Current Measurement – Minimal Implementation of Current Loop Receiver

current measurementreceiversensor

I'm looking for a minimal design of a current loop receiver with the following characteristics:

  • I_in 0-20mA
  • U_in (-24) 0 – 24V.
  • U_out 0 – 3.3V

The precision doesn't need to be very high – an output voltage with a precision of +-5% proportional to the input current may be sufficient.

I was therefore looking first for a solution from a manufacturer and several forums. Unfortunately, I was not very satisfied with the solutions I found.

Either the presented solution was fully integrated within an IC and very expensive or it was too simple or too expensive, in terms of functionality:

1) Simple Shunt

The most frequently found solution was a simple shunt resistor with the high-side connected to an ADC or connected to a buffer and then to an ADC.
This solution is for me too error-prone in terms of failed connection or overvoltage.

2) Fully Integrated IC

This solution may be promising, but I'm looking for a more general solution. Maybe even cheaper.

I thought about an AC-input optocoupler, with the following benefits:

  1. I don't have to share a common ground.
  2. Isolated input.
  3. Connection order doesn't matter.

Possible Problems with this solution:

  1. Current-transfer-ratio seems to be in a wide range with low-cost optocouplers.
  2. Current-transfer-ratio may be asymmetric.

Can someone recommend a possible solution or add some thoughts about my idea?

EDIT #1
After the first answers and comments I came up with this conclusion:
Possible States

I draw this picture, representing the mentioned states.

  1. Normal operation.
  2. Overvoltage, the voltage gets internally clamped and the protection resistor drops most of the overvoltage.
  3. Pin swapped, the voltage gets internally clamped and the protection resistor drops most of the reverse voltage.

As I understand a TSV diode in parallel with the resistor would now only protect against fast voltage change if the internal circuit is not fast enough to react.
Therefore my next point is, what if the connection to the low side of the shunt resistor has a higher potential than the ground potential on my board. Wouldnt that lead to a ground loop carrying some current?

Best Answer

I_in 0-20mA U_in (-24) 0 - 24V. U_out 0 - 3.3V

That's basically a shunt resistor of value 165 Ω. No power supply needed. This is the minimal implementation. It's a totally viable and accurate solution.

The most frequently found solution was a simple shunt resistor with the high-side connected to an ADC or connected to a buffer and then to an ADC. This solution is for me to error-prone in terms of falls connection or overvoltage.

Well, given that any solution will use a shunt resistor, all you have to do is ensure that it is protected. Protected from what is the next consideration and you haven't said but, given that any solution will use a current shunt resistor you face the same problem whatever route you choose.

Consider adding a 6.8 volt TVS in parallel with the resistor shunt then work out what the threat is in terms of surge energy and what level of protection you need. A reminder: you will have to do this on all and every solution.

As for protecting the ADC or op-amp or DAQ that may follow it, use a 10 kΩ series resistor and check that the follow-up circuit won't fry with a maximum input current of 6.8/10000 = 0.68 mA. OK, with a really big indirect lightning surge, you might need an SMC package device and the peak voltage during the surge might become 10 volts. But that is still only a silicon current of 1 mA and virtually any op-amp or ADC will survive this.

A simple shunt resistor is needed on all solutions so, not using a silicon interface and not using an interface that requires a power supply is fundamentally the most accurate solution.

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