TCA9800: Power Consumption

For when both sides are high:

During stand by, the power consumption can be calculated by P = I*V where I is the total standby current and V is the voltage of operation.

Example1 typical: at VccA=VccB=1.8V, the total stand by current is 18uA. P = 18uA*1.8V or 32.4uW

Example2 typical: VccA=2.5V, VccB = 3.6V, the total stand by current is 0.2uA and 24uA for VccA and VccB seperately. Total power would be = (2.5V*0.2uA+3.6V*24uA) = 86.9uW

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For 'dynamic' (When one side is driving low):

The worst case occurs when A side is being driven because B side has its own current source which it must sink and we can assume that the B side has no external pull up resistors. The majority of the power is going to go across the FET's RDSON. We will use P = (V^2)/Rdson to provide the power consumption from there. Worst case IoL of current source on B side is 0.54mA (typical) x 1.25%(tolerance) = 0.675mA. Rdson = 0.22V/0.54mA = 407 ohms.

Power from B side FETs = (0.26V)^2 / 407 = 166uW. Recall this is for one channel, we have two (SDAB and SCLB) so we should multiply this by 2.

2*166uW= 332uW.

This value will be used whether VccB is 1.8V or 3.6V as the current source and Rdson should not change.

IccB current at maximum is around 1.6mA and we calculated around 0.675mA per channel (we have two channels) so the difference will give us the power across the rest of the device. 1.6mA-2*(0.675mA) = 0.25mA. So I assume we can use the 0.25mA against the  supply voltage. so if the VccB was 2.5V then we would have around 2.5V*250uA = 625uW.

The total power from B side would be around 1mW (625uW+332uW).

A side current should be much lower than 1 mW that it probably wouldn't make much difference if we did calculate it.

-Bobby