# How to calculate the switching loss of an OR Gate?

This is the datasheet of the OR gate. Current consumption as per the datasheet is 25 uA. (Electrical Characteristics table - Page 4 Input current Icc = 25 uA).

I am wondering how power consumption will change if the gate is working, for example Assume OR gate is switching a clock of 3.3 V at 24.576 MHz.

Intention is find extra power consumption, if there is any. I have gone through the datasheet, but not found such any information.

• What formula have you studied for dynamic power consumption? The information you need to apply it is clearly in the datasheet. – Spehro Pefhany Feb 19 '18 at 6:52
• @SpehroPefhany, = ( (Vcc)^2 ) * f_sw * C_load. Just need to do this calculation? – vt673 Feb 19 '18 at 7:00
• You got it. The value of C_load includes the internal capacitances. The latter is shown in the datasheet. – Spehro Pefhany Feb 19 '18 at 7:16
• See TI's FAQ. – CL. Feb 19 '18 at 8:15
• consider the input capacitance Is that capacitance charged and discharged by the OR gate? Then yes, it is part of the load. – Bimpelrekkie Feb 19 '18 at 9:48

Every time the OR-gate output rises to "1" you can regard the capacitor (in the table above) being charged-up to the supply voltage. This takes energy ($CV^2/2$) and that energy is unrecoverable because when the OR output falls to "0" the capacitor is discharged. A small amount of heat is produced.