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Power Management ICs Forum Discussions

Sudheer
Level 1
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Level 1

Hey,

Good Day!

 

Can PCB Cu traces be used as a current sense resistor in XDPP 1100 controller? If yes, then how to calculate trace resistance and what layout guidelines needs to be followed?

 

Thank you!

1 Solution
SuhasBobade
Moderator
Moderator 10 solutions authored 25 sign-ins 10 replies posted
Moderator

Hi VinayKumar,

 

Thanks for posting.

 

The high resolution of IADC allows the XDPP1100 sense output current through very small PCB copper shunt,

saving power loss and cost of precision sense resistor and Op-Amp. Use the following equation to calculate

copper trace resistance.

 

𝑅𝑐𝑜𝑝𝑝𝑒𝑟 = 𝜌 ∙(𝐿/(𝑇 ∙ 𝑊))∙ (1 + 𝑡𝑐 ∙ (𝑡𝑒𝑚𝑝 − 25))

 

ρ: resistivity of copper, 17 ∙ 10−6 𝛺 𝑚𝑚

L: Length of the copper trace

W: Width of the copper trace

T: Thickness of the copper trace

tc: temperature coefficient, 3.9 ∙ 10−3 /˚C

temp: trace temperature, unit ˚C

 

The thickness of copper trace is usually rated in ounces and represents the thickness of 1 ounce of copper

rolled out to an area of 1 square of foot. 1 oz. copper has a thickness of 1.4 mils or 0.0356 mm. Here is a design

example of copper shunt: 130 mil x 100 mil (L x W), top layer, 4 Oz copper, trace resistance is 0.158 mΩ.

 

It is recommended to layout the copper trace shunt in the first mid layer, so that the XDPP1100 controller could

be placed right on top of the shunt trace for the shortest routing distance. Also put a ground shielding layer

next to the copper sense layer to reduce stray inductance for better current sense accuracy. please refer attached image.

 

Kindly let me know if you need any other information.

Thanking you.

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SuhasBobade
Moderator
Moderator 10 solutions authored 25 sign-ins 10 replies posted
Moderator

Hi VinayKumar,

 

Thanks for posting.

 

The high resolution of IADC allows the XDPP1100 sense output current through very small PCB copper shunt,

saving power loss and cost of precision sense resistor and Op-Amp. Use the following equation to calculate

copper trace resistance.

 

𝑅𝑐𝑜𝑝𝑝𝑒𝑟 = 𝜌 ∙(𝐿/(𝑇 ∙ 𝑊))∙ (1 + 𝑡𝑐 ∙ (𝑡𝑒𝑚𝑝 − 25))

 

ρ: resistivity of copper, 17 ∙ 10−6 𝛺 𝑚𝑚

L: Length of the copper trace

W: Width of the copper trace

T: Thickness of the copper trace

tc: temperature coefficient, 3.9 ∙ 10−3 /˚C

temp: trace temperature, unit ˚C

 

The thickness of copper trace is usually rated in ounces and represents the thickness of 1 ounce of copper

rolled out to an area of 1 square of foot. 1 oz. copper has a thickness of 1.4 mils or 0.0356 mm. Here is a design

example of copper shunt: 130 mil x 100 mil (L x W), top layer, 4 Oz copper, trace resistance is 0.158 mΩ.

 

It is recommended to layout the copper trace shunt in the first mid layer, so that the XDPP1100 controller could

be placed right on top of the shunt trace for the shortest routing distance. Also put a ground shielding layer

next to the copper sense layer to reduce stray inductance for better current sense accuracy. please refer attached image.

 

Kindly let me know if you need any other information.

Thanking you.

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