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TIDA-00366: +-50A Trace Width to Current Shunt Resistors

Prodigy 70 points

Replies: 8

Views: 156

Part Number: TIDA-00366

Hi,

I am designing a +-50A current monitoring board intended for a similar application as the TIDA-00366 design. The shunt resistor i am using is the WSLP2726, which is a 4-terminal resistor with a power rating of 7W and has a value of 1mOhm. The voltage drop across this resistor will be fed to an AMC1300B isolation amplifier. Now my issue is with the pcb trace thickness for such high current. I used the digikey trace with calculator as seen on the image below and even with 4oz of copper thickness, the required trace with is still high. How can I reduce this required trace width as you have done in the TIDA-00366 design?

Kind Regards,

Bright C

  • Hello Bright,

    thank you for your interest in our TIDA-00366 reference design and TI products.

    The shunt on the TIDA-00366 is 5mOhm, with the AMC1300 input voltage range of +/-250mV the maximum full-scale input current is 50A (peak), which
    equals a 35Arms maximum phase current. 

    With regards to the TIDA-00366 PCB, we didn't design the PCB to support continuous currents of 35Arms at maximum ambient operating temperature of 85C.
    I need to check with the designer of this PCB, what safe operating area the PCB supports. 

    I have a question on the parameters you chose of the Digikey too: Why would you limit the temperature increase of the copper trace to 10 degree only?
    I'd expect you can go much higher, e.g. 40 degree or even more depending on the maximum ambient temperature you run the PCB.  

    Thanks. 

    Regards,

    Martin Staebler

    .

  • In reply to Martin Staebler:

    Hie Martin,

    Firstly, thank you for your quick reply, and yes those are the same maximum input current values I want my PCB to monitor up to. I also wanted to use a 5mOhm resistor to use the maximum input range of the AMC1300 as you mentioned but chose the 1mOhm resistor as it has a lower power dissipation. 

    Okay, I didn't know what sort of temperature rise to expect so I just put 10 degrees, but yes by changing the temperature rise from 10 degrees to 40 degrees, the required trace width is now 4.36mm instead of 16.55mm. 

    Just to confirm, should I enter 50A or 35A in the digikey trace width calculator?

    Kind Regards,

    Bright C

  • In reply to Bright Chitura:

    Hello Bright,

    yes, for thermal calculation you'd take the maximum continuous phase current root mean square (rms), in your case I understood this is 35 Arms.

    Since you chose a 1 mOhm shunt, a 35 Arms maximum phase current, will lead to a +/-50mV maximum voltage across the shunt. In that case I recommend you use our isolated amplifier AMC1302 which has a +/-50mV input range: http://www.ti.com/product/AMC1302

    Regards,

    Martin Staebler

  • In reply to Martin Staebler:

    Hie Martin,

    Unfortunately my board has 4 channels so I have already ordered 4 of the AMC1300B isolated amplifiers. Does this heavily compromise the accuracy or I should be fine still? 

    Kind Regards,

    Bright C

  • In reply to Bright Chitura:

    Hello Bright,

    if you only use +/-50mV from an isolated amplifier with +/-250mV full-scale input range you will not get the optimum resolution and accuracy. 
    The signal-to-noise ratio will be a factor of 5 lower, and you SNR will be around 14dB lower. If you translate this into resolution in bits, assuming you drive and ADC and won't use the full-scale range you'd loose around 2.2 bits of resolution.

    In case you did not order the shunt yet: Why not go for a 5mOhm 7W rated shunt? The worst case shunt losses are 35Arms^2*5mOhm = 6.12W. .

    Regards,
    Martin Staebler

  • In reply to Martin Staebler:

    Hie Martin,

    Yes I still hadn't ordered the shunt yet, the only 5mOhm, 7W rated shunt I could find was the 17FPR005E, however on it's datasheet https://au.mouser.com/datasheet/2/303/res_10-1265495.pdf, the maximum current is 22Amps and it's price will go beyond my budget.

    In terms of the output and accuracy, my required output will be a voltage of 0 to +/-10V, with 35Arms on the input, equating to 10V on the output or a value within 1% of +/-10V which will be between 9.9V and 10.1V. With my current circuit (image below), the +/-50mV on the input of the isolated amplifier becomes 0.41V (since it has a fixed gain of 8.2),  and then the 0.41V will be amplified to 10V by a conditioning amplifier. 

    Is there any other 5mOhm, 7W rated shunts you can recommend or should I try to return the AMC1300B isolated amplifiers and get the AMC1302 ones instead? 

    Kind Regards,

    Bright C

  • In reply to Bright Chitura:

    Hello Bright,

    thank you for feedback.

    I won't be able to recommend you a suitable 4-wire 5mOhm, 7W current shunt. I'd recommend you check with the corresponding vendors of current shunts.

    It's excellent you use TINA for simulation and you can use to compare the impact on signal-to-noise ratio and bandwidth/settling time using AMC1300 versus AMC1302 with a corresponding lower gain setting of the differential to single-ended amplifier. You may use to estimate this impact and take a decision from the simulation results. 

    Whether you can return the AMC1300B versus AMC1302 you'd need to check with the distributor you've got your TI samples.

    Regards,

    Martin

  • In reply to Bright Chitura:

    Hi Bright,

    Thanks for the questions. The trace width depends on the max. current, maximum ambient temperature of PCB operation, layer location and PCB material. Standard FR-4 can handle a maximum of 130 deg. C. To stay within safe margins, it is recommended to keep PCB temperatures within 100 deg. C for standard FR-4. The max. allowable temperature rise in PCB traces is calculated depending on your max. ambient operating temperature. In this example, if 50 deg. C is the max. ambient operating temperature, the temperature rise of PCB trace is (100-50) = 50 deg. C. You can use this value in the Digikey calculator to estimate the required PCB thickness. As Martin mentioned, the current value in the calculator will be RMS and not max/peak value.

    Hope this helps.

    Regards

    Siva

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