If your 4.5V is being supplied from three 1.5V batteries, you already
have the 1.5V available. All you have to do is make the connection on
the other side of the battery closest to the "minus" end of the output,
and you're done. Total cost: zero.
If your setup isn't like that, and you just need a steady 1.5V, and
aren't drawing any load current (or a very small amount) from your 1.5V
source, you can just use a resistive voltage divider, like this (view
in fixed font or M$ Notepad):
| .--------.
| | |
| | .-.
| | | |2K
| | | |
| +|4.5V '-'
| --- | 1.5V
| - o-----o
| | |
| | .-.
| | | |1K
| | | |
| | '-'
| | |
| '--------'
(created by AACircuit v1.28.6 beta 04/19/05
www.tech-chat.de)
If you need to draw some current (even 1mA) from the 1.5V source,
though, that will mess up the voltage divider and lower your voltage.
By using lower value resistors (again, in the 2:1 ratio), you can
increase the current you have available a little, at an enormous
increase in the power you'll need to waste. Total cost: $0.02 to $0.20
USD.
If you can live with something close (say, 1.3V to 1.5V), you can just
use two diodes and a series resistor to get the lower voltage, like
this:
|
| .-------.
| | |
| | .-.
| | | |33 ohm 1/2 watt
| | | |
| | '-'
| +|4.5V | 1.5V
| --- o------o
| - |
| | V 1N4001
| | -
| | |
| | V 1N4001
| | -
| | |
| '-------'
|
(created by AACircuit v1.28.6 beta 04/19/05
www.tech-chat.de)
This will give you a fairly steady 1.4V or so with a load current
between zero and 90mA. It's probably the simplest way to get a lower
battery voltage, but it's very wasteful of power. You will draw 90mA
from your 4.5V source no matter what your 1.5V load. Of course, if you
need less current, you can bump up the 33 ohm resistor to a higher
value. If all you'll need is a maximum of 30mA. you can make the
resistor a 100 ohm 1/4 watt resistor and save 2/3 of the current.
Total cost: $0.20 to $0.75 USD.
But again, this is very wasteful of the 4.5V power. A more complex and
somewhat more expensive way to do this is to use the LM317 linear
voltage regulator IC, along with a small capacitor and two resistors.
This setup will only waste 6mA, and will provide a steady, precise 1.5V
for a load current of zero to 1/4 amp, and over 1 amp if you use a heat
sink on the IC. Total cost: $1.00 to $4.00 USD.
To determine which is best, you could provide more information. Is the
4.5V from a battery, or a series of three batteries? What kind of load
current do you need from the 1.5V "bypass"? If you don't know, could
you at least describe the load or what you're doing?
Cheers
Chris
