# 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 – KyranF Sep 26 '14 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! – KyranF Sep 26 '14 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. – George Herold Sep 26 '14 at 11:49