If you want to
use this circuit, you really should go to Radio Shack and purchase a
10uF (microfarad) cap, and put it in parallel with the 150 ohm resistor
like this:
|
| .--------. .
| | |
| | .-.
| | 750| |
| | | |
| | '-'
|9V | |
| +| o-----o-----.
| --- | | |
| - .-. | |
| | 150| | +| _
| | | | --- / \
| | '-' --- ( T )
| | | 10uF| \_/
| | | | |
| | | | |
| '--------o-----o-----'
|
(created by AACircuit v1.28.5 beta 02/06/05
www.tech-chat.de)
Be sure to attach the lead of the cap marked "+" on the side of the
circuit where the 750 ohm and 150 ohm resistors meet. This will work
for a fleapower load. But there are two problems here.
First, if your little clock has a piezo alarm, current draw on the 1.5V
is going to go way up when the alarm goes off. That will bring the
voltage way down, and probably stop your clock. Even worse, if you've
got one of those quartz clocks, their average current draw is more like
10mA, which is completely unacceptable, alarm or not. Not only that,
but the quartz clock draws current in spikes at the moment the hands
move, which will ruin things for sure.
The second problem is also a biggie. For the above circuit, you're
burning 10mA to get a useful 1mA at 1.5V with 10% regulation. In other
words, you're using 90 milliwatts of power (9V * 10mA) of power to get
a useful 1.5 milliwatt (1.5V * 1mA). Your batteries aren't going to
last nearly as long as they could.
If this is just an experiment, you should be OK here. If you're
planning on running this permanently and don't own stock in Ray O Vac,
you can get an IC from Radio Shack that will make your voltage
regulator more efficient. This IC is required if you've got a quartz
clock -- you might be able to make the ohms of the resistors a lot
lower to get better regulation for a larger load, but the 9V battery
can't crank that much current.
If you'd like, please feel free to post again for more advice. Be sure
to describe what kind of timer you're using, how much current it needs
if you know, and also if it has an alarm output and what type.
Good luck
Chris