# 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

This gives you a guideline: - 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.

Along comes another "1" and you put energy into the capacitor only to lose that energy when the output goes low. So, if your Vcc is 3.3 volts then the energy injected into the capacitor is 114 pico joules.

However, you do that 24,576,000 times a second hence the power is 2.81 mW or, the average current taken from the 3.3 volt DC supply is 0.85 mA.

It's a rough guide.

• If you are happy with this answer please consider formally accepting it or raise a comment if there's something you need help with understanding. – Andy aka Mar 2 '18 at 10:03