resistors
If I have a parallel resistor network with different values for each resistor, how do I calculate the total resistance?
Sorry I dont have enough rep to post pics so I'll try and explain.
Lets say I have a matrix of 9 resistors total in a 3x3x3 grid with
3 different values for the 9 resistors, say 10 ohm, 20 ohm and 30 ohm.
So that makes 3 sets of 10, 20 and 30 ohms.
I need to calculate the total resistance of the 9 resistor network.
Well you can certainly get op-amps that do work well from a 5V rail but ignoring that part of the question and concentrating on the two 0V points on the first circuit, if these were at 1V then there would be no problem. However, the outputs from the MCU (shown as switches connected to either 0 volts or positive rail) cannot be disentangled from 0V.
But, if you used a bunch of analoge switches (as you have basically drawn in the diagram) this would work fine. The two triangular shaped ground points could connect to any abitrarily low voltage (such as 2V or 1V) and the range of your DAC would stretch from whatever voltage you choose upon for the "ground" up to whatever voltage you choose for the other supply to the analogue switches.
There are suitable parrts that run from 5V and give very low impedance for "on" and beyod-the-rails capability. I'll try and find one to recommend.
Here's a single device (one-eigth of the way) that does SPDT but I'm sure with a bit of searching a quad device could be found (half way there): -
In this case I would use Y-delta transformation. As shown in the schematic below:
simulate this circuit – Schematic created using CircuitLab
This should give you one unique solution for an equivalent resistance. It's probably an enormous equation, so luckily you have a CAS.
Best Answer
You add the conductances (reciprocal of resistance) and then convert back to resistance. For example, lets say we have 500, 1500, and 2500 ohm resistors in parallel. The total resistance is:
1 / ((1/500) + (1/1500) + (1/2500)) = 326.1 ohms.
For 2 resistors in parallel you can also use (R1 * R2) / (R1 + R2) e.g. for 500 and 700 ohms:
(500 * 700) / (500 + 700) = 291.7 ohms.
For your example of the 10, 20 and 30 ohms in a 3 x 3 network, you can either work out each row then add the results, or multiply each value by three and calculate directly:
1 / ((1/30) + (1/60) + (1/90)) = 16.3636... ohms.
If the example is like this (in/out at top and bottom of matrix):
then the result is 20 ohms (row 1 is 3.33... row 2 is 6.66... row three is 10 - added together is 20) which agrees with the answer you have.