Part 1: Fundamental Concepts: Simplifying Circuits

When solving for the currents and voltages that appear in a circuit, two techniques have been explored. The first involved combining component values and then using the result to find the specific value of interest. The second approach, using Kirchhoff’s Laws to develop a set of equations, allows you to finds all unknown values in the circuit. There is another technique that is also useful when dealing with circuits. You can, in many cases, simplify the circuit you are interested in to find the value we are after. Take, for example, the circuit shown in Figure 1. If you are interested in knowing the value of Vout, the schematic can be redrawn to eliminate most of the components so that the solution can be easily found. Circuit simplification is also useful for developing circuit equivalents so that the effects of external loading can be explored. This will be covered in the next section.

Figure 1. Circuit used to demonstrate simplification.

The strategy is to repeatedly combine the components to find their equivalent values until the circuit is simplified and the solution found. A good first step is to eliminate the parts of the circuit that aren’t involved with the problem. What can we get rid of? R2 seems like a good candidate. Removing it does not change Vout because its presence does not alter Vin. Another immediate observation is that R3 and R4 would act like a single resistor if they were combined. There are no circuit branches connected to them, so combining them does not alter the circuit. These changes are shown in Figure 2.

Figure 2. Circuit after first simplification.

With the first simplification done, things are beginning to look considerably simpler. One thing that sticks out in the simplified circuit is that there are three series resistors in a row. Eliminating them by combining their values does not alter the circuit because again, there are no branches from them, and they are electrically the same as a single resistor. Figure 3 illustrates this.

Figure 3. Circuit after second simplification.

There are several ways to proceed from here. If possible, it is best to take advantage of obvious features in the circuit. One obvious thing is that R6 and R8 are equal values. Vout is taken between them. If we knew the value of the node that connects R6 and R7, labeled as V5 in Figure 4, then the answer would be half of that (if R6 and R8 were not equal what we are about to do would still work but it involves another calculation). So, how do we find V5? One way is to combine R6 and R8 as series resistor then take that series combination in parallel with R7. This makes one value out of the three and allows to make a simple voltage divider with the resistor connected to Vin. This is shown below in Figure 4.

Figure 4. Circuit after third simplification.

With this done the value of V5 can be found by treating the two resistors as a voltage divider which yields a value if 3.116V for V5. We are almost done because Vout is simply half the value of V5. If R6 and R8 were not equal in value we would still do this step but we would have to calculate the voltage divider ratio for R6 and R8 rather than knowing it was 1/2 due to the equal values.

The last step to finding the answer is shown in the following figure. The schematic on the left is shown in a partially simplified state and the one on the right gives the final solution. Notice that, up to this point, we didn’t care what Vin is. Now we do because we are looking for a numeric answer. The answer could also be written as a ratio to Vin if that fits your needs.

Figure 5. Final solution.

It is important to remember that this output voltage is only valid when the output is unloaded (connected to an open, or very high resistance, circuit). If you connect a load to it, another resistor for example, its value will change. If you are designing a circuit where this is the case, where the load is likely to vary, then the use of a buffer amplifier (discussed in a later chapter), or something similar, is needed.