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DC voltage converter 0-110mV to 0-5V

I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's.
 
G

Genome

Jan 1, 1970
0
I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's.

Given your 'experience' of 'electronics' I would suggest that this is a
question better asked in a Linux newsgroup. I am have saved you some time
with a little crosspost. In the meantime you might try the man page.

HTH

DNA
 
J

John Fields

Jan 1, 1970
0
I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's

---
Since:

5V
------ = 50,
0.1V

you'll need a gain of 50, and you don't need an inversion, so you
can do this:



Vin>------|+\
| >--+-->Vout
+--|-/ |
| |
+--[R2]--+
|
[R1]
|
GND>---+----------->GND


An easy way to work out the resistors is to decide how much current
you want in the divider and then choose R1 to drop the maximum
voltage on the + input with Vout at its maximum output also.

In your case, if you wanted to have 1mA in the divider, then with
0.1V across R1 it would have to have a value of:

E 0.1V
R = --- = -------- = 100 ohms
I 0.001A

and R2 would have a value of:

E 4.9V
R = --- = -------- = 4900 ohms
I 0.001A

Looking at the divider like this:


E1
|
[R2]
| 0.1V
+----E2
|
[R1]
|
GND

and choosing 1% resistors for R1 and R2, the opamp output voltage
with 0.1V on the + input will be:


E2 (R1+R2) 0.1V * 4970R
E1 ------------ = -------------- = 4.97V
R1 100R


So, you've got a 30 millivolt error.

If you diddle around with the resistors you can probably do better
than that, but the easy way out is just to use a decent multi-turn
trimpot, put 100mV on the + input, and crank away until you get
precisely 5.00V on the output.

BTW, the classical formula for the circuit is:

R1 + R2
Vout = -------- Vin
R1
 
J

Jim Thompson

Jan 1, 1970
0
I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's

---
Since:

5V
------ = 50,
0.1V

you'll need a gain of 50, and you don't need an inversion, so you
can do this:



Vin>------|+\
| >--+-->Vout
+--|-/ |
| |
+--[R2]--+
|
[R1]
|
GND>---+----------->GND


An easy way to work out the resistors is to decide how much current
you want in the divider and then choose R1 to drop the maximum
voltage on the + input with Vout at its maximum output also.

In your case, if you wanted to have 1mA in the divider, then with
0.1V across R1 it would have to have a value of:

E 0.1V
R = --- = -------- = 100 ohms
I 0.001A

and R2 would have a value of:

E 4.9V
R = --- = -------- = 4900 ohms
I 0.001A

Looking at the divider like this:


E1
|
[R2]
| 0.1V
+----E2
|
[R1]
|
GND

and choosing 1% resistors for R1 and R2, the opamp output voltage
with 0.1V on the + input will be:


E2 (R1+R2) 0.1V * 4970R
E1 ------------ = -------------- = 4.97V
R1 100R


So, you've got a 30 millivolt error.

If you diddle around with the resistors you can probably do better
than that, but the easy way out is just to use a decent multi-turn
trimpot, put 100mV on the + input, and crank away until you get
precisely 5.00V on the output.

BTW, the classical formula for the circuit is:

R1 + R2
Vout = -------- Vin
R1

± 50X the offset voltage of the OpAmp

...Jim Thompson
 
G

Genome

Jan 1, 1970
0
Jim Thompson said:
I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's

---
Since:

5V
------ = 50,
0.1V

you'll need a gain of 50, and you don't need an inversion, so you
can do this:



Vin>------|+\
| >--+-->Vout
+--|-/ |
| |
+--[R2]--+
|
[R1]
|
GND>---+----------->GND


An easy way to work out the resistors is to decide how much current
you want in the divider and then choose R1 to drop the maximum
voltage on the + input with Vout at its maximum output also.

In your case, if you wanted to have 1mA in the divider, then with
0.1V across R1 it would have to have a value of:

E 0.1V
R = --- = -------- = 100 ohms
I 0.001A

and R2 would have a value of:

E 4.9V
R = --- = -------- = 4900 ohms
I 0.001A

Looking at the divider like this:


E1
|
[R2]
| 0.1V
+----E2
|
[R1]
|
GND

and choosing 1% resistors for R1 and R2, the opamp output voltage
with 0.1V on the + input will be:


E2 (R1+R2) 0.1V * 4970R
E1 ------------ = -------------- = 4.97V
R1 100R


So, you've got a 30 millivolt error.

If you diddle around with the resistors you can probably do better
than that, but the easy way out is just to use a decent multi-turn
trimpot, put 100mV on the + input, and crank away until you get
precisely 5.00V on the output.

BTW, the classical formula for the circuit is:

R1 + R2
Vout = -------- Vin
R1

± 50X the offset voltage of the OpAmp

...Jim Thompson

And the input bias current?

DNA
 
L

linnix

Jan 1, 1970
0
Genome said:
Given your 'experience' of 'electronics' I would suggest that this is a
question better asked in a Linux newsgroup. I am have saved you some time
with a little crosspost. In the meantime you might try the man page.

What does "voltage multipler" have to do with linux? Am I talking to a
machine?
 
Thank you John. You are the man. What op-amp would you recommend that
would preferably not need to be mounted on circuit board's.



I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's

Since:

5V
0.1V

you'll need a gain of 50, and you don't need an inversion, so you
can do this:



Vin>------|+\
| >--+-->Vout
+--|-/ |
| |
+--[R2]--+
|
[R1]
|
GND>---+----------->GND


An easy way to work out the resistors is to decide how much current
you want in the divider and then choose R1 to drop the maximum
voltage on the + input with Vout at its maximum output also.

In your case, if you wanted to have 1mA in the divider, then with
0.1V across R1 it would have to have a value of:

E 0.1V
R = --- = -------- = 100 ohms
I 0.001A

and R2 would have a value of:

E 4.9V
R = --- = -------- = 4900 ohms
I 0.001A

Looking at the divider like this:


E1
|
[R2]
| 0.1V
+----E2
|
[R1]
|
GND

and choosing 1% resistors for R1 and R2, the opamp output voltage
with 0.1V on the + input will be:


E2 (R1+R2) 0.1V * 4970R
E1 ------------ = -------------- = 4.97V
R1 100R


So, you've got a 30 millivolt error.

If you diddle around with the resistors you can probably do better
than that, but the easy way out is just to use a decent multi-turn
trimpot, put 100mV on the + input, and crank away until you get
precisely 5.00V on the output.

BTW, the classical formula for the circuit is:

R1 + R2
Vout = -------- Vin
R1
 
G

Genome

Jan 1, 1970
0
linnix said:
What does "voltage multipler" have to do with linux? Am I talking to a
machine?

I apologise. If you come to SED then you can see the original post. Headers
suggest...

User-Agent: slrn/0.9.8.1pl1 (Linux)

Obviously I thought you lot could help out because things will get tricky
when he wants to plug it into DAQView.... however I've just noticed a top
post so it's OK.

DNA
 
M

martin griffith

Jan 1, 1970
0
Jim Thompson said:
On Sat, 15 Apr 2006 19:28:23 GMT, [email protected] wrote:

I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's

---
Since:

5V
------ = 50,
0.1V

you'll need a gain of 50, and you don't need an inversion, so you
can do this:



Vin>------|+\
| >--+-->Vout
+--|-/ |
| |
+--[R2]--+
|
[R1]
|
GND>---+----------->GND


An easy way to work out the resistors is to decide how much current
you want in the divider and then choose R1 to drop the maximum
voltage on the + input with Vout at its maximum output also.

In your case, if you wanted to have 1mA in the divider, then with
0.1V across R1 it would have to have a value of:

E 0.1V
R = --- = -------- = 100 ohms
I 0.001A

and R2 would have a value of:

E 4.9V
R = --- = -------- = 4900 ohms
I 0.001A

Looking at the divider like this:


E1
|
[R2]
| 0.1V
+----E2
|
[R1]
|
GND

and choosing 1% resistors for R1 and R2, the opamp output voltage
with 0.1V on the + input will be:


E2 (R1+R2) 0.1V * 4970R
E1 ------------ = -------------- = 4.97V
R1 100R


So, you've got a 30 millivolt error.

If you diddle around with the resistors you can probably do better
than that, but the easy way out is just to use a decent multi-turn
trimpot, put 100mV on the + input, and crank away until you get
precisely 5.00V on the output.

BTW, the classical formula for the circuit is:

R1 + R2
Vout = -------- Vin
R1

± 50X the offset voltage of the OpAmp

...Jim Thompson

And the input bias current?

DNA
Jim's devices dont have input bias current problems


martin
 
G

Genome

Jan 1, 1970
0
martin griffith said:
Jim Thompson said:
On Sat, 15 Apr 2006 15:34:21 -0500, John Fields

On Sat, 15 Apr 2006 19:28:23 GMT, [email protected] wrote:

I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's

---
Since:

5V
------ = 50,
0.1V

you'll need a gain of 50, and you don't need an inversion, so you
can do this:



Vin>------|+\
| >--+-->Vout
+--|-/ |
| |
+--[R2]--+
|
[R1]
|
GND>---+----------->GND


An easy way to work out the resistors is to decide how much current
you want in the divider and then choose R1 to drop the maximum
voltage on the + input with Vout at its maximum output also.

In your case, if you wanted to have 1mA in the divider, then with
0.1V across R1 it would have to have a value of:

E 0.1V
R = --- = -------- = 100 ohms
I 0.001A

and R2 would have a value of:

E 4.9V
R = --- = -------- = 4900 ohms
I 0.001A

Looking at the divider like this:


E1
|
[R2]
| 0.1V
+----E2
|
[R1]
|
GND

and choosing 1% resistors for R1 and R2, the opamp output voltage
with 0.1V on the + input will be:


E2 (R1+R2) 0.1V * 4970R
E1 ------------ = -------------- = 4.97V
R1 100R


So, you've got a 30 millivolt error.

If you diddle around with the resistors you can probably do better
than that, but the easy way out is just to use a decent multi-turn
trimpot, put 100mV on the + input, and crank away until you get
precisely 5.00V on the output.

BTW, the classical formula for the circuit is:

R1 + R2
Vout = -------- Vin
R1

± 50X the offset voltage of the OpAmp

...Jim Thompson

And the input bias current?

DNA
Jim's devices dont have input bias current problems


martin

You mean he has multiple output devices?

That must keep the wife happy.

DNA
 
J

Jim Thompson

Jan 1, 1970
0
Jim Thompson said:
On Sat, 15 Apr 2006 19:28:23 GMT, [email protected] wrote:

I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's

---
Since:

5V
------ = 50,
0.1V

you'll need a gain of 50, and you don't need an inversion, so you
can do this:



Vin>------|+\
| >--+-->Vout
+--|-/ |
| |
+--[R2]--+
|
[R1]
|
GND>---+----------->GND


An easy way to work out the resistors is to decide how much current
you want in the divider and then choose R1 to drop the maximum
voltage on the + input with Vout at its maximum output also.

In your case, if you wanted to have 1mA in the divider, then with
0.1V across R1 it would have to have a value of:

E 0.1V
R = --- = -------- = 100 ohms
I 0.001A

and R2 would have a value of:

E 4.9V
R = --- = -------- = 4900 ohms
I 0.001A

Looking at the divider like this:


E1
|
[R2]
| 0.1V
+----E2
|
[R1]
|
GND

and choosing 1% resistors for R1 and R2, the opamp output voltage
with 0.1V on the + input will be:


E2 (R1+R2) 0.1V * 4970R
E1 ------------ = -------------- = 4.97V
R1 100R


So, you've got a 30 millivolt error.

If you diddle around with the resistors you can probably do better
than that, but the easy way out is just to use a decent multi-turn
trimpot, put 100mV on the + input, and crank away until you get
precisely 5.00V on the output.

BTW, the classical formula for the circuit is:

R1 + R2
Vout = -------- Vin
R1

± 50X the offset voltage of the OpAmp

...Jim Thompson

And the input bias current?

DNA

Most modern OpAmps don't have enough to fret over.

...Jim Thompson
 
G

Genome

Jan 1, 1970
0
Jim Thompson said:
Jim Thompson said:
On Sat, 15 Apr 2006 15:34:21 -0500, John Fields

On Sat, 15 Apr 2006 19:28:23 GMT, [email protected] wrote:

I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's

---
Since:

5V
------ = 50,
0.1V

you'll need a gain of 50, and you don't need an inversion, so you
can do this:



Vin>------|+\
| >--+-->Vout
+--|-/ |
| |
+--[R2]--+
|
[R1]
|
GND>---+----------->GND


An easy way to work out the resistors is to decide how much current
you want in the divider and then choose R1 to drop the maximum
voltage on the + input with Vout at its maximum output also.

In your case, if you wanted to have 1mA in the divider, then with
0.1V across R1 it would have to have a value of:

E 0.1V
R = --- = -------- = 100 ohms
I 0.001A

and R2 would have a value of:

E 4.9V
R = --- = -------- = 4900 ohms
I 0.001A

Looking at the divider like this:


E1
|
[R2]
| 0.1V
+----E2
|
[R1]
|
GND

and choosing 1% resistors for R1 and R2, the opamp output voltage
with 0.1V on the + input will be:


E2 (R1+R2) 0.1V * 4970R
E1 ------------ = -------------- = 4.97V
R1 100R


So, you've got a 30 millivolt error.

If you diddle around with the resistors you can probably do better
than that, but the easy way out is just to use a decent multi-turn
trimpot, put 100mV on the + input, and crank away until you get
precisely 5.00V on the output.

BTW, the classical formula for the circuit is:

R1 + R2
Vout = -------- Vin
R1

± 50X the offset voltage of the OpAmp

...Jim Thompson

And the input bias current?

DNA

Most modern OpAmps don't have enough to fret over.

...Jim Thompson

So it's a violin thing.

DNA
 
J

John Fields

Jan 1, 1970
0
Thank you John. You are the man.

---
That's very kind. Thank you. :)
---
What op-amp would you recommend that
would preferably not need to be mounted on circuit board's.

---
I don't have a clue about that but, more importantly, Jim Thompson
and Genome have brought up some good points which need to be
addressed, and to do that we'll need to know what power supply(ies)
you're planning to use and what kind of accuracy you're looking for.
Also, please bottom post since it makes it much easier for all of us
to follow the chronology of the thread.

Thanks,
 
J

John Fields

Jan 1, 1970
0
Jim Thompson said:
On Sat, 15 Apr 2006 19:28:23 GMT, [email protected] wrote:

I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's

---
Since:

5V
------ = 50,
0.1V

you'll need a gain of 50, and you don't need an inversion, so you
can do this:



Vin>------|+\
| >--+-->Vout
+--|-/ |
| |
+--[R2]--+
|
[R1]
|
GND>---+----------->GND


An easy way to work out the resistors is to decide how much current
you want in the divider and then choose R1 to drop the maximum
voltage on the + input with Vout at its maximum output also.

In your case, if you wanted to have 1mA in the divider, then with
0.1V across R1 it would have to have a value of:

E 0.1V
R = --- = -------- = 100 ohms
I 0.001A

and R2 would have a value of:

E 4.9V
R = --- = -------- = 4900 ohms
I 0.001A

Looking at the divider like this:


E1
|
[R2]
| 0.1V
+----E2
|
[R1]
|
GND

and choosing 1% resistors for R1 and R2, the opamp output voltage
with 0.1V on the + input will be:


E2 (R1+R2) 0.1V * 4970R
E1 ------------ = -------------- = 4.97V
R1 100R


So, you've got a 30 millivolt error.

If you diddle around with the resistors you can probably do better
than that, but the easy way out is just to use a decent multi-turn
trimpot, put 100mV on the + input, and crank away until you get
precisely 5.00V on the output.

BTW, the classical formula for the circuit is:

R1 + R2
Vout = -------- Vin
R1

± 50X the offset voltage of the OpAmp

...Jim Thompson

And the input bias current?

---
The non-inverting input is going to be fed from a shunt, so that
smacks of way less than an ohm, and the inverting input is going to
be fed from what looks like a 5mA source, so even with something
horribly gross, like a 5µA bias current requirement, we're still
looking at only about a tenth of a percent of error in the output,
due to that, no matter what.
 
J

Joerg

Jan 1, 1970
0
Thank you John. You are the man. What op-amp would you recommend that
would preferably not need to be mounted on circuit board's.
Depends on the supply voltages you can provide. I usually take the old
LM324 for this but it's pretty high in offset, a few mV. If you need
better precision look for a rail-to-rail opamp with less offset. TI,
National etc.

Regards, Joerg
 
S

Stanislaw Flatto

Jan 1, 1970
0
I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's.
Given that the input is from a current sensing shunt, it pays to feed it
into differential amplifier to nulify the voltage present on the sensor.
Very few current sensors connect directly to some zero reference point.
Depending on your expertise such can be made from a set of dual input
NAND CMOS gates. _VERY low power_!

Good luck

Stanislaw
Slack user from Ulladulla.
 
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