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Part Number: INA169
Hi Team,Would you please help provide your comment for the INA16 schemtic review.The common mode is from 5V ~20V. Max current is 5A.here is the schematic.https://txn.box.com/v/ina
previous discusstion mail thread
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In reply to Peter Iliya:
In reply to SHH:
I see this now, sorry for the confusion. I would say the design is “doable”, but customer needs to understand the error calculation and see if it meets their overall design requirements. Customer needs to understand what the lowest current they need to sense accurately. Maybe this current is 5A and that is all customer cares about, but I cannot know based upon limited information.
Please consider watching our training videos which can you give you an overall understanding as to analyze the circuit and predict error.
Anyway to briefly cover the schematic you have sent: V+ = supply = 5V and VCM can vary from 5V ~ 20V. Max current is 5A. The shunt resistor is 10mΩ, 1% and the load resistance is 54.5kΩ, 1%. Given these conditions we can calculate total error. We will ignore error due to temperature for now.
First let’s calculate INA169 inherent error.
At 12-V VCM, the maximum input offset (Vos) is ±1mV. This matches testing condition.
This means offset error at 5.5A is 100*1mV/55mV = 1.818%
At 5-V V+, the offset due to PSRR is 0mV since this matches testing condition.
Max Gain error (or transconductance error) is ±1%.
Max nonlinearity error is ±0.1%.
The total error at Vsense = 100mV and RL = 25kΩ is ±2%, but we can ignore this for now and focus on individual error sources.
The ideal shunt voltage at 5.5A is 55mV.
So if use a root-sum-squares (RSS) error calculation we see that total RSS error for INA169 in your system is:
Error_169 = SQRT( (1.818%)^2 + (1%)^2 + (0.1%) ) = 2.077%
Now let’s consider the error from your shunt and load resistors:
Error_resistors = SQRT( (1%)^2 + (1%)^2 ) = 1.41%
Now consider error from the voltage detector (RT9818G-30GVL). According to its datasheet, for the 3.0V threshold variant, the supply current will be 1.75µA at around 2.7V input voltage, so right before the voltage threshold is met.
This current will be diverted away from the load resistors of INA169 and directly contribute error. So if the ideal IOUT is 55µA at 5.5A load current across 10mΩ shunt resistor, then this will mean an error of 100*1.75µA/55µA = 3.182%.
Error_RT9818G = -3.182%
Since the RT91818G will only siphon (sink) current away from INA169, you know that it will always contribute negative error to total error calculation. So you could approximate a worst-case total error by assuming the errors from INA169 and resistors are all negative as well.
So total RSS error without the voltage detector is
Error_total = SQRT( 2.077%^2 + 1.41%^2) = ±2.51%
If all error sources are negative and we assume worst-case total RSS system error then,
Error_system = -2.51% - 3.182% = -5.692%
It is up to customer to decide if this error is satisfactory or doable. They can greatly reduce error by replacing the voltage detector with a comparator or by calibrating out the voltage detector current. Hope this helps.
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