Resistance and Resistor
CHAPTER 33 Resistor and Resistance All conductors, PCB traces, wire bonds have finite length. If we apply a voltage V across this finite length, a current I flows through it. If we double the voltage across it, the current doubles. Resistance can be defined by the Ohm’s law equation. If by applying a voltage V across a conductor, a current I flows across it, then the resistance of the conductor is given by
R = V/I
The above equation governs the behavior of a resistance. If a current I is flowing across a resistance, then the voltage drop across it is given by
V = IR
The higher the resistance, the more is the voltage drop across it, and, therefore, more is the signal attenuation. This ohmic voltage drop happens at all frequencies. From a signal integrity perspective, a longer trace will cause more attenuation than a smaller trace.
It is therefore important that we understand the dependence of resistance on the physical structure of trace or wire.
The resistance of a conductor with uniform cross section of area A and length d is given by
where, d = length of the conductor in inches
A = Area of the cross section of the conductor in sq inch
ρ = Bulk resistivity of the conductor in Ohm - inches
R = Resistance of the conductor
Figure 3.1 - A resistance formed by a uniform cross section area.
Example 3.1 - Copper has bulk resistivity of 1.58 micro ohms-cm. What is its bulk resistivity in Ohm-inches. Find the resistance of 5 inches of copper trace that has a width of 4 mils and thickness of 1 mil.
Solution – The bulk resistivity in Ohm-inch is given by
or, R = 0.777 Ohms
So, this 5 inch of copper trace has resistance of 0.777 Ohms. If there is a 50 mA current flowing through this wire, there will be a voltage drop of 38 mV.
Doubling the trace length will double the trace resistance. In the above example, if we increase the trace length from 5 inches to 10 inch, the trace resistance will increase from about 0.78 ohm to 1.56 ohm. This doubles the voltage drop across the trace from 38 mV to 76 mV for 50 mA current. Further, if we reduce the trace thickness from 1 mil to 0.5 mils, the cross section area is halved and the trace resistance will increase to 3.12 ohm. The voltage drop across the conductor in that case will further increase to 152 mV for 50 mA current. The 152 mV drop can be significant for a 3.0 V signal and may erode voltage margin of the signal. Do not ignore the ohmic IR drop for thin and long traces. If it happens to be part of the power supply net, it might carry substantially higher current resulting in greater IR drop.
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