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I have 3 resistors of varying values connected in a delta configuration. I have access to all 3 nodes for measurements. I know that if I measure resistance across 2 points, I'm measuring 1 resistor in parallel with the sum of the other 2 resistors. Is there a way to determine each of the resistor values as if the circuit were opened (not connected in parallel) without any guarding and without destructively opening the circuit? I'm trying to verify my manufactured resistors in-house prior to sending them out for laser trim.

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  • \$\begingroup\$ There's nothing in your circuit but high-impedance devices connected to these resistors? \$\endgroup\$ – The Photon Dec 7 '15 at 19:57
  • \$\begingroup\$ Ever heard of simultaneous equations? \$\endgroup\$ – Andy aka Dec 7 '15 at 21:01
  • \$\begingroup\$ I'm still in school and am a novice in the field. Yes, I've certainly heard of simultaneous equations, but I have no experience using them. \$\endgroup\$ – Alex Maxwell Dec 7 '15 at 21:18
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The equations are easy to write down, but cumbersome to solve.

When you have a triangle of 3 resistors (a, b, c), and nodes A, B, C. Resistor a is connected between nodes B & C; resistor b between A & C, and c between A & B. Make 3 measurements -- AC, BC, and AB between nodes A&C, B&C and A&B.

Mathematica gives this:

$$a->\frac{(-AB^2+2\cdot AB \cdot AC-AC^2+2\cdot AB\cdot BC+2\cdot AC\cdot BC-BC^2)}{2 \left(AB+AC-BC\right)}$$ $$b->\frac{(-AB^2+2\cdot AB\cdot AC-AC^2+2\cdot AB\cdot BC+2\cdot AC\cdot BC-BC^2)}{2 (AB-AC+BC)}$$ $$c->\frac{(AB^2-2\cdot AB\cdot AC+AC^2-2\cdot AB\cdot BC-2\cdot AC\cdot BC+BC^2)}{2 (AB-AC-BC)}$$

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You have three nodes two measure from, so you can make two independent measurements to start from. For example, you can measure R1 || (R2 + R3) and R2 || (R1 + R3).

To get a third independent measurement you can (for example) short out R3 and measure R1 || R2.

With some algebra, you can now calculate each of the individual resistances.

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  • \$\begingroup\$ Thanks, Photon. I received a similar answer from an old professor of mine. That would work, but I'm trying to setup a formula so that test operators can simply probe across each of the 3 resistors and then have the 3 resistor values shown to them. Shorting across two nodes will not be easy to implement considering node access. I've spent hours and hours on this: simulating and manipulating any values that I can...all to no avail. \$\endgroup\$ – Alex Maxwell Dec 7 '15 at 20:17
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    \$\begingroup\$ You've only got three accessible nodes, so you can only make 2 independent resistance measurements. So you can't determine 3 unknowns without doing something to modify the circuit. Without knowing more about your circuit, I can't tell you any more than that. \$\endgroup\$ – The Photon Dec 7 '15 at 20:18
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    \$\begingroup\$ Why can't you make 3 measurements? Give the nodes names of A, B, C. Measure resistance AB, then AC, finally BC. That should allow you to create 3 simultaneous equations. \$\endgroup\$ – Dwayne Reid Dec 7 '15 at 20:24
  • \$\begingroup\$ Dwayne, that's the goal I'm after: Probe AB, AC, then BC and use an equation to determine values. I just can't figure the equation out, or even if it is possible without knowing out-of-circuit values of any of the 3 resistors. \$\endgroup\$ – Alex Maxwell Dec 7 '15 at 20:34
  • \$\begingroup\$ I'm not near my computer, but I normally use a software package called "TK Solver" to do this kind of work. Simply create 3 equations, one for each of the terminal pairs, then tell the software to solve. I used to be able to do this stuff when I was younger but excellent software has made me lazy. \$\endgroup\$ – Dwayne Reid Dec 8 '15 at 0:30
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First you can find the equivalent star. Since we have a simple equation to solve (eg \$ R_1+R_2=x, R_2+R_3 = y, R_3+R_1=z \$) from the results of \$R_1, R_2 and R_3 \$we can find the equivalent delta it is one of the best way without the need to remember all formulas

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