Calculating Power for all IC's on a Breadboard

  • Thread starter HD555
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Hi,

I have a few breadboards with many IC's - analog and digital. How can I estimate the total power consumption? I know P = IV = I^2/R. Is it just my overall supplied source * total current consumption?

I'm currently supplying my breadboards with a single power source, and using voltage regulators to knock the voltage levels down. (I'm given a source > 20V. I'm using +15, +5 regulators to power my ICs). I see that the oscilloscope sources approximately 500 mA of current.

Thanks.
 

Answers and Replies

berkeman
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The current consumption will be a combination of the quiescent (DC) consumption, plus the switching current (AC).

The quiescent consumption is the sum of the Iq for each gate or IC. This is usually given in the datasheet for each part, as a min/typ/max range. You would have to choose which value is pertinant to your situation.... For example, you would use the max numbers if you were making sure your voltage regulator was sized big enough for the worst-case current consumption.

The dynamic switching current comes from the switching activity of the circuit, with I = C dV/dt being the relevant equation. You need to know the capacitance of each driven net, and the slew rate of the logic. When the gate output drives a net high, it is charging the capacitance of that net, which requires current drawn from the + voltage rail. When the gate pulls a net low, it is dischargning that capacitance by pulling current out of the net to ground. The charging part of the AC cycle is what draws extra current from the + rail, and is what gets added to the Iq to give you the total current drawn from the regulator.
 

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