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Voltage Divider Help Please.

R

royalmp2001

Jan 1, 1970
0
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
 
P

peterken

Jan 1, 1970
0
royalmp2001 said:
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

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" ?
 
R

Robert Monsen

Jan 1, 1970
0
royalmp2001 said:
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.

--
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.
 
J

John Popelish

Jan 1, 1970
0
royalmp2001 said:
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

How accurate must the 0.25 volts be? Might a schottky diode
paralleled with a capacitor under the 555 do? A rail to rail opamp
and voltage reference chip could be combined to make a 0.25 volt
regulator with a few milliamps capability, but it is getting
complicated. Can you elaborate on why the output needs this offset?
There may be a better solution from that end.
 
J

John Fields

Jan 1, 1970
0
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

---
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
 
R

Robert Monsen

Jan 1, 1970
0
Robert said:
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.

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.
 
J

John Fields

Jan 1, 1970
0
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 ^^^^
0.25
 
R

royalmp2001

Jan 1, 1970
0
The project is called a ZAPPER pioneered by Dr Hudda Clark and is a
pseudo-medical experimental device that supplies a 39KHz square wave
with 0.25V positive offset to two handhold electrodes, which the
patient holds for a certain duration.

The claim is that parasites, bacteria, molds, and viruses in the body
are killed by this operation.

The circuit is a simple 7555 producing the required waveform from a 9V
battery, so there is no additional circuitry after the 7555 output.

PLEASE PLEASE...How do I add biasing .25V to the output of the 7555?
Please tell....
Thanks
 
R

Roger Johansson

Jan 1, 1970
0
royalmp2001 said:
How do I add biasing .25V to the output of the 7555?

Put a big capacitor in series with the output. Now you have a new output
which is not locked in any way DC-wise.

Put two resistors in series between your positive and negative supply
lines, choose the resistor so that the DC voltage in the point between
the resistors is .25V.

Connect the DC-less output with the 0.25V DC "generator" you you built
from two resistors. The result is an output signal with a certain AC
signal, and a certain DC average level.

Both signals can be controlled separately from each other and are added
to each other at the output.
 
P

peterken

Jan 1, 1970
0
Roger Johansson said:
Put a big capacitor in series with the output. Now you have a new output
which is not locked in any way DC-wise.

Put two resistors in series between your positive and negative supply
lines, choose the resistor so that the DC voltage in the point between
the resistors is .25V.

Connect the DC-less output with the 0.25V DC "generator" you you built
from two resistors. The result is an output signal with a certain AC
signal, and a certain DC average level.

Both signals can be controlled separately from each other and are added
to each other at the output.

that's one way, BUT dependant of supply level unless the voltage divider is
created using a V-reference
 
B

Bob Masta

Jan 1, 1970
0
The project is called a ZAPPER pioneered by Dr Hudda Clark and is a
pseudo-medical experimental device that supplies a 39KHz square wave
with 0.25V positive offset to two handhold electrodes, which the
patient holds for a certain duration.

The claim is that parasites, bacteria, molds, and viruses in the body
are killed by this operation.

Sorry, I can't let this go past.

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.

Consider that the signal is going between the two hands,
ie right across the heart. Do you want to be responsible
for any unfortunate outcomes? Consider that this design
was "pioneered" by an obvious quack with little knowledge
of biology, let alone medicine. Not the sort of person to
trust with safety or health issues.

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com
 
R

Roger Johansson

Jan 1, 1970
0
peterken said:
that's one way, BUT dependant of supply level unless the voltage
divider is created using a V-reference

Yes, of course the "DC generator" can be improved, if needed, through
lower resistors to give lower impedance, or by buffering it with an
opamp, or by creating a voltage which is better regulated, if needed,
etc..
 
J

Jonathan Kirwan

Jan 1, 1970
0
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.

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
 
J

JeffM

Jan 1, 1970
0
The project is called a ZAPPER pioneered by Dr Hudda Clark
This is quack medicine at its finest.
Bob Masta

It seems he's aware of that.
We can hope that it's royalmp2001's entry into the science fair
with the goal being to prove it's junk science.

OTOH, maybe he's an entrepreneur who's familiar with Barnum.

I google Dr. Dean Edell
and was surprized when this didn't send up flares.
http://www.google.com/search?&q=site:healthcentral.com+Hudda-Clark
 
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