# 1.5V bypass

Discussion in 'Electronic Basics' started by [email protected], Jan 8, 2007.

1. ### Guest

I have a 4.5V power supply, and I need make a 1.5 V bypass.
How can I do this ?

Regards.

2. ### Charles SchulerGuest

I am not familiar with your use of the word bypass. Do you mean a second
output at 1.5 volts? If so, a simple shunt regulator using two silicon
rectifiers in series might do it.

3. ### JeffMGuest

Yup. Using more words to describe the concept
would have been useful.
You've used words clumbsily as well.
You think that a guy who has this much trouble
clearly describing what he needs can decypher "shunt"?

The diodes are in series WITH EACH OTHER
but are in PARALLEL with the load.

There is also a resistor between the original supply and the diodes.
With no further details about what the OP is doing,
the resistor's value can not be stated.

View in a monospaced font like Courier.
(Click "Fixed font" at the top of the Google Groups page.)

---/\/\--+--- OUT
¦
_V_
¦
_V_
¦
---------+---

4. ### PeteSGuest

I'll admit the biggest impediment to the transfer of knowledge around
here is the inability to write clearly and concisely.

Cheers

PeteS

5. ### Charles SchulerGuest

Yup. Using more words to describe the concept
would have been useful.
You've used words clumbsily as well.
You think that a guy who has this much trouble
clearly describing what he needs can decypher "shunt"?

Clumbsily? Googling for "shunt regulator" yields a ton of information that
I have little interest in regurgitating here.

The diodes are in series WITH EACH OTHER
but are in PARALLEL with the load.

There is also a resistor between the original supply and the diodes.
With no further details about what the OP is doing,
the resistor's value can not be stated.

Oh, you are indeed a sage among mere strugglers!

Piss off and PLONK!

6. ### ChrisGuest

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
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

7. ### ehsjrGuest

"clumbsily" ???

pot/kettle

8. ### defaultGuest

This guy is posting through Google Groups. If one is used to
AOL,Earthlink, etc., there's a tendency to treat Usenet like some AOL
bulletin board - or chat room - or myspace.

Google, for their part, does nothing to educate or dissuade one from
that conclusion - no mention of what Usenet is, its history, or
etiquette. They even encourage you to put in a personal profile -

Half the time, the original poster can't remember what group he posted
to, or was asking a question he had only a passing interest in -
attention span limitations, or is too clueless to understand the

9. ### Rich GriseGuest

Yes, it makes the sentence clumbersome. ;-)

Cheers!
Rich

10. ### Guest

Chris escreveu:
Thanks a lot, Chris. I'm using your circuit below with total success !

| .--------.
| | |
| | .-.
| | | |2K
| | | |
| +|4.5V '-'
| --- | 1.5V
| - o-----o
| | |
| | .-.
| | | |1K
| | | |
| | '-'
| | |
| '--------'

Regards !