d123
Advanced Member level 5

Hi.
I'm really bad at sums, it's "depressing" and embarrassing. Could some-one tell me if my calculations are - or at least look - correct, please.
I have a current shunt monitor which has 100V/V gain, it can be supplied from +3 to +18V. In the bread-boarding phase (now), for low currents it provides an acceptably accurate measurement and readout on an ADC.
If it matters, the shunt resistor is 0.02R (+0.12R lead resistance) and an ADC 10K/4K divider - which apparently works give or take a milliamp of accuracy.
Using a multi-meter I see that:
20mv opamp output. On input side of ADC resistor voltage divider, 5mV = 5mA.
33mV opamp output. On input side of ADC resistor voltage divider, 9mV = 9mA.
47mV opamp output. On input side of ADC resistor voltage divider, 13mV = 13mA.
Great, whatever...
So extrapolating based on these results, the opamp output should be: 50mA = 200mV, 500mA = 2000mV, 1000mA = 4000mV, 2000mA = 8000mV
On a +5V rail the opamp can't output the ADC full range, only up to (an optimistic) about 4.9V
I see the V+ supply to the opamp needs to be on an at least 10 i.e. 12V rail to definitely swing up to the needed 8V output (= 2000mV ADC input).
By my calculations, I think on the opamp Vin+ to Vin- side for 5mA load and 20mV output the input is 0.0002mV. 0.0002mV - that's as good as my maths skills get.
so 8000mV output should be, hopefully, if I was able to calculate anything here right:
80mV input max. for 2A shunt sense. This would be well within the opamp's -5/+5V input range.
Is (any of) this right, or is my maths so "special" I should go into "flexible accounting" or statistical analysis, please?
I'm really bad at sums, it's "depressing" and embarrassing. Could some-one tell me if my calculations are - or at least look - correct, please.
I have a current shunt monitor which has 100V/V gain, it can be supplied from +3 to +18V. In the bread-boarding phase (now), for low currents it provides an acceptably accurate measurement and readout on an ADC.
If it matters, the shunt resistor is 0.02R (+0.12R lead resistance) and an ADC 10K/4K divider - which apparently works give or take a milliamp of accuracy.
Using a multi-meter I see that:
20mv opamp output. On input side of ADC resistor voltage divider, 5mV = 5mA.
33mV opamp output. On input side of ADC resistor voltage divider, 9mV = 9mA.
47mV opamp output. On input side of ADC resistor voltage divider, 13mV = 13mA.
Great, whatever...
So extrapolating based on these results, the opamp output should be: 50mA = 200mV, 500mA = 2000mV, 1000mA = 4000mV, 2000mA = 8000mV
On a +5V rail the opamp can't output the ADC full range, only up to (an optimistic) about 4.9V
I see the V+ supply to the opamp needs to be on an at least 10 i.e. 12V rail to definitely swing up to the needed 8V output (= 2000mV ADC input).
By my calculations, I think on the opamp Vin+ to Vin- side for 5mA load and 20mV output the input is 0.0002mV. 0.0002mV - that's as good as my maths skills get.
so 8000mV output should be, hopefully, if I was able to calculate anything here right:
80mV input max. for 2A shunt sense. This would be well within the opamp's -5/+5V input range.
Is (any of) this right, or is my maths so "special" I should go into "flexible accounting" or statistical analysis, please?