LTC6106 ADC to Current Conversion

[chandlerbing65nm]

Hi folks,

I'm doing a project that involves reading a current from a supply line. To read the current using PIC16F1789, I used its ADC function (8-bit @ 5V) and LTC6106 for current sensing. However, I don't know how to convert the ADC reading to actual current sensed.

This is the 1st configuration: 3v3 Supply

of which I read HEX:16

This is the 2nd configuration: 5v0 Supply

of which I read HEX:07

I followed this formula and schematic in the datasheet.

For someone that can help, I am very thankful that you shared your knowledge.

Regards,
Chandler

 

[stenzer]

Hi,

for the 5 V setup one Bit corresponds to 5 V / (\[{2}^{8 }\]-1) = 19.6 mV.


According to the datasheet the transimpedance is given by

V_out / I_sense = R_sense \[\cdot\] R_out/ R_in,

which is given in V/A.

For your 5 V case this means

0.025 \[\Omega\] \[\cdot\] (2 \[\cdot\] 4900 \[\Omega\]) / 100 \[\Omega\] = 2.495 V/A.

In other words, a voltageof 2.495 V at the output pin of the LTC6106 corresponds to 1 A.


And in your case the measured ADC value is V_ADC = V_out/2, as V_out is divided by 2 by the attached resistor divider.

So for the 5 V case the current should be

I_out = 0.0196 V \[\cdot\] 2 \[\cdot\] 7 / 2.495 V/A = 109.98 mA

Does this sound familiar?


This calculation does not consider the input impedance of your ADC as well as the interaction with e.g. an anti-aliasing filter.

BR

 

[KlausST]

Hi,

Why did you use 2 output resistors for the 5V measurement?

I don't want to make the calculations for you until yout showed what you did so far.
So please show your calculations and we well correct them.
And you need to say what is your input value (volt, current), and what is your output value (volt range, current range, LSB...) of the formula.

Klaus

 

[FvM]

Factor 2 in 5V case is applied incorrectly in post #2. The output related voltage gain is setup only by R24 as 4.99k/0.1k = 49.9. R23 isn't affecting the gain but limiting the maximum output.

 

[stenzer]

Hi,

Factor 2 in 5V case is applied incorrectly in post #2. The output related voltage gain is setup only by R24 as 4.99k/0.1k = 49.9. R23 isn't affecting the gain but limiting the maximum output.

indeed, it's a current output!

Thx