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4-20mA using XTR111

PRIYADHARSHINI

Feb 6, 2014
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Hi ..
i am using xtr111 to produce 4-20mA with 0.5-2v input.
i have fixed 20mA output for 2V
(i.e) for 4 mA have to give 0.4V as an input [2/20mA*4mA]
I don t know how to reduce the 0.5v to 0.4v.I have gone through the Data sheet.
but its quit confusing for me to understand..can any one give the suggesstion to reduce 0.5v to 0.4v

thanks regards....
 

Arouse1973

Adam
Dec 18, 2013
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You normally use a 250 Ohm resistor to give you 1 to 5 volts. Why do uou want 0.4 volts?
Adam
 

PRIYADHARSHINI

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yes exactly.i have already gone through the Datasheet which you mentioned.but i dont know how to reduce the 0.5 as 0.4..
 

PRIYADHARSHINI

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KrisBlueNZ

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Here's what I suggest. I don't have much experience with instrumentation so someone else may be able to suggest some improvements, but this is the basic idea.

epoint 269944 suggested input stage.png

This is a non-inverting amplifier based on a Texas Instruments OPA376 op-amp. It has a gain of ((2 - 0.4) / (2 - 0.5)) and a fixed offset voltage. It converts an input signal (VIN) in the range 0.5~2.0V into an output voltage in the range 0.4~2.0V. It requires a 4.0V supply which comes from the XTR111 on pin 5. The trimpot connected to pin 4 must be adjusted for exactly 4.000V on pin 5 otherwise the voltage offset won't be exactly right. The resistors must be 1% or better. You can change the actual values of the resistors as long as you keep them in the right ratio.
 

PRIYADHARSHINI

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Here's what I suggest. I don't have much experience with instrumentation so someone else may be able to suggest some improvements, but this is the basic idea.

View attachment 14738

This is a non-inverting amplifier based on a Texas Instruments OPA376 op-amp. It has a gain of ((2 - 0.4) / (2 - 0.5)) and a fixed offset voltage. It converts an input signal (VIN) in the range 0.5~2.0V into an output voltage in the range 0.4~2.0V. It requires a 4.0V supply which comes from the XTR111 on pin 5. The trimpot connected to pin 4 must be adjusted for exactly 4.000V on pin 5 otherwise the voltage offset won't be exactly right. The resistors must be 1% or better. You can change the actual values of the resistors as long as you keep them in the right ratio.
thank u..
i will try it..
1.is this condition surely requires an addtional circuit?
2.can t we reduce the voltage using that voltage divider network which i mentioned previously?
3.and what is the reason for choosing the instrumentation amplifier for reducing the voltage?
4.or any other circuit can be used here to reduce the voltage in better way...sorry i can t understand ..thats why i m asking doubts
 

KrisBlueNZ

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1. Yes.
2. No. A voltage divider will reduce a voltage range but it cannot expand a voltage range. You have a voltage range of (2.0 - 0.5) which is 1.5V and you want a voltage range of (2.0 - 0.4) which is 1.6V. This requires gain. A voltage divider cannot do this. If you put your 0.5V~2.0V voltage through a voltage divider to reduce the 0.5V to 0.4V, it will also reduce the 2.0V down to 1.6V.
3. I don't understand the question. Why did I choose the instrumentation amplifier as opposed to what?
4. Not that I know of.

If you had presented your question with a lot more information I might have been able to suggest something. But you just said you have "a voltage" and you want to convert it into another voltage. What does this voltage come from? Give me some more information and I might be able to suggest something that doesn't require an IC.
 

PRIYADHARSHINI

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1. Yes.
2. No. A voltage divider will reduce a voltage range but it cannot expand a voltage range. You have a voltage range of (2.0 - 0.5) which is 1.5V and you want a voltage range of (2.0 - 0.4) which is 1.6V. This requires gain. A voltage divider cannot do this. If you put your 0.5V~2.0V voltage through a voltage divider to reduce the 0.5V to 0.4V, it will also reduce the 2.0V down to 1.6V.
3. I don't understand the question. Why did I choose the instrumentation amplifier as opposed to what?
4. Not that I know of.

If you had presented your question with a lot more information I might have been able to suggest something. But you just said you have "a voltage" and you want to convert it into another voltage. What does this voltage come from? Give me some more information and I might be able to suggest something that doesn't require an IC.
oops !..my 3rd question was wrong
apology
the input voltage(0.5-2v) comes from ADC in controller.we give the RTD as an input & the output is voltage from an ADC
1.Is it non inverting opamp right?
 
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KrisBlueNZ

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That circuit is not a differential amplifier. It has only one input, which is referenced to the 0V rail. That means it is single-ended, not differential.

I assume you can't control the output voltage range from the ADC? You can't modify the controller to use a different output voltage range?
 

PRIYADHARSHINI

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That circuit is not a differential amplifier. It has only one input, which is referenced to the 0V rail. That means it is single-ended, not differential.

I assume you can't control the output voltage range from the ADC? You can't modify the controller to use a different output voltage range?
ya its correct.
and according to the circuit
v out=(1+Rf/Rin)vin
=(1+4k/240k)0.5
=0.5v only.it Does not provide 0.4V
or i m Doing wrong in my calculation?
and for 2V INPUT it provides 2v output
could u please explain me through calculations?
sorry i need it urgently
 
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KrisBlueNZ

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It works. I simulated it.

The two 120k resistors across a 4V supply are equivalent to a 60k resistor to a 2V supply (see Thevenin equivalent on Wikipedia for more information). A non-inverting amplifier with 4k feedback resistor and 60k to ground has a gain of 64/60 or 1.06666667. If you multiply your input voltage range, which is (2.0V - 0.5V) = 1.5V by 1.066666667 you will get 1.6V which is your output voltage range (2.0V - 0.4V).

At 0.5V input, the circuit's output is 0.4V.
At 2.0V input, the circuit's output is 2.0V.
 

Arouse1973

Adam
Dec 18, 2013
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Here's what I suggest. I don't have much experience with instrumentation so someone else may be able to suggest some improvements, but this is the basic idea.

View attachment 14738

This is a non-inverting amplifier based on a Texas Instruments OPA376 op-amp. It has a gain of ((2 - 0.4) / (2 - 0.5)) and a fixed offset voltage. It converts an input signal (VIN) in the range 0.5~2.0V into an output voltage in the range 0.4~2.0V. It requires a 4.0V supply which comes from the XTR111 on pin 5. The trimpot connected to pin 4 must be adjusted for exactly 4.000V on pin 5 otherwise the voltage offset won't be exactly right. The resistors must be 1% or better. You can change the actual values of the resistors as long as you keep them in the right ratio.

Very neat trick Kris, although I think your formula works out that it has gain, how does this work?
Cheers
Adam
 

PRIYADHARSHINI

Feb 6, 2014
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thanks a lot.....it works very nice.i too have simulated it,
cheers
sorry for my delay..
1.how did u choose that the opamp can solve this issue?
i simply blind before ask this doubt to u
and in data sheet we don t have any other ways to solve this issue right?
they providing the details in data sheet for
0-20mA for 0.1v to 4.096v [4.096/20*0=0v]but here 0.1v(positve offset)
4-20mA for 0 to 5v [5/20*4=1v]but here 1v(positve offset)
but we need 4-20mA for 0.5 to 2.00v [2/20*4=0.4v]but here -0.1v(negative offset)
 
Last edited:

KrisBlueNZ

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1.how did u choose that the opamp can solve this issue? i simply blind before ask this doubt to u
You wanted to convert a voltage range of (2.0V - 0.5V) = 1.5V into a voltage range of (2.0V - 0.4V) = 1.6V. Converting a 1.5V voltage variation to a 1.6V voltage variation requires gain, so you can't do it with just voltage dividers. Op-amps are the usual way to provide accurately controlled gain. The circuit characteristics can be defined very accurately by just the external components (resistors). The "op" in "op-amp" stands for "operational" which comes from the fact that op-amps were originally used to perform arithmetic "operations" - multiplication, addition, subtraction, division etc. So multiplying by (1.6/1.5) with a reference voltage of 2V is easily done with an op-amp.

I chose that specific op-amp because of the requirements for low input offset voltage (needed for accuracy), and low-voltage operation with the inputs and output able to swing close to the supply rails. I originally planned to use a 3V supply, not 4V, so I eliminated the Maxim MAX4236 (http://www.digikey.com/product-detail/en/MAX4236BEUA+/MAX4236BEUA+-ND/1702410) which is a bit cheaper than the OPA376 I specified. Then I realised I had to use a supply voltage of at least 3.15V because of tolerances in the XTR111, so the MAX4236 is also suitable.
and in data sheet we don t have any other ways to solve this issue right?
Yes, your persistence in asking the same question for a third time has paid off - I've come up with a simpler solution.

epoint 269944 simplified design reduced.png

This design applies the adjustment using the op-amp inside the XTR111 so no external op-amp is used. Adjust the multi-turn trimpot for exactly 4V on pin 5 and use good quality components for the two 30kΩ resistors and the 1kΩ resistor.
Very neat trick Kris, although I think your formula works out that it has gain, how does this work?
Thanks Adam :)

I don't have time to explain now. Both of these designs just use gain and offset adjustment methods that can be understood in terms of Thevenin's and Kirchhoff's laws.
 
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