Simple DC-DC voltage reduction

[Simon Morden]

Hello again. I'm the bloke who was on a few months ago worry about his
frequency-voltage converter. Which now works magnificently, and we have
one wind tunnel up and running.

So - on to Mk II. I intend to put all the components on one circuit
board, with one power source: a regulated 9v DC supply with a max power
of 350mA.

The IR LED/sensor works at 4.5v, the f-v chip at 9v. Is there anything
to stop me from slapping down a variable resistor and taking the 4.5v
from it?

If not, how ought I calculate the value of R I need?
If yes, how should I do it?

Simon Morden
--
Visit the *all new* Book of Morden (www.bookofmorden.co.uk)
"I haven't had that much fun with a novel for a while." - Bookbag
The Lost Art - from David Fickling Books

 

[petrus bitbyter]

"Simon Morden" <simon.morden@spamtastic.blueyonder.co.uk> schreef in bericht
news:aLA0l.1997$wp1.804@newsfe09.ams2...

Hello again. I'm the bloke who was on a few months ago worry about his
frequency-voltage converter. Which now works magnificently, and we have
one wind tunnel up and running.

So - on to Mk II. I intend to put all the components on one circuit board,
with one power source: a regulated 9v DC supply with a max power of 350mA.

The IR LED/sensor works at 4.5v, the f-v chip at 9v. Is there anything to
stop me from slapping down a variable resistor and taking the 4.5v from
it?

If not, how ought I calculate the value of R I need?
If yes, how should I do it?

Simon Morden
--
Visit the *all new* Book of Morden (www.bookofmorden.co.uk)
"I haven't had that much fun with a novel for a while." - Bookbag
The Lost Art - from David Fickling Books

You forgot an important piece of information: The current required by the IR
LED/sensor. So I can only guess it to be some tens of mA with some
variations depending on the status of the components and the environment.

A resistor is not the way to go as it cannot provide a reliable, stable
voltage. A variable resistor is worse as the common types, trimpots, can
dissipate only a few mW so you may fry it pretty fast. The quick and easy
way is using an 7805 type voltage regulator and one or two capacitors.

petrus bitbyter

 

[Jamie]

Simon Morden wrote:

Hello again. I'm the bloke who was on a few months ago worry about his
frequency-voltage converter. Which now works magnificently, and we have
one wind tunnel up and running.

So - on to Mk II. I intend to put all the components on one circuit
board, with one power source: a regulated 9v DC supply with a max power
of 350mA.

The IR LED/sensor works at 4.5v, the f-v chip at 9v. Is there anything
to stop me from slapping down a variable resistor and taking the 4.5v
from it?

If not, how ought I calculate the value of R I need?
If yes, how should I do it?

Simon Morden
if that is a integrated IR sensor, using a R to drop the voltage

may make it unstable.
Try using a 78L05 regulator with a small diode in series on the
output to give you a ~0.5V drop and place a small cap like a 0.1 uf
across the diode..


http://webpages.charter.net/jamie_5"

 

[Simon Morden]

Jamie wrote:

if that is a integrated IR sensor, using a R to drop the voltage
may make it unstable.
Try using a 78L05 regulator with a small diode in series on the
output to give you a ~0.5V drop and place a small cap like a 0.1 uf
across the diode..

Many thanks to yourself and Petrus - it's not just working out what the
question is, it's working out whether there's a question in the first place!

My local supplier doesn't do the National 78Lxx, but does do the TS78Lxx
series. The IR sensor draws 50mA, but will work just as fine off 5v as
it will 4.5v - different resistor required is all.

Again, thanks for all your help.

Simon Morden
--
Visit the *all new* Book of Morden (www.bookofmorden.co.uk)
"I haven't had that much fun with a novel for a while." - Bookbag
The Lost Art - from David Fickling Books

 

[Robert Monsen]

On Fri, 12 Dec 2008 21:28:39 +0000, Simon Morden
<simon.morden@spamtastic.blueyonder.co.uk> wrote:

Hello again. I'm the bloke who was on a few months ago worry about his
frequency-voltage converter. Which now works magnificently, and we have
one wind tunnel up and running.

So - on to Mk II. I intend to put all the components on one circuit
board, with one power source: a regulated 9v DC supply with a max power
of 350mA.

The IR LED/sensor works at 4.5v, the f-v chip at 9v. Is there anything
to stop me from slapping down a variable resistor and taking the 4.5v
from it?

If not, how ought I calculate the value of R I need?
If yes, how should I do it?

Simon Morden

If you use a trimmer, follow it with an NPN transistor like this:


9V -------o-------.
| |
| |
| |
.-. |/
pot | |<---| NPN
| | |>
'-' |
| '---o sensor supply +
|
|
Gnd ------o-----------o sensor supply -
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

This is called a voltage follower configuration. The base is
connected to the pot, the collector to the 9V supply, and the
emitter is connected to the positive (4.5V) supply input for
the sensor. The current you can pull from the emitter is like
100x the current you can pull from the pot without perturbing
the voltage significantly. You can adjust the pot to the appropriate
voltage.

One problem is that if the 9V supply is not regulated, you can get a
proportional droop in the output, so using a little regulator like
petrius mentioned could be better. You might also use a 5.1V zener
like this:

9V ---------o-----.
| |
.-. |
| | |
1k | | |
'-' |
| |/
o---|
| |>
5.1V z |
A '---o Sensor supply +
|
|
Gnd -------o---------o Sensor supply -
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

That will keep the output voltage very close to 4.5V.

Regards,
Bob Monsen

 

[Varactor]

On Dec 14, 5:04 am, Simon Morden
<simon.mor...@spamtastic.blueyonder.co.uk> wrote:

Jamie wrote:

 if that is a integrated IR sensor, using a R to drop the voltage
 may make it unstable.
   Try using a 78L05 regulator with a small diode in series on the
output to give you a ~0.5V drop and place a small cap like a 0.1 uf
 across the diode..

Many thanks to yourself and Petrus - it's not just working out what the
question is, it's working out whether there's a question in the first place!

My local supplier doesn't do the National 78Lxx, but does do the TS78Lxx
series. The IR sensor draws 50mA, but will work just as fine off 5v as
it will 4.5v - different resistor required is all.

Again, thanks for all your help.

Simon Morden
--
Visit the *all new* Book of Morden (www.bookofmorden.co.uk)
"I haven't had that much fun with a novel for a while." - Bookbag
The Lost Art - from David Fickling Books

Try a 4.5 V 300mW Zener to drop the voltage?

Cheers