420mA current output using XTR111 with offfset input voltages

[PRIYADHARSHINI PALANISAMY]

4-20mA current output using XTR111 with offfset input voltages

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....

 

[esp1]

hi,
I would consider using a voltage divider on the 0.5V thru 2V input to reduce it to 0.4V thru 1.9V and then set the Span with Rset to give 4mA thru 20mA for that 0.4V thru 1.9V input.
E

 

[FvM]

Adjusting the span has essentially the same effect as scaling the input voltage. But you need to apply a negative offset in output current respectively a postive level shift for the input range. Data sheet figure 44 shows how to. The exact calculation is up to you.

 

[Gorgon]

Can you show the schematic of your setup. From the datasheet there should be no problem to change the input to what you want, but a look at how you have soleved it so far, is necessary to understand your problem.

 

[FvM]

The solution is essentially simple:
calculate Rset for intended scale: Rset = 10*(2-0.5)/(20m-4m)
determine the necessary offset resistor for 4 mA @ 0.5 V input

 

[PRIYADHARSHINI PALANISAMY]

The solution is essentially simple:
calculate Rset for intended scale: Rset = 10*(2-0.5)/(20m-4m)
determine the necessary offset resistor for 4 mA @ 0.5 V input

you told that fig 44 can solve this problem.but it refers positive level shift of the input voltage range.but i need to shift my input voltage range to negative shift( 0.5v to 0.4v)
And according to ur calculation
Rset=1.5/16*10=0.935k
offset resistor R4 should be calculated for 4mA @0.5v or 4mA @0.4v?

- - - Updated - - -

hi FvM
i understood following things from data sheet figs:
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 i need 4-20mA for 0.5 to 2.00v [2/20*4=0.4v]but here -0.1v(negative offset)

 

[FvM]

I believe that you are confused about the offset sign. But it's as I said, you get the right signa with a offset resistor according to figure 44.

If you choose Rset as suggested, you get 5.33 mA at 0.5 V and 21.33 mA at 2 V input. So you need a negative shift of output current by 1.33 mA. This is achieved by injecting +0.133 mA into the Rset node.

 

[Gorgon]

You will not get a linear +0.133mA by using a single resistor. A resistor of 33.75k should work at 0.5V input, from the 5V regulator output. If you use this for the 2V input it will be down to 0.0888mA, giving an output fault.

What you could do to get around this is to use a constant current source of 0.1333 mA, that is independant of the resulting voltage over the Rset. It should be possible to make this from the 5V with 3V overhead over the input.

 

[PRIYADHARSHINI PALANISAMY]

K.if i choose the R4=22.56k and Rset=0.935k for 3v regulator output i get 0.119 voltage drop which is the offset for 0.5-0.4=0.1
right?

 

[Gorgon]

K.if i choose the R4=22.56k and Rset=0.935k for 3v regulator output i get 0.119 voltage drop which is the offset for 0.5-0.4=0.1
right?

What is the drop on the other end of the scale, at 2V?

 

[FvM]

You will not get a linear +0.133mA by using a single resistor. A resistor of 33.75k should work at 0.5V input, from the 5V regulator output. If you use this for the 2V input it will be down to 0.0888mA, giving an output fault.

Please reconsider. This is a linear circuit with a transfer function y = ax + b. Any combination of resistors will result in a linear characteristic, within the voltage and current limits of the circuit. You definitely don't need a current source to implement the intended characteristic.

It's however true that the offset resistor modifies the effective Rset value (parallel circuit). The bottom resistor has to be increased to about 0.975 k.

K.if i choose the R4=22.56k and Rset=0.935k for 3v regulator output i get 0.119 voltage drop which is the offset for 0.5-0.4=0.1
right?

Basically yes, but the Rset value has to be corrected.

 

[Gorgon]

Please reconsider. This is a linear circuit with a transfer function y = ax + b. Any combination of resistors will result in a linear characteristic, within the voltage and current limits of the circuit. You definitely don't need a current source to implement the intended characteristic.

It's however true that the offset resistor modifies the effective Rset value (parallel circuit). The bottom resistor has to be increased to about 0.975 k.

Basically yes, but the Rset value has to be corrected.

What happens to the 0.133 mA offset at the top of the scale then? As long as you reduce the voltage drop over the pullup resistor, the correction current will be reduced. The 'b' in your formula is not constant then, is it?
If I'm wrong here I'm sorry, but I'm curious about this.

 

[FvM]

There are two resistors in play. It's not necessary that the current through Roffset stays constant to achieve the intended offset.

If you have difficulties in intuitive understanding of the circuit's operation, it's probably time for a calculation.

 

[PRIYADHARSHINI PALANISAMY]

What is the drop on the other end of the scale, at 2V?

i have considered that for 20mA -2V input
if i reduced my input voltage 0.5 as 0.4 then it will give 20mA for 2V

 

[Gorgon]

I believe that you are confused about the offset sign. But it's as I said, you get the right signa with a offset resistor according to figure 44.

If you choose Rset as suggested, you get 5.33 mA at 0.5 V and 21.33 mA at 2 V input. So you need a negative shift of output current by 1.33 mA. This is achieved by injecting +0.133 mA into the Rset node.

Here you say the offset need to be 0.133mA over the whole range.

There are two resistors in play. It's not necessary that the current through Roffset stays constant to achieve the intended offset.

If you have difficulties in intuitive understanding of the circuit's operation, it's probably time for a calculation.

Here you say not.

I'm just refering to your first post saying is has to be constant. You then refer to a calculation to offset the zero point(fig44), that will not make an offset linear over the range, as I've already calculated in post #8.

 

[FvM]

"This is achieved by injecting +0.133 mA into the Rset node."
Here you say the offset need to be 0.133mA over the whole range.

Not actually. But I apologize for impreciseness.

The solution of the riddle is superposition theorem. The current must be 0.133 mA when the Rset node is shorted.

But instead of guessing about what I said, just calculate the circuit, or use a simulator.
Here is my equivalent circuit:

 

[PRIYADHARSHINI PALANISAMY]

Please reconsider. This is a linear circuit with a transfer function y = ax + b. Any combination of resistors will result in a linear characteristic, within the voltage and current limits of the circuit. You definitely don't need a current source to implement the intended characteristic.

It's however true that the offset resistor modifies the effective Rset value (parallel circuit). The bottom resistor has to be increased to about 0.975 k.


Basically yes, but the Rset value has to be corrected.

sorry to ask this again.
what is the effect of increasing the R set value in the circuit..
for 22.56 and 0.9375k produce 0.12v offset
for 22.56 and 0.9375k produce 0.124v offset

 

[Gorgon]

After some testing I found that Rset=974ohm and Rpullup=23000ohm to 3V makes a good combination.

 

[FvM]

After some testing I found that Rset=974ohm and Rpullup=23000ohm to 3V makes a good combination.

Yes, within 0.1 mA error bounds for an ideal part. The 3V reference has maximum 5% error, resulting in a similar deviation.

Or use E96 values 22.6 k and 0.976 k.

 

[Gorgon]

Even better is E96 values 23.2k and 0.976k