# Formula for voltage drop vs PCB trace width, temperature, current, and trace length

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?

• 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 Sep 26, 2014 at 9:05

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 ;) )

• I haven't checked the numbers, but your theory is right! Sep 26, 2014 at 10:33
• 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. Sep 26, 2014 at 11:49