# using ADC on the current output of a sensor

My project uses a LEM LAH 50-P current sensor. Current will vary between +/- 22 Amps. The project uses a Texas Instruments TMS320F28027 MCU which operates at 3.3 V. The output of the sensor is a current equal the sensed current / 2000; the output will vary between +/- 0.011 A.

I'm trying to implement the resistor network by following this post which describes a summing amplifier. The post describes the input as a voltage source with all resistors of the same value. Also a comment to the post notes that the attenuation resistors can be incorporated into the circuit thereby saving on parts. Would the following design work? How to design for V1, R1, and R2?

simulate this circuit – Schematic created using CircuitLab

• (1) Are you trying to scale the +/-11 mA to 0 to 3.3 V and bias it to 1.65 V at zero current? (2) How are you powering the LEM and with what voltages? – Transistor Feb 8 '18 at 23:24
• (1) Yes, exactly. (2) IZ2415S duel output source – rur2641 Feb 8 '18 at 23:50
• IZ2415S at +/- 15 V – rur2641 Feb 8 '18 at 23:55

I've studied the LEM datasheet and had a quick look at the referenced answer so I may have missed some detail but I think you can simplify it.

Figure 1. Extract from the LEM datasheet showing the maximum $R_M$ the device can drive under various conditions.

Let's go with 100 Ω for now. You can increase it if it suits. This will drop 1.1 V at 11 mA which is good for our application - but bear in mind that a current spike or surge could raise this beyond your ADC's maximum.

simulate this circuit – Schematic created using CircuitLab

Figure 2. Biasing the $R_M$ voltage reading to mid-ADC range.

Fortunately for what we want to do next $R_M$ is a low value and the ADC has (he said without checking) a high input impedance.

• If $V_M$ (measured voltage on $R_M$ = 0 then we want half supply voltage on the input. We can do this with a 1:1 potential divider. We'll pick 47k resistors as these are not so high as to introduce noise susceptibility but 500 times higer than $R_M$ and so should introduce a measurement error of less than 0.2%.
• If $V_M$ were to rise to 3.3 V the ADC input would also be 3.3 V. (This would be 33 mA output.)
• If $V_M$ were to fall to -3.3V the ADC input would be held at the midpoint of +3.3 and -3.3 V = 0 V.
• At $V_M$ = 0 the ADC input will be 1.65 V.
• At +11 mA the ADC input will go up 1/3 of the way from midpoint to +3V3. At -11 mA the input will go down 1/3 of the way from midpoint to 0 V.