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In a circuit, when two ideal components are used in parallel instead of using just one (same component in both cases), does it have any effect on any of the temperature values? junction / case / ambient (inside the box containing the circuit)

Ideal meaning that the components can somehow share the current 1/2 - 1/2. Also, consider a very long time running the circuit in an environment that does not change in time.

---EDIT---

I was a bit tired when I posted; obviously the datasheet params will not change. I am interested in how the temp values differ. Junction, case and surrounding temperatures inside the circuit box. Sorry for that

Is there thermal resistance between them?

This is the gist of it. Both cases TBD in the answer please.

So, with one component: I, U with two parallel and ideal components: I/2, U through / across each.

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  • \$\begingroup\$ What kind of components? Are they linear or no? Is there thermal resistance between them? Is the total current the same? \$\endgroup\$
    – Tim Williams
    Commented Jun 7 at 0:17
  • \$\begingroup\$ Your asking for a general rule of thumb, but the quantitative thermals are highly application-specific. Not limited to things like airflow, heatsinking, and other thermal geometries. The analysis for one device will be exactly the same as it is for 2 devices. It's how these parts are thermally coupled that governs the behavior. That's the part that's missing. For example, 2 devices tightly coupled to a common heatsink will have a greater temp rise than if the devices were spaced apart (all other factors staying equal). \$\endgroup\$
    – MOSFET
    Commented Jun 7 at 15:00

2 Answers 2

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There will be no effect on the temperature rise parameters - the temperature rise parameters (junction to case, junction to ambient) are device parameters and are independent of the number of devices used. Those parameters are intrinsic to the components.

If the components are ideal, and their environmental conditions are the same (same temperature, humidity, airflow, ect.), then the devices with split the current and thus the heat load equally. Considering a long time, short time, or any other transient time the circuit is running does not matter - the current will be shared equally between devices. The situation is time-invariant.

Even ideal components with positive temperature coefficients will share current equally in all cases and all circuit durations as there is no bifurcation and therefore no thermal run away.

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    \$\begingroup\$ The idea that the components will "share" the load depends on the components not being fed by a constant voltage source. With a constant voltage source, the load will simply be doubled. \$\endgroup\$
    – Math Keeps Me Busy
    Commented Jun 7 at 1:12
  • \$\begingroup\$ @MathKeepsMeBusy Whether the components are current-sourced or voltage source doesn't matter. The current drawn will be 1/2 -1/2 between components. But I see what you're saying - it would be more accurate to say that the components share the "source" rather than share the "load". \$\endgroup\$
    – MOSFET
    Commented Jun 7 at 1:20
  • \$\begingroup\$ They also don't mention if they are linear components (current and voltage shared) or nonlinear (current hogged, or voltage ~independent of current). They could be resistors, capacitors, inductors, MOSFETs, IGBTs, CC or CV condition, and without having defined those... a good answer must consider all these possibilities. \$\endgroup\$
    – Tim Williams
    Commented Jun 7 at 2:17
  • \$\begingroup\$ @TimWilliams The answer would still be the same for all those cases. Could you provide a counter example in which two of the same ideal components placed in parallel wouldn't split the current equally? \$\endgroup\$
    – MOSFET
    Commented Jun 7 at 2:32
  • \$\begingroup\$ It seems that the author wanted to write that the components share the source and they are not connected in parallel on positive and negative voltage rails. Does adding more components like that reduce heat build up? This may reduce the amplification of the circuit though. \$\endgroup\$
    – Amit M
    Commented Jun 7 at 2:34
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So, with one component: I, U with two parallel and ideal components: I/2, U through / across each.

If a voltage U is applied to one component, and that component draws current I, then if the same voltage U is applied to each of two identical components, then each of these components will draw a current of I, not I/2!

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  • \$\begingroup\$ Would whoever downvoted this answer like to share their disagreement? The OP, appears to be working under a misapprehension about "load sharing", and perhaps the down-voter has a similar misapprehension. Two identical light bulbs connected in parallel to the mains consume twice the power as one. Perhaps an open discussion can clear this up. \$\endgroup\$
    – Math Keeps Me Busy
    Commented Jun 8 at 13:10

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