INA169: Asking for the INA169 current sensing schematic review

Hello SHH,

I do have a couple comments. Why is the 30-V voltage detector (U5301) necessary? You have this sensing the output and it seems to be there to alert the host processsor if there is a over-current protection (OCP) event. This seems to mean that if VOUT of INA169 exceeds 30 Volts, then U5301 will alert the processor, but this simply cannot happen. The INA169 is powered with a 5-V rail and thus its ouput can't exceed {5V-0.7V-(Vsense)}.

Additionally, during normal conditions the voltage detector on the output is going to increase current measurement error. The device is specified with an operating current of 3µA to 8µA. With your circuit when the source current (Is) is 5.5A, the IOUT current of INA169 is:
IOUT = Is*Rs/1kΩ = 5.5A*0.01Ω/1kΩ = 55µA.

This means the voltage detector's operating current will divert current aways from the output current and directly add anywhere from 5.45% to 14.54% error when measuring 5.5A.

Hope this helps.
Sincerely,
Peter Iliya
Current Sensing Applications

Hi Peter,
could you please explain where the 30V is form? I am confused about this.
This seems to mean that if VOUT of INA169 exceeds 30 Volts, then U5301 will alert the processor, but this simply cannot happen.

for 2nd concern about the precision, what's the suggestion to improve the precision?

BR,
SHH

Hey SHH,

I hope I am not mistaken, but the '30V' is simply from the inclusion of voltage detector U5301 (Part Number: RT981G-30GVL) in your schematic. You have the output of INA169 connected to the VDD pin of the voltage detector. Could you please explain what this part is used for in you design? I can't really help out with the precision of your design without removing this U5301 from the design or finding a way to work around it.

Sincerely,
Peter Iliya
Current Sensing Applications

Hi Peter,
part number "30" means the VDD is 3.0V for the reset IC.
so is this design doable?
BR,
SHH

Hey 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.

https://training.ti.com/getting-started-current-sense-amplifiers

 

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.

 

Sincerely,

Peter Iliya

Current Sensing Applications

Hi Peter,
Thanks for reply.
Would you please help review customer's layout?
txn.box.com/.../oiayx7f3htt9trpe3ju4f0ingc4dhmkz


BR,
SHH

Hi SHH,

It's been over two weeks since your last post, so it seems like your issue has been resolved. I will close the thread for now, but if you have any further questions you can reply to re-open it or start a new thread with a new question.

Best regards,

Ian Williams
Applications Manager
Current & Magnetic Sensing