Component Power Computation

At least one exterior node must have a current flowing out of it.

The voltages at the boundaries cannot be held at constant potential. U1 will quickly charge up to the point where no current flows.

If the currents don't balance, you get a lightning bolt.

If there's an actual device, the currents do average to zero. Power is just

sigma(Vn*In)

which is what N power supplies would provide if you connect them to the chip. Some can of course provide negative power.

The slight short cut is to make one of the supplies zero volts, namely ground, and reduce n by one.

Huh? U1 is presumed to have content... maybe a PMOS device with

Maybe you need to rethink circuit fundamentals? ...Jim Thompson

Huh? U1 is presumed to have content... maybe a PMOS device with

Maybe you need to rethink circuit fundamentals?

The right answer _is_ "sigma(Vn*In)", but no one said some of the currents can't be exiting, only _by_convention_ was it stated that positive current is _into_ the node.

"If the currents don't balance, you get a lightning bolt." ????

Remember the "hint": Kirchhoff's Laws, specifically Igozinta == Igozouta ?>:-}

The voltages are known by the simulator (or other measuring device), there is _no_ "float". ...Jim Thompson

Sigma Vn*In.

Tim

-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website:

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

It's a black box. The power that crosses the boundary doesn't know what's inside.

Just below you say that my answer is right, so why would I need to rethink anything?

Actually, Conservation of Energy is more dependable than "circuit fundamentals." I assume that there's not a battery or a diesel generator inside; if there were, net consumed power could be negative, until you run out of fuel.

Connect a current source to an insulated anvil. You'll eventually get a lightning bolt. The schoolroom version of K's Law assumes zero node capacitances.

I'm working on a board right now that has a bunch of parts riding on a floating common node, with an ADUM201 isolator and a dc/dc converter coming up from the low-side stuff. That local common can slew 50 volts/ns relative to earth ground. So if a node has 1 pF to the universe, it gets 50 mA stuffed into it. That would break things. So I sandwiched most of the traces between two PCB pours that are the floating common, essentially making a Farady cage. The surface-mount chips are directly above one of those pours, not near any ground planes. So Kirchhoff mostly still works.

The arrows are just to indicate the polarity of the measurement.

This is probably fall-out from the "Free energy!" thread.

Power input is sum of all the I*V; don't know about 'dissipated' , though, it could be charging batteries or electroplating baby shoes or pushing a spacecraft.

All my DVMs have a sign character on their display to indicate current direction. I suppose Jim's don't.