# dropping a variable voltage by 1v

R

#### Ray Manning

Jan 1, 1970
0
I'd like to offset a variable voltage by 1 volt. What I have is an LM34
temperature sensor that outputs a linear voltage based on temperature
(+10mV/F). I'd like to change that to (+10mV/F - 1V). What's the easiest way
to drop the LM34's output by 1V?

I tried a potentiometer but my understanding is that a pot is a divider so
it doesn't give me the result I'm looking for. I also tried a diode that was
supposed to have a 1V drop but that didn't seem to work either (not sure
why). Is there a simple solution?

Thanks,
Ray

J

#### John Popelish

Jan 1, 1970
0
Ray said:
I'd like to offset a variable voltage by 1 volt. What I have is an LM34
temperature sensor that outputs a linear voltage based on temperature
(+10mV/F). I'd like to change that to (+10mV/F - 1V). What's the easiest way
to drop the LM34's output by 1V?

I tried a potentiometer but my understanding is that a pot is a divider so
it doesn't give me the result I'm looking for. I also tried a diode that was
supposed to have a 1V drop but that didn't seem to work either (not sure
why). Is there a simple solution?

Thanks,
Ray
An opamp with 4 resistors configured as a subtracter (difference
amplifier) outputs the difference of two voltages plus a third
voltage. A 2 opamp version has two of the inputs with high impedance,

http://www.national.com/an/AN/AN-20.pdf
This opamp might be a good choice:
http://www.national.com/an/AN/AN-116.pdf

C

#### Chris

Jan 1, 1970
0
Ray said:
I'd like to offset a variable voltage by 1 volt. What I have is an LM34
temperature sensor that outputs a linear voltage based on temperature
(+10mV/F). I'd like to change that to (+10mV/F - 1V). What's the easiest way
to drop the LM34's output by 1V?

I tried a potentiometer but my understanding is that a pot is a divider so
it doesn't give me the result I'm looking for. I also tried a diode that was
supposed to have a 1V drop but that didn't seem to work either (not sure
why). Is there a simple solution?

Thanks,
Ray

Hi, Ray. Get an op amp and four equal value resistors (match values
with your ohmmeter -- the closer they are, the better), and make a
difference amplifier like this (view in fixed font or M\$ Notepad):
___
.-|___|-.
| 22K |
| |
___ | |
Vin o--|___|---o VCC |
22K | |\| |
'-|-\ |Vout =
___ | >--o----o
1VDCo--|___|-o---|+/ |
22K | |/| |
.-. GND .-.
22K| | 22K| |(not matched)
| | | |
'-' '-'
| |
| |
=== ===
GND GND
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

This circuit will do the job for you. If you don't need anything below
about +50mV DC output from the op amp, you can use a single supply and
half of an LM358, or 1/4 of an LM324 (available at Radio Shack and many
other places). If you need a negative voltage output, you'll need a
dual supply (say, +/-5V) and you can use an LM741 or any other standard
op amp.

Good luck
Chris

B

#### Bob Masta

Jan 1, 1970
0
I'd like to offset a variable voltage by 1 volt. What I have is an LM34
temperature sensor that outputs a linear voltage based on temperature
(+10mV/F). I'd like to change that to (+10mV/F - 1V). What's the easiest way
to drop the LM34's output by 1V?

I tried a potentiometer but my understanding is that a pot is a divider so
it doesn't give me the result I'm looking for. I also tried a diode that was
supposed to have a 1V drop but that didn't seem to work either (not sure
why). Is there a simple solution?

Sure, you just need an op-amp difference circuit. One input
is your signal, the other is 1V, and the output is the difference.
You will need a known voltage source, but it doesn't have to be
1.00 V or even close, since you can scale it with the input resistors
or a separate divider. You can even use a negative reference and
sum it with the LM34 to get a difference.

Best regards,

Bob Masta

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Home of DaqGen, the FREEWARE signal generator

J

#### John Fields

Jan 1, 1970
0
I'd like to offset a variable voltage by 1 volt. What I have is an LM34
temperature sensor that outputs a linear voltage based on temperature
(+10mV/F). I'd like to change that to (+10mV/F - 1V). What's the easiest way
to drop the LM34's output by 1V?

C

#### Chris

Jan 1, 1970
0
Ray said:
I'd like to offset a variable voltage by 1 volt. What I have is an LM34
temperature sensor that outputs a linear voltage based on temperature
(+10mV/F). I'd like to change that to (+10mV/F - 1V). What's the easiest way
to drop the LM34's output by 1V?

I tried a potentiometer but my understanding is that a pot is a divider so
it doesn't give me the result I'm looking for. I also tried a diode that was
supposed to have a 1V drop but that didn't seem to work either (not sure
why). Is there a simple solution?

Thanks,
Ray

Oops. Wrong labels -- try this:

___
.-|___|-.
| 22K |
| |
___ | |
1VDCo--|___|---o VCC |
22K | |\| |
'-|-\ |Vout =
___ | >--o----o
Vin o--|___|-o---|+/ |
22K | |/| |
.-. GND .-.
22K| | 22K| |(not matched)
| | | |
'-' '-'
| |
| |
=== ===
GND GND
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

Inputs backwards -- sorry.

Chris

R

#### Ray Manning

Jan 1, 1970
0
John,

Are you suggesting that if I applied -1v to the GND terminal of the LM34,
the output would be offset by -1 volt? If I understand correctly, this would
work because the output would still be referenced against the original
ground. If thats the case, can I produce the -1V using the input +Vs (12v in
this case) and the orginal ground? Wouldn't that solution (offseting the
input) be the same as offseting the output or am I missing something?

Thanks,
Ray

J

#### John Fields

Jan 1, 1970
0
John,

Are you suggesting that if I applied -1v to the GND terminal of the LM34,
the output would be offset by -1 volt?
---
Yes.
---

If I understand correctly, this would
work because the output would still be referenced against the original
ground.

---
If by that you mean that the thing connected to the LM34 has as it's
ground, 0V, then yes.
---
If thats the case, can I produce the -1V using the input +Vs (12v in
this case) and the orginal ground? Wouldn't that solution (offseting the
input) be the same as offseting the output or am I missing something?

---
You're missing that the LM34 doesn't care what the supply voltage is
(within limits) since it only uses that to run the chip. Its
output, however, is referenced to the "holy" ground pin. That means
that at 0°F the output will be at whatever voltage the ground
terminal is at and will go more positive by 10 millivolts for every
degree F change in ambient temperature. That is, if the ground
terminal is at 0V, the output will be at 0V and will go to 1V for
the change from 0°F to 100°F, but if the ground terminal is at -1V
when the temperature is 0° the output will be at -1V and will rise
to 0V as the temperature goes to 100°F. _If_ you measure that
voltage with a voltmeter referenced to the junction of the two
supplies.

Look at it like this: (in Courier)

+------------------+
| |+
| [12V]
| |
[LM34]--[VOLTMETER]--+
| |+
| [1V]
| |
+------------------+

R

#### Ray Manning

Jan 1, 1970
0
John Fields said:
Are you suggesting that if I applied -1v to the GND terminal of the LM34,
the output would be offset by -1 volt?
---
Yes.
---

If I understand correctly, this would
work because the output would still be referenced against the original
ground.

---
If by that you mean that the thing connected to the LM34 has as it's
ground, 0V, then yes.
---
If thats the case, can I produce the -1V using the input +Vs (12v in
this case) and the orginal ground? Wouldn't that solution (offseting the
input) be the same as offseting the output or am I missing something?

---
You're missing that the LM34 doesn't care what the supply voltage is
(within limits) since it only uses that to run the chip. Its
output, however, is referenced to the "holy" ground pin. That means
that at 0°F the output will be at whatever voltage the ground
terminal is at and will go more positive by 10 millivolts for every
degree F change in ambient temperature. That is, if the ground
terminal is at 0V, the output will be at 0V and will go to 1V for
the change from 0°F to 100°F, but if the ground terminal is at -1V
when the temperature is 0° the output will be at -1V and will rise
to 0V as the temperature goes to 100°F. _If_ you measure that
voltage with a voltmeter referenced to the junction of the two
supplies.

Look at it like this: (in Courier)

+------------------+
| |+
| [12V]
| |
[LM34]--[VOLTMETER]--+
| |+
| [1V]
| |
+------------------+

I understand what you are saying but still don't see how to accomplish it.
The curcuit that is driving the LM34 hands me the ground reference (0V) and
a +12V source. How do I produce the negative voltage to connect to the LM34
to change its reference point?

It would seem to me that I would have to use the same difference amplifier
others have suggested for dropping the output by 1V to create the negative
1V for the "holy" ground pin. Are you saying there is an easier way?

Thanks,
Ray

J

#### John Fields

Jan 1, 1970
0
John Fields said:
Are you suggesting that if I applied -1v to the GND terminal of the LM34,
the output would be offset by -1 volt?
---
Yes.
---

If I understand correctly, this would
work because the output would still be referenced against the original
ground.

---
If by that you mean that the thing connected to the LM34 has as it's
ground, 0V, then yes.
---
If thats the case, can I produce the -1V using the input +Vs (12v in
this case) and the orginal ground? Wouldn't that solution (offseting the
input) be the same as offseting the output or am I missing something?

---
You're missing that the LM34 doesn't care what the supply voltage is
(within limits) since it only uses that to run the chip. Its
output, however, is referenced to the "holy" ground pin. That means
that at 0°F the output will be at whatever voltage the ground
terminal is at and will go more positive by 10 millivolts for every
degree F change in ambient temperature. That is, if the ground
terminal is at 0V, the output will be at 0V and will go to 1V for
the change from 0°F to 100°F, but if the ground terminal is at -1V
when the temperature is 0° the output will be at -1V and will rise
to 0V as the temperature goes to 100°F. _If_ you measure that
voltage with a voltmeter referenced to the junction of the two
supplies.

Look at it like this: (in Courier)

+------------------+
| |+
| [12V]
| |
[LM34]--[VOLTMETER]--+
| |+
| [1V]
| |
+------------------+

I understand what you are saying but still don't see how to accomplish it.
The curcuit that is driving the LM34 hands me the ground reference (0V) and
a +12V source. How do I produce the negative voltage to connect to the LM34
to change its reference point?

It would seem to me that I would have to use the same difference amplifier
others have suggested for dropping the output by 1V to create the negative
1V for the "holy" ground pin. Are you saying there is an easier way?

R

#### Ray Manning

Jan 1, 1970
0
The LM34 is driven by a circuit that converts the LM34's voltage to a
frequency that is read by a remote computer. The computer supplies the
circuit with a ground and +12V and receives a frequency on a third wire. The
circuit in turn supplies the LM34 with the same ground and 12v supply and
monitors the output voltage of the LM34.

- Ray

J

#### Jasen Betts

Jan 1, 1970
0
I'd like to offset a variable voltage by 1 volt. What I have is an LM34
temperature sensor that outputs a linear voltage based on temperature
(+10mV/F). I'd like to change that to (+10mV/F - 1V). What's the easiest way
to drop the LM34's output by 1V?

drop it's the supply and "ground" by one volt.
I tried a potentiometer but my understanding is that a pot is a divider so
it doesn't give me the result I'm looking for. I also tried a diode that was
supposed to have a 1V drop but that didn't seem to work either (not sure
why). Is there a simple solution?

or something with an op-amp

in ----[100k]--.
| +|\
0V-----[100k]--+--| \
| >----+--- out
1V-----[100k]--+--| / |
| -|/ |
| |
`--[100k]--'

a silicon diode ansd two germanium diodes in series will
give approx 1V drop but it's terperature dependant...

J

#### John Fields

Jan 1, 1970
0
The LM34 is driven by a circuit that converts the LM34's voltage to a
frequency that is read by a remote computer. The computer supplies the
circuit with a ground and +12V and receives a frequency on a third wire. The
circuit in turn supplies the LM34 with the same ground and 12v supply and
monitors the output voltage of the LM34.

---
OK. So what you have is a voltage-to-frequency converter and an
LM34 wired like this:

CPU+12>---------+-------------------+
| |
+-----+-----+ +-----+-----+
| +12 | | +12 |
CPU fIN<--|fOUT Vin|<------|Vout TEMP| <~~~
| GND | | GND |
+-----+-----+ +-----+-----+
| VFC | LM34
CPU GND>--------+-------------------+

Where the LM34 outputs 10mV/F° starting with 0V at 0F, and what you
want is to offset the output of the LM34 to start at -1V for 0F and
rise at a rate of 10mV/F°?

If that's the situation, then you can either lower the ground of the
LM34 to -1V or raise the VFC's ground to +1V. Raising the VFC's
ground to 1V would be easier, but since its output would also be
referenced to the 1V "ground" it would probably be offset as well,
so that may not be a good idea. Can you tell us what you're using
for a VFC or post its spec's?

If you have -5V or -12V available at your CPU and you've got an
extra wire in the cable, you could generate the -1V like this:

CPU+12>---------+-------------------------+
| |
+-----+-----+ +-----+-----+
| +12 | | +12 |
CPU fIN<--|fOUT Vin|<------------|Vout TEMP| <~~~
| GND | | GND |
+-----+-----+ +-----+-----+
| VFC | LM34
| |
CPU GND>--+-----+------+-------+-[0.1µF]--+
| | | |
| [0.1µF] [R1] |
| | | |
[0.33µF] +-------+ |
| | | |
| | [R2] |
| +---+---+ | |
CPU-12>---+--------|IN OUT|---+----------+
+-------+
79L05

If you don't, then you could use something like a 7555 to build a
charge pump followed up with a regulator, like this:

CPU+12>------------+------------+------------+--+12
| | |
+-----+-----+ | +-----+-----+
| +12 | | | +12 |
CPU fIN<-----|fOUT Vin|<-----|------|Vout TEMP| <~~~
| GND | | | GND |
+-----+-----+ | +-----+-----+
| VFC | | LM34
CPU GND------------+ | |
| | |
GND | |
| |
+---[C1]---+------|------+ |
|K | | | |
[1N4001] [Rt] +---+---+ | |
| | | Vcc | | |
+----+ +--|TH OUT|--+ |
|- | | |___ _| |
[C2] | +-O|DIS R|O--+12 |
| | | | GND | |
GND | [Ct] +---+---+ |
| | |7555 |
| GND GND |
| |
| |
| 79L05 |
| +-------+ |
+-----|IN OUT|---+-------+
| +---+---+ | |
| | [R1] |
| | | |
[0.33µF] +-------+ [0.1µF]
| | | |
| [0.1µF] [R2] |
| | | |
+---------+-------+-------+
|
GND

If you're interested in either of these approaches I'll be happy to
post the component values.

Hmmm... After looking at the spec's for the 79L05 I think the lowest
output available from it is 1.2V, and I couldn't find any 1V shunt
regulators or adjustable negative regulators with outputs which went
below 1.2V, (In all fairness, I didn't look that hard ) so in
order to keep from having to use a Zener and a voltage divider
(which might be OK depending on how tight your -1V spec is) it looks
like the charge pump and an opamp with a reference and its output
forced to -1V with proper biasing will work. Depending on the
current requirement for the LM34's output it might be possible to
use a chip with an opamp, a comparator, and a reference in it to
build the whole thing.

In particular, what are you using for your voltage-to-frequency
converter?

J

#### Jamie

Jan 1, 1970
0
Ray said:
The LM34 is driven by a circuit that converts the LM34's voltage to a
frequency that is read by a remote computer. The computer supplies the
circuit with a ground and +12V and receives a frequency on a third wire. The
circuit in turn supplies the LM34 with the same ground and 12v supply and
monitors the output voltage of the LM34.

- Ray
Hmm, you could use a dual osc, one that generates your freq to the
computer and the other to generate a small low power - source for the
Vee (common of the LM34).
using a couple of caps and 2 diodes and a divider for the common to
have the common of the LM34 connect to this -1 output.
since there is very little current involved here i think you could get
away with it.
i don't know if your using a mpu or what but, either an extra osc or
a clocked output can generate the constant you need for the - voltage
source.

J

#### Jasen Betts

Jan 1, 1970
0
The LM34 is driven by a circuit that converts the LM34's voltage to a
frequency that is read by a remote computer. The computer supplies the
circuit with a ground and +12V and receives a frequency on a third wire. The
circuit in turn supplies the LM34 with the same ground and 12v supply and
monitors the output voltage of the LM34.

it may be possible to modify the VCO (voltage to frequency circuit)
by raising its ground by one volt

OTOH can you account for the 1V offset by modifying the software to read the
frequencies differently?

J

#### Jamie

Jan 1, 1970
0
John said:
"John Fields" <[email protected]> wrote in message
---
OK. So what you have is a voltage-to-frequency converter and an
LM34 wired like this:

CPU+12>---------+-------------------+
| |
+-----+-----+ +-----+-----+
| +12 | | +12 |
CPU fIN<--|fOUT Vin|<------|Vout TEMP| <~~~
| GND | | GND |
+-----+-----+ +-----+-----+
| VFC | LM34
CPU GND>--------+-------------------+

Where the LM34 outputs 10mV/F° starting with 0V at 0F, and what you
want is to offset the output of the LM34 to start at -1V for 0F and
rise at a rate of 10mV/F°?

If that's the situation, then you can either lower the ground of the
LM34 to -1V or raise the VFC's ground to +1V. Raising the VFC's
ground to 1V would be easier, but since its output would also be
referenced to the 1V "ground" it would probably be offset as well,
so that may not be a good idea. Can you tell us what you're using
for a VFC or post its spec's?

If you don't, then you could use something like a 7555 to build a
charge pump followed up with a regulator, like this:

CPU+12>------------+------------+------------+--+12
| | |
+-----+-----+ | +-----+-----+
| +12 | | | +12 |
CPU fIN<-----|fOUT Vin|<-----|------|Vout TEMP| <~~~
| GND | | | GND |
+-----+-----+ | +-----+-----+
| VFC | | LM34
CPU GND------------+ | |
| | |
GND | |
| |
+---[C1]---+------|------+ |
|K | | | |
[1N4001] [Rt] +---+---+ | |
| | | Vcc | | |
+----+ +--|TH OUT|--+ |
|- | | |___ _| |
[C2] | +-O|DIS R|O--+12 |
| | | | GND | |
GND | [Ct] +---+---+ |
| | |7555 |
| GND GND |
| |
| |
| 79L05 |
| +-------+ |
+-----|IN OUT|---+-------+
| +---+---+ | |
| | [R1] |
| | | |
[0.33µF] +-------+ [0.1µF]
| | | |
| [0.1µF] [R2] |
| | | |
+---------+-------+-------+
|
GND

If you're interested in either of these approaches I'll be happy to
post the component values.
not to criticize because i know your a smart guy and all but shouldn't
the Diode be going to ground on the K side and A of the diode to the
C1 and maybe another Diode in series to C2 to isolate the 0.7 average
ripple ? or have i over looked something?

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