Voltage Divider Help Please.

R

royalmp2001

Guest
I need to build a voltage divider for a 9v battery that will produce
0.25v on one leg and the rest on the other leg.

The 0.25V must not change even if battery voltage drops, so a pure
resistive divider is not suitable.... and I don't think there are 0.25V
Zeners available. Perhaps something with an ordinary diode will do the
trick but that is still 0.7V I think.

Any help??? It must be as simple and cheap as possible. Total load
current drawn from the configuration will be no more than 4mA.

My application is a CMOS 555 set up as an oscillator but I need the
square waveform to be offset from ground by 0.25V. So I guess some kind
of divider is necessary so that the chip runs on 8.75V supply from the
divider, leaving the ground rail 0.25V below this.
Thanks....hope this all makes sense
 
"royalmp2001" <royalmp2001@hotpop.com> wrote in message
news:1106857818.991823.183350@c13g2000cwb.googlegroups.com...
I need to build a voltage divider for a 9v battery that will produce
0.25v on one leg and the rest on the other leg.

The 0.25V must not change even if battery voltage drops, so a pure
resistive divider is not suitable.... and I don't think there are 0.25V
Zeners available. Perhaps something with an ordinary diode will do the
trick but that is still 0.7V I think.

Any help??? It must be as simple and cheap as possible. Total load
current drawn from the configuration will be no more than 4mA.

My application is a CMOS 555 set up as an oscillator but I need the
square waveform to be offset from ground by 0.25V. So I guess some kind
of divider is necessary so that the chip runs on 8.75V supply from the
divider, leaving the ground rail 0.25V below this.
Thanks....hope this all makes sense
Click to expand...
Why would you need the zero-offset of .25V ?
Is the trailing circuit instable if it's input is below .25V ?
Then solve that first!

If not, ever thought of "adding" (biasing) .25V to the output of the 555
itself instead of creating the more difficult solution having the 555
"floating" ?
 
royalmp2001 wrote:
I need to build a voltage divider for a 9v battery that will produce
0.25v on one leg and the rest on the other leg.

The 0.25V must not change even if battery voltage drops, so a pure
resistive divider is not suitable.... and I don't think there are 0.25V
Zeners available. Perhaps something with an ordinary diode will do the
trick but that is still 0.7V I think.

Any help??? It must be as simple and cheap as possible. Total load
current drawn from the configuration will be no more than 4mA.

My application is a CMOS 555 set up as an oscillator but I need the
square waveform to be offset from ground by 0.25V. So I guess some kind
of divider is necessary so that the chip runs on 8.75V supply from the
divider, leaving the ground rail 0.25V below this.
Thanks....hope this all makes sense
Click to expand...
An easy way is to use a voltage divider on the 555 output, and then a
voltage follower. This will give you a relatively stiff voltage, but it
won't be very accurate.

VCC
+
|
555 |
__ |
-o|1 |o-----o
-o| |o- |
.-----o| |o- |
| -o|__|o- |
| |
| |
| |
.-. |/
| |<------------|
| | |>
'-' | Output
| '---------------
|
===
GND
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

Adjust the pot so that the output is 0.25 when the output is high. Use a
pot that is reasonably low resistance... a 1k pot makes sense. Then, the
output impedance will be at most 1/10 of that, maybe less.

A better buffer would be the following:

VCC
+
|
555 |
__ |
-o| 1|o-----o----------.
-o| |o- | |
.-----o| |o- | |
| -o|__|o- | | Output
| | |
| |\| |
| .---|-\ |/
.-. | | >-------|
| |---------)---|+/ |>
| | | |/| |
'-' | | |
| '-----)----------o
| | |
=== === '-------------
GND GND

(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

This one is nicer because it doesn't depend on the Vbe of the
transistor, which will change due to load, temperature, etc.

The opamp should be a 'single supply' opamp that can get down to near 0,
like an LM324. The LM324 might be too slow, though. Check the datasheet.

This time, the POT can be higher resistance, maybe 100k.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
 
On 27 Jan 2005 12:30:19 -0800, "royalmp2001" <royalmp2001@hotpop.com>
wrote:

I need to build a voltage divider for a 9v battery that will produce
0.25v on one leg and the rest on the other leg.

The 0.25V must not change even if battery voltage drops, so a pure
resistive divider is not suitable.... and I don't think there are 0.25V
Zeners available. Perhaps something with an ordinary diode will do the
trick but that is still 0.7V I think.

Any help??? It must be as simple and cheap as possible. Total load
current drawn from the configuration will be no more than 4mA.

My application is a CMOS 555 set up as an oscillator but I need the
square waveform to be offset from ground by 0.25V. So I guess some kind
of divider is necessary so that the chip runs on 8.75V supply from the
divider, leaving the ground rail 0.25V below this.
Thanks....hope this all makes sense
Click to expand...
---
Depending on what your load looks like, you might be able to do
something like this:


+9V
|
[R1]
|
+-----+-- 2.5V
| |
[REF1] |
| [R2]
GND |
|
555 OUT>--[1N4148>]--+
|
[LOAD]
|
GND

That way you wouldn't have to run the 555 above ground and the
reference would keep the low input to your load at 2.5V while the
battery aged

--
John Fields
 
Robert Monsen wrote:
royalmp2001 wrote:

I need to build a voltage divider for a 9v battery that will produce
0.25v on one leg and the rest on the other leg.

The 0.25V must not change even if battery voltage drops, so a pure
resistive divider is not suitable.... and I don't think there are 0.25V
Zeners available. Perhaps something with an ordinary diode will do the
trick but that is still 0.7V I think.

Any help??? It must be as simple and cheap as possible. Total load
current drawn from the configuration will be no more than 4mA.

My application is a CMOS 555 set up as an oscillator but I need the
square waveform to be offset from ground by 0.25V. So I guess some kind
of divider is necessary so that the chip runs on 8.75V supply from the
divider, leaving the ground rail 0.25V below this.
Thanks....hope this all makes sense


An easy way is to use a voltage divider on the 555 output, and then a
voltage follower. This will give you a relatively stiff voltage, but it
won't be very accurate.

VCC
+
|
555 |
__ |
-o|1 |o-----o
-o| |o- |
.-----o| |o- |
| -o|__|o- |
| |
| |
| |
.-. |/
| |<------------|
| | |
'-' | Output
| '---------------
|
===
GND
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

Adjust the pot so that the output is 0.25 when the output is high. Use a
pot that is reasonably low resistance... a 1k pot makes sense. Then, the
output impedance will be at most 1/10 of that, maybe less.

A better buffer would be the following:

VCC
+
|
555 |
__ |
-o| 1|o-----o----------.
-o| |o- | |
.-----o| |o- | |
| -o|__|o- | | Output
| | |
| |\| |
| .---|-\ |/
.-. | | >-------|
| |---------)---|+/ |
| | | |/| |
'-' | | |
| '-----)----------o
| | |
=== === '-------------
GND GND

(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

This one is nicer because it doesn't depend on the Vbe of the
transistor, which will change due to load, temperature, etc.

The opamp should be a 'single supply' opamp that can get down to near 0,
like an LM324. The LM324 might be too slow, though. Check the datasheet.

This time, the POT can be higher resistance, maybe 100k.
Click to expand...
Sorry, I wasn't paying attention. This is obviously answering the wrong
question.

Use a 2V voltage regulator to power the C555. You can get low dropout
regulators that only require 150mV or so headroom, like the sanyo S-812C
series. A 317 will do in a pinch, but only if your input voltage is
above 5.5V.

Then, use a voltage divider into a buffer, like either of the above. The
pass transistor doesn't have to go through the regulator.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
 
On Sat, 29 Jan 2005 14:12:25 GMT, NoSpam@daqarta.com (Bob Masta) wrote:

This is quack medicine at its finest. I'll bet that it also cures
cancer and HIV, as well as neutralizing fat while you sleep.
Click to expand...
And it may even be harmless, for all we know. Like crystal therapy. As well as
being just as useless.

It's probably written up so that folks have to buy some exact product or else
"it won't work right." Scientologists use a special galvanic meter of sorts,
for example. Completely worthless claims for it. But they say that if you
don't use the exact and precise method they do, it won't provide the desired
information about your 'clarity.' So folks either have to pay for a service
using a "sanctified device" or else buy one of their own. Either way, the
controlling group gets the revenue.

That .25V component may be just that "special something" that makes it very
different from what you can pick up off the shelf. And if you make your own
design that provides this .25V, since they don't specify it too precisely, those
claimants can always say that you didn't provide the "right kind" of .25V
offset! (Or the right kind of 39kHz!)

Oh, well.

Jon
 
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