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By 8Volts
#198172
Hi folks,

I am looking into using the INA169 part to make a current measuring circuit... heres the link to the hook uo guide:

https://learn.sparkfun.com/tutorials/i ... 481576911

I was just wondering if I am calculating RS and RL properly?

As RS I need approximately (0-35ma) to 1 Amp measuring range.

And an RL that proportionally depicts a 0 to 3.0VDC as an output.

Here's the proposed formula:

IS =(VOUT x 1K ) / (RS x RL)

filling in the blanks:

1A = (3V x 1K) / (RS x 10K)

Solving for RS:

RS = 0.3 ohms at 1/2 watt, 2% precision

Is this correct???

Thank you all for your help

8V
By paulvha
#198173
you have a very wide range to cover and on the low current it will fall below the minimum input to be accurate.
Page 13 of the data-sheet : For most applications, best performance is attained with an RS value that provides a full-scale shunt voltage of 50 mV to 100mV; maximum input voltage for accurate measurements is 500mV.

50mA and 0.3 ohm = U = I * R = 50 * 0.3 = 15 mV (that is low)
150mA and 0.3 ohm = 150 * 0.3 = 45mV (just about on the low side)
500mA * 0.3 = 150mV good.
1000mA * 0.3 = 300mV good

On the Sparkfun schematics they look to work in the 35mV to 350mV range and advice
10 ohm 3.5mA - 35mA
1 ohm 35mA - 350mA
0.1 ohm 350mA - 3.5A

On the 1/2 watt (if you stick to the 0.3 ohm) : P = I * I * R = 1 * 1 * 0.3 = 330mW. I would take at least 1W. Then the resistor stays relative cool and within it resistor accuracy boundary (the price difference is peanuts..)
By 8Volts
#198193
Hello paulvha,

Thank you for your reply!

So if I understand correctly.... I will not be able to monitor the full 0-1A range!

Given the the lowest shunt voltage allowable of 50mv would give me arespective current of:

@50mv >>> I=V/R = 50mv/0.3ohms = ~166ma

Applying this to their formula to get my voltage output range, we can say:

Is=(Vo X 1K) / (RS x RL)

Isolating for Vo, becomes:

Vo = (IS x RS x RL) / 1K

Therefore:
Vo min = (166ma x 0.3 x 10K) / 1k
Vo min = 0.5 v

and:

Vo max = (1A x 0.3 x 10K) / 1K
Vo max = 3V

So my output voltage range = 0.5V to 3V proportionally for a given current range of 166ma to 1A

right?

Thank you for your help

8V
By paulvha
#198194
Yep.... with one remark.... you can monitor... but it is expected to be less accurate in the lower range. Sparkfun schematics is starting at 35mV instead of 50mV, so starting at 35mv/ 0.3 ~100mA. I don't know what the expected "most common" load will be, but I would concentrate the resistor calculation on that.
By 8Volts
#198206
Ok thank you very much Paulvha

Your help is very appreciated
8V