Making sense of test circuits with Kirchhoff’s laws: part 1

You can avoid solving simultaneous equations in multiple unknowns by identifying series and parallel combinations of resistors.

When you buy test instruments, you hope they’ll have the flexibility to provide the necessary stimulus to the device under test (DUT) and acquire and process the response. Occasionally, however, you’ll need to design an external network to get exactly what you want, or perhaps worse, you’ll have to deal with a custom interface network that somebody else built.

Figure 1. This “rat’s nest” of resistors comes from a Heathkit oscilloscope, where all the components were hand-soldered. (Image: Rick Nelson)

Q: You mean like an external network of resistors?
A: Yes, something such as the rats’ nest in Figure 1. The result is similar to what you might get if you gave a handful of random components to an engineer and asked for an interface between a test system and DUT without using a circuit board.

Q: So basically, is it just a combination of resistors in parallel and series to get the necessary values?
A: Right. Figure 2 shows a specific schematic, but with fewer resistors than Figure 1.

Figure 2. You might encounter resistors connected in series and parallel as part of a test interface. (Image: Rick Nelson)

Q: How can we analyze this?
A: If you have the actual circuit and a multimeter, you can disconnect the voltage source and measure the equivalent impedance. If you have a schematic such as Figure 2 or can derive one, you can use Ohm’s law and Kirchhoff’s current or voltage law and write node or loop equations, respectively, allowing you to calculate currents through each resistor and the voltages at each node. But we can avoid solving simultaneous equations in multiple unknowns if we can identify series and parallel combinations of resistors, as shown in Figure 3.

Figure 3. We can arrange the network of Figure 1 into series and parallel combinations of resistors. (Image: Rick Nelson)

We know that for two resistors in series, we simply add the resistances, and for two resistors R₁ and R₂ in parallel, the resistance is the product over the sum:

For N resistors in parallel, you can take them two at a time, computing the successive products over sums, or you might find it more convenient to compute the inverse of the sum of the reciprocals:

For the three parallel resistors in the light blue box on the right, we can determine that 820 Ω in parallel with 360 Ω equals 250 Ω, which in turn is in parallel with 1 kΩ and equals 200 Ω. Or we can simply calculate the inverse of 1/1,000 Ω plus 1/820 Ω plus 1/360 Ω, which also equals 200 Ω. That 200 Ω is in series with 1.8 kΩ, so in the dark blue block on the right, we have a total of 2 kΩ.

Figure 4. The Figure 3 circuit simplifies down to this simple voltage divider. (Image: Rick Nelson)

You can perform similar calculations on resistors in the other blocks shown in Figure 3, and get the voltage-divider network shown in Figure 4. This divider could be useful for a DUT that requires a supply voltage V, and that also has two high-impedance inputs requiring voltages of 50% and 20% of V. The resistor approach costs less than buying two additional programmable power supplies to generate the 50% and 20% levels. It can be faster than a multiplexer-based approach, where you would use a single supply to sequentially apply the required 100%, 50%, and 20% levels. And if your DUT requires that all three voltages be present simultaneously, the multiplexer won’t work at all. Just remember that the voltages must go to high-impedance loads to prevent loading down the circuit.

Q: Can all resistor networks be simplified into series and parallel combinations?
A: No. Consider the circuit on the left of Figure 5. Although it’s simple, consisting of just five resistors, none are in series or parallel with any of the others (unless R₅ is either zero or infinite). This particular circuit is very important in test and measurement. If I redraw it as shown on the right, you might recognize it.

Figure 5. The circuit on the left, equivalent to the Wheatstone bridge on the right, cannot be rearranged into a simple network of series and parallel resistors. (Image: Rick Nelson)

Q: It looks like the Wheatstone bridge, but isn’t R₂ variable, and isn’t a galvanometer used in place of R₅? A:
A: Yes, it’s the Wheatstone bridge, and in the original Wheatstone bridges, you’re right about R₂ and the galvanometer. In addition, R₄ was the unknown value that the bridge was designed to measure. Next time, we’ll comment on why the variable resistor and galvanometer are not often used in modern applications. We’ll see how we can analyze voltages and currents in this circuit using Ohm’s law and Kirchhoff’s current law, and we’ll see why this configuration is useful in strain gauges and other sensitive measurement devices.