IR Drop


In the latest deep submicron CMOS technologies, the problem of the voltage drop on the integrated circuit supply rails has become significant.

The supply voltage has dropped from 5V common with 0.6µm technology down to 1.2V at 0.13µm technology. The power consumption has remained about the same, or is even higher than it used to be, because integrated circuits have more gates and run at higher frequencies.

This means, for the same power, that the currents flowing in the power supply are proportionately higher:
up by a factor of 4.17 as the voltage has scaled from 5V down to 1.2V.

At the same time, the voltage drop which can be tolerated along the supply rails has also decreased. This is typically expressed as a percentage of the power supply, often 5%. A 5V power supply could then tolerate a 250mV drop, but a 1.2V power supply only 60mV. This drop is ΔV=IR, where R is the supply rail resistance, which must then scale proportionately with the supply voltage.

This means that for the same supply current, the power supply resistance going from 5V to 1.2V needs to be 4.17× smaller.

But since the current is already 4.17× bigger, the power supply resistance needs to be 17.4× smaller in a 0.13µm chip than it was in a 0.6µm one.

IR drops from external pin to centre of core

Picture showing the resistances between the external power supply and a cell at the centre of a chip. IR drop is the sum of the voltage drop across Rvdd and Rvss.

The reduced power supply resistance is achieved

  • by increasing the width of the power rails
  • by increasing the number of metal layers carrying the power
  • by making the top metal layer extra thick for increased conductivity

It's worth noting too that most 0.6µm chips had oversized power supplies, so the full 17.4X reduction in resistance isn't always necessary. This paper shows how to determine the correct amount of metal needed for the power supplies.