Author Topic: How do I model an ideal load?  (Read 686 times)

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Offline bri

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How do I model an ideal load?
« on: September 01, 2017, 04:34:50 pm »
I’m very new to circuit modeling, and I’m trying to design a circuit in a Spice-powered designer. I want to insert a black-box model of an ideal 12A@5v load.

I know a “current source” is a “truly ideal” model, but (presumably because voltage would equal infinity) the simulator throws an error when the current source does not have something attached to both ends. I want to be able to “switch off” my load, so this doesn’t work.

Is the solution simply using Ohm’s law to calculate a resistor value and then use a resistor as my load? 5/12=0.41666… so would a 0.4166? resistor suffice to simulate a back box that draws 12A at 5v?

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Online Ian.M

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Re: How do I model an ideal load?
« Reply #1 on: September 01, 2017, 05:06:32 pm »
It depends what sort of load its powering - a resistor is of course fine for modelling a resistive load, but many loads aren't resistive.  e.g. a load with a linear regulator  at its input looks more like a current sink, with the load on the linear regulator setting its input current, and a load with a buck or boost converter or other switching regulator at its input typically looks like a constant power load ove a significant range of voltages and currents.

The constant current load is fairly easy to model - just orient a current source to sink current, and put an ideal diode (or at least as near ideal as your simulator will tolerate) across it so it cant take the output off towards negative infinity if your simulation doesn't supply enough current for the load.   

Constant power loads are trickier, and the details differ depending on the simulator.  Here's how LTspice handles it:


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