How can I convert +-5V to a 4 to 20mA loop current signal?

Hi,

I have an electromechanical seismometer with a +-5V voice coil signal output representing velocity, and I want to signal condition the line by converting the signal to a 4 - 20mA output. I recently experimented with the EVAL-CN0314-EB1Z circuit eval board that features the AD8420 amplifier and it looks to be fairly promising. However it's only configured for a single-ended input. My signal is bidirectional. Is there a good way to add some offset to the circuit ? I was thinking of substituting a 6.8K resistor into R4 to add some DC offset from the reference onto the input signal.

A concern I have is how these low resistance values affect my sensor. I have not seen anything within the literature for the AD8420 on how input resistance affects noise on the input, but the voltage divider used on this board is going to load my sensor. What are the tradeoffs of moving R3/R4 and R5 into the Mohm range? Such as setting R3 = 1.1M, R5 = 100K, R4 = 680K. This would reduce the loading of my signal coil to a manageable level. However I am unsure of the compromise in terms of noise or accuracy.

• Hi Daniel,

In the CN0314 circuit note, kindly refer to the "selection of scaling resistor" section of the circuit note. There's a formula (equation 1 and 2), which was derived using nodal analysis, that you can use to compute for the values of R3, R4 and R5 resistors for a corresponding minimum and maximum input voltage, which in your case that would be -5V and 5V respectively. We also need to maintain the differential input voltage from 195mV - 990mV to achieve the loop current of 4-20mA for a sense resistor value of 50 ohms. I already computed the exact values of R3 and R4 that you can use, R4=6.768K and R3=10.088K. You can use the nearest standard resistor value for R3 and R4.

Other option, instead of using large resistance values for R3, R4 and R5, is to buffer the output of your sensor before connecting it to the input of the circuit.

Regards,

Jay

• Thank you Jay,

I will take another look at the formulas in the circuit note.

I need to know that the tradeoffs are to moving up to a 100K value for R5. I could probably use a buffer on the input, but my concern is in the increasing of input noise. Which path is the quietest path for preserving signal-to-noise ratios? Simple circuitry via resistor voltage divider, or replacement of the voltage divider altogether with a second stage of amplifier?

That's what I am struggling with.

-Dan

• Hi Dan,

When you use a 100K value for R5, you will be needing a 1M resistor value for R3 and 676.8K for R4. Using large resistance values would contribute higher noise. As a matter of fact, if the total source impedance is in mega-ohms range, thermal and current noise could add up a significant amount of noise. If the total source impedance is around 1M, the 80fA/rt-Hz current noise density would turn into 80nV/rt-Hz noise, plus the 128nV/rt-Hz thermal noise from the resistor itself while the AD8420 part noise contribution is 55nV/rt-Hz. In addition to that, offsets due to bias current becomes larger. You may try to use larger resistor values, but it is not guaranteed that it will be suitable to your purposes.

Regards,

Jay