A simple way to convert 0-20ma into 0-10volt

[Marco T.]

Hallo,
I should acquire an analog signal in current and convert it in voltage.

The signal is in 0-20ma current range.

I thoguht to use the following circuit:

Voltage source 5volt
|
R1
input 0-20ma------|------------ output about 0-10v
R2
|
ground

R1 = R2 = 1k ohm

Any suggestion?

Many Thanks
Marco Toschi

 

[Marco T.]

Ops... R1 = R2 = 500 ohm

 

[Frank Bemelman]

Hallo,
I should acquire an analog signal in current and convert it in voltage.

The signal is in 0-20ma current range.

I thoguht to use the following circuit:

Voltage source 5volt
|
R1
input 0-20ma------|------------ output about 0-10v
R2
|
ground

R1 = R2 = 1k ohm

Any suggestion?

One resistor is enough:

input 0-20ma------|------------ output about 0-10v
R2 = 500
|
ground

 

[John Woodgate]

One resistor is enough:

input 0-20ma------|------------ output about 0-10v
R2 = 500
|
ground

One resistor is never enough. Use thousands.

And a PIC.

And a 555. (;-)

 

[Marco T.]

One resistor is enough:

input 0-20ma------|------------ output about 0-10v
R2 = 500
|
ground

I'm trying to simulate the circuit using spice. I have used a current
generator to simulate the input, but in this way if I don't use also R1 and
voltage source I don't obtain the desired output.

Which circuit should I use to simulate the 0-20ma input?

Many Thanks
Marco

 

[Frank Bemelman]

I'm trying to simulate the circuit using spice. I have used a current
generator to simulate the input, but in this way if I don't use also R1
and voltage source I don't obtain the desired output.

Which circuit should I use to simulate the 0-20ma input?

0-20mA suggests an industrial sensor of some sort. These give a
DC current.

Are you using AC or DC current? And what is the desired
voltage output, AC or DC?

 

[Frank Bemelman]

One resistor is never enough. Use thousands.

And a PIC.

And a 555. (;-)

Perhaps Bill can scribble something that uses a 741 as well

 

[Marco T.]

0-20mA suggests an industrial sensor of some sort. These give a
DC current.

Are you using AC or DC current? And what is the desired
voltage output, AC or DC?

I'm using a current generator with current fixed to 20ma.

I would obtain a dc voltage of 10volt.

Many Thanks
Marco

 

[Frank Bemelman]

I'm using a current generator with current fixed to 20ma.

DC current?

I would obtain a dc voltage of 10volt.

And so you should. 20mA @ 500 ohm is 10V.

 

[Marco T.]

DC current?


And so you should. 20mA @ 500 ohm is 10V.

Yes, DC current.

 

[Roger Hamlett]

One resistor is enough:

input 0-20ma------|------------ output about 0-10v
R2 = 500
|
ground

However _beware_ that there may be a 'caveat' in the drive capabilities of
the 4-20mA source. Most have a 'maximum resistance' specification, into
which they can maintain drive. For many of the systems, it is something
like 420R. The reason is that the drive itself is often only powered from
12v, and with transistor drops etc., 10v may be 'pushing things'.

Best Wishes

 

[default]

Hallo,
I should acquire an analog signal in current and convert it in voltage.

The signal is in 0-20ma current range.

I thoguht to use the following circuit:

Voltage source 5volt
|
R1
input 0-20ma------|------------ output about 0-10v
R2
|
ground

R1 = R2 = 1k ohm

Any suggestion?

Many Thanks
Marco Toschi

Lots of industrial controls use 4-20 ma for the two wire sensor. 4 is
used instead of zero because that gives a way to detect an open
connection to the sensor or malfunctioning/out of range sensor. A
shorted wire/out of range sensor indicates a greater than 20 ma
output. The idea is sometimes called a "supervised loop."

The control consists of a battery or power supply to put 10-30 VDC on
the line. To convert to a voltage you just supply a single fixed
resistor and put it on the return of the sensor to ground and measure
the voltage at the junction of the return and ground. You want ten
volts? use ohms law to find the resistor that will take.

In theory, a current source will output whatever voltage it needs to
to maintain the current in the loop. In practice the power supply
voltage is the limiting factor.

Many sensors today use four wires with a two wire current loop - the
extra pair supplys power to the sensor so that it can amplify and
massage the signal at the sensor head.

 

[linnix]

Perhaps Bill can scribble something that uses a 741 as well

I agree, an op-amp is need. How about this:


+------ 10K -----+
| |
| |\ |
| | \ |
IN----+------|- \----+---- OUT
1K | /
+------|+ /
| | /
| /
GND

 

[John Woodgate]

I agree, an op-amp is need. How about this:


+------ 10K -----+
| |
| |\ |
| | \ |
IN----+------|- \----+---- OUT
1K | /
+------|+ /
| | /
| /
GND

20 mA in gets you 200 V out. Some op-amp! All the 20 mA going in MUST
flow through the 10 k; the inverting input is a 'virtual earth' point.
The 1 k does nothing.

If an op-amp is working as an amplifier, there is, to a good
approximation, NO voltage between its inputs. This is because of the
huge open-loop gain. For any output voltage V, the voltage between the
inputs is V/huge, which is very nearly zero.

 

[Phil Hobbs]

In message <[email protected]>,



20 mA in gets you 200 V out. Some op-amp! All the 20 mA going in MUST
flow through the 10 k; the inverting input is a 'virtual earth' point.
The 1 k does nothing.
very nearly zero.

Well, it does force the amp to work at a noise gain of at least 10,
which will help stability some. Hanging a wire on a summing junction is
not guaranteed to produce good behaviour.

Cheers,

Phil Hobbs

 

[John Woodgate]

Well, it does force the amp to work at a noise gain of at least 10,
which will help stability some. Hanging a wire on a summing junction
is not guaranteed to produce good behaviour.

It usually produces the local pop music radio stations, for your
entertainment while you wonder why. (;-)

 

[linnix]

20 mA in gets you 200 V out.

No, it will max out at Vcc.

Some op-amp! All the 20 mA going in MUST
flow through the 10 k;

Most of the current will flow to ground via the input R, not the
feedback R.

the inverting input is a 'virtual earth' point.
The 1 k does nothing.

OK, we need the split the input current to avoid saturating the op-amp
too much. The values are just approxmation (in ratios), you have to
figure-out the exact values.

+------ 10K -----+
| |
| |\ |
| | \ |
IN----+------|- \----+---- OUT
| 1K | /
100 +------|+ /
| 1K | /
| | /
GND

If an op-amp is working as an amplifier, there is, to a good
approximation, NO voltage between its inputs. This is because of the
huge open-loop gain. For any output voltage V, the voltage between the
inputs is V/huge, which is very nearly zero.

But this is not open-loop.

 

[jasen]

Hallo,
I should acquire an analog signal in current and convert it in voltage.

The signal is in 0-20ma current range.

I thoguht to use the following circuit:

Voltage source 5volt
|
R1
input 0-20ma------|------------ output about 0-10v
R2
|
ground

R1 = R2 = 1k ohm

Any suggestion?

for 0..20mA -> 0..10V

r2 = 500 ohms, r1 not used.

Bye.
Jasen

 

[Paul Hovnanian P.E.]

One resistor is never enough. Use thousands.

And a PIC.

And a 555. (;-)

With an FPGA for 'logic glue'.

 

[mc]

With an FPGA for 'logic glue'.

Back up and debate PIC versus AVR, and end up with a 68HC11.