Unusual resistor divider configuration and equation - Page 1
Forum section: electronics / projects

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 21, 2023, 04:54:56 pm
content: Was doing a thermal resistance problem and found this worth sharing. Not likely useful enough to make a calculator on my site, but something to archive somewhere.Suppose you have the following setup:Suppose V represents device junction temperature (TJ), and there are two heatsinks attached. Examples could be a DFN or D(2)PAK transistor, or thermal pad or BGA IC, etc., which is partly sunk through the board (pours + vias) and partly through a thermal pad on top. (Which might be much more complicated scenarios, for example if the top thermal pad smooshes down to board level and therefore intercepts some of the heat going to the backside pad/sink; apply good judgement before applying this formula to a given problem.) Now, obviously you can't probe TJ (V) directly; maybe you have alternative means to measure it in testing (like pulsed body diode measurement), maybe not. So maybe the 'V' node is hidden, and we just have V1 and V2 measurements (tabs, heatsinks, heat spreaders, something representative in there).So, as a complete problem statement: suppose we have unequal thermal resistances R1 and R2, from a device, to two known heatsinks of thermal resistance R3 and R4.Suppose further, the heatsinks are equal, R3 = R4 = R.Suppose finally, we don't know R, they're heatsinks pulled from the box -- or maybe we have some confidence in their values, but notexactly, because we've not measured them in the exact orientation and everything they're in during the test. (So, we replace two knowns, with one unknown and a symmetry. We could also have one resistance known, and the other completely unknown -- for a different expression, of course.)Working through things a bit, we find a relation for the value of R:$$ R = \frac{V_1 + V_2}{I_1}$$And R1 in terms of the possible value of R2:$$ R_1 = R_2 \frac{V_2}{V_1} - \frac{{V_1}^2 - {V_2}^2}{V_1 I_1} $$and since the problem is symmetrical, swap subscripts for the opposite case.This doesn't give us an exact resistance, but a relation to find one in terms of the other. In general, we can't know how much higher V is above V1 and V2, just that the resistances must be mismatched by some amount given by the current divider ratio. Indeed, notice that we could assume R2 is zero, and thus we get the minimum differential resistance between nodes V1 and V2. That differential resistance will be higher of course if the absolute value of these resistances is higher.In the event that you can measure TJ, you can measure R1 and R2 directly; or if you have one (say because R2 = RthJC is given, and V2 is the case, or close to it), maybe just representationally but not exactly measured (e.g. typical RthJC from the datasheet), then you can use that assumption to get a similarly representative value for R1. We can use a simplification to apply this to such problems. Suppose there is some resistance between V1 and V2 (representing the mutual resistance between heatsinks, due to heat flow through the board and other "sneak paths"). R1, R2 and the bridging resistance, let's call it Rb, constitute a delta network, which can be transformed into a wye network. A refresher:https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-resistor-circuits/a/ee-delta-wye-resistor-networksAfter transforming to wye, note that I1 is in series with a single resistor, so V --> V + I1 * Ry1, and treat the new (imaginary midpoint) wye node as 'V'. The bottom legs of the wye (call them Ry2 and Ry3) become R1 and R2.