I need to find out the voltage drop in the PCB traces with respect to the possible factors like: copper thickness, trace length, trace width, temperature, etc.

I found some calculators available at:

Since all the calculators are providing different values for same input, I am not sure which calculator is giving the correct value. Is there any formula so I can calculate the voltage drop in the PCB traces?

  • \$\begingroup\$ why don't you model the copper trace as a parasitic resistor and then simulate current through it (use PSpice or other form of circuit simulator). You might be able to model temperature effects but it will not be easy \$\endgroup\$
    – KyranF
    Commented Sep 26, 2014 at 9:05

1 Answer 1


I'm going to have a stab at some maths :)

The DC resistance of a conductor - any conductor - is calculated as:

\$R_{DC} = \frac{{\rho}l}{A}\$

Where \$\rho\$ is the resistivity of the conductor in \$\Omega/m\$, \$l\$ is the length in meters, and \$A\$ is the cross-sectional area in m².

The thickness of 1oz copper is \$0.000034798m\$. Say you have a 3mm (or 0.003m) wide trace. The cross-sectional area is (approximately, assuming a perfectly rectilinear cross-section) \$0.000034798 × 0.003 = 0.000000104m^2\$. Resistivity of copper is \$1.68×10^{−8}\$ at 20C, and your trace is 100mm long (0.1m).

\$R_{DC} = \frac{1.68×10^{−8} × 0.1}{0.000000104} = 0.016153846\Omega\$ at 20C.

Ok, now for the tricky bit. The temperature co-efficient (\$\alpha\$) for copper is 0.003862.

\$R(T) = R(T_o)(1+\alpha{\Delta}T)\$

So for a temperature of 30C we have a \${\Delta}T\$ of 10C, or 10K (30 - 20 = 10, K = C + 272.15).

So \$R(30) = R(20)(1+0.003862×10) = 0.016153846×1.03862 = 0.016777708\Omega\$

So now solve Ohm's Law for voltage. Say you have 100mA flowing through the trace. That's \$V=RI\$, so \$0.016777708×0.1 = 0.001677771\$ or \$1.678mV\$ dropped across the trace at 30C.

Who says you need online calculators?

(Now, it's been about 20 years since I did this kind of thing at college, so I may be completely wrong ;) )

  • \$\begingroup\$ I haven't checked the numbers, but your theory is right! \$\endgroup\$
    – KyranF
    Commented Sep 26, 2014 at 10:33
  • \$\begingroup\$ Your theory looks fine (+1). In practice if the traces are on the outer layers of the pcb, then the thickness is rarely 1 oz. The few times I'v measured the resistance I've found the outer traces to be thicker. (For 1 oz I think they start with 1/2 oz and plate it up.) And of course at high frequency there will be the skin effect. \$\endgroup\$ Commented Sep 26, 2014 at 11:49

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