power - What is the input current of the 74HC595?

I need to calculate the power consumption of my circuit, but I can't find any information about the input current of the 74HC595.

I drive the 74HC595 with the atmega8 on 5v, but what is the input resistance of the 74HC595?

The darlington uln2803 clearly states that it's input resistance is 2.7k therefore I can calculate its input current 5/2.7k. But there is no such information for the 74HC595. How do I calculate its input current?

Answer

The static input current is trivially low compared to the input current if you are toggling the input at some high frequency. This is due to the input capacitance and may well be several orders of magnitude above the static dc current.

In this document, on page 8 it states that input current is +/- 1uA and this is the static current. On page 9 it states that input capacitance is typically 3.5 pF and each time you charge that capacitor up you consume some energy. That energy is related to the supply voltage: -

Energy is \$\dfrac{CV^2}{2}\$

For 5V that energy is about 44 pico joules. If you are toggling this input at 10 MHz then you are consuming this energy 10 million times per second and, in terms of watts this is 0.44 mW.

From a 5V supply this is equivalent to a power current of 88 uA i.e. significantly higher than the static dc current taken by the input.

You should also consider the "power dissipation capacitance" within the chip itself. On page 10 this is stated as being 115 pF i.e. a lot higher than the input capacitance and, not surprisingly will produce a power dissipation of about 15mW at a toggle frequency of 10 MHz. This is in addition to the input capacitance and, on its own is equivalent to a current draw (wasteage) from a 5V supply of about 3 mA.