Ten Essential Skills for Electrical Engineers

1

HOW TO DESIGN RESISTIVE CIRCUITS

Chances are good that every schematic diagram you’ve seen contained at least a few resistors. This component is an electrical workhorse, commonly used for establishing bias voltages, programming gain, summing signals, attenuating signals, and numerous other functions. Ideal resistors dutifully follow Ohm’s law, which has no frequency dependence, so it is easy to believe that designing with resistors is a simple task. This is probably the most common reason candidates are caught off guard and fail when asked to design simple resistive networks in interviews. This chapter will show you how to design and analyze practical resistive networks that solve problems you’ll encounter in interviews and in the workplace.

The resistor was probably the first component you studied in school. At that time, it was the only component in your toolset, so the problems you solved were limited to finding voltages and currents in DC networks. Doing these problems taught you valuable skills such as nodal and mesh analysis, but the problems were not particularly practical and perhaps not very interesting. As a graduating engineer your knowledge of circuit elements has broadened significantly and you have better computer tools to help with the mathematical manipulations. The examples and problems in this chapter should be much more interesting because they represent practical design problems; they should be more enjoyable because, after setting up the problems, we will rely on the computer for the brunt of the manipulations.

This chapter begins with the commonly asked interview problem of creating a voltage source with a specified Thevenin resistance. Since this is such a common problem, we derive equations so that you can easily compute the resistor values when you encounter it. Next, we design a coupling circuit with specific design requirements. Many experienced engineers design this circuit using an op-amp and numerous resistors, but you’ll see that a network with only three resistors fulfills the design requirement. We then design a 50 Ω bidirectional attenuator that is commonly found in RF circuits. This is not an easy problem, but it provides a good example of converting design requirements into solvable equations and then calculating the components. Since the attenuator design is a difficult problem, we check our result using mesh analysis, and you’ll see that analyzing resistive networks is simply a matter of writing the correct equations and then letting a matrix solver compute the solution.

1.1 DESIGN OF A RESISTIVE THEVENIN SOURCE

Designing circuits involves taking inventory of what is required, what is known, what is unknown, and what is available. When designing resistive networks we must often solve simultaneous nonlinear equations as shown in the following example:

Example 1.1. Given a 3.3 V power supply, design a resistive circuit that provides 1.8 V with a source resistance of 1.5 kΩ.

Solution.   This is a practical problem encountered frequently when designing bias circuits. We immediately note the desired output voltage is less than the supply voltage so a resistive circuit will suffice. We also recognize that the requirement leads us to a Thevenin source.

Knowing that two design parameters (Thevenin voltage and resistance) must be simultaneously satisfied, it makes sense1 to try the circuit with two resistors as shown in Figure 1.1b.

Figure 1.1 Required Thevenin source (a) and proposed resistive network (b) for Example 1.1.

The design procedure is to compute R1 and R2 so that the Thevenin equivalent of the proposed network in Figure 1.1b is the Thevenin source of Figure 1.1a. The Thevenin resistance of the proposed network is the resistance seen at the terminals with the voltage source shorted, so our first equation is the formula for parallel resistances:

(1.1)

The Thevenin voltage is the voltage at the terminals of the circuit of Figure 1.1b when it is unloaded, so we use voltage division to write the second equation:

(1.2)

These two equations are nonlinear and cannot be simultaneously solved using a simple matrix. Instead, we solve Equation 1.1 for R1 which gives

(1.3)

Then, substitute R1 from Equation 1.3 into Equation 1.2 and solve for R2:

(1.4)

Finally substitute Equation 1.4 into Equation 1.3 to compute R1:

(1.5)

To check the result, substitute these resistances into Equations 1.1 and 1.2.

1.2 DESIGN OF A COUPLING CIRCUIT

A common application for resistive circuits is the coupling circuit of Figure 1.2 that conditions a signal from integrated component IC1 and feeds it to integrated component IC2. The signal of interest, or the signal that carries useful information, is an alternating current (AC) signal with no DC component. However, DC offsets play an important role in this application because they keep the input and output signals within the working range of the amplifiers.

Figure 1.2 Coupling circuit.

The required functionality of this circuit is to

This functionality is provided by the circuit of Figure 1.3. Our design strategy is to first analyze the circuit and then use the analysis results to compute the component values.

Figure 1.3 This coupling circuit provides a specified load impedance for IC1, attenuates the signal from IC1 to IC2, and provides a specified DC bias for IC2.

The focus of this chapter is resistive circuits, so the capacitor in Figure 1.3 deserves explanation. It is called a “blocking capacitor” because its primary function is to prevent or block the DC voltage at the output of IC1 so that it does not appear at the input of IC2. The capacitor is selected so that it approximates a short circuit to the signal of interest. Chapter 3 shows how to select the correct value of this capacitor.

Prior to analyzing the circuit of Figure 1.3, we digress briefly to clarify the concept of “AC ground.” This powerful tool is often a source of confusion for both students and experienced engineers. The confusion arises from not fully grasping the difference between resistance and impedance. Resistance is the ratio of the DC voltage across a device to the DC current passing through it. Impedance is the frequency-dependent, complex ratio of the AC voltage to the AC current in a device or circuit. One way to remember this is to think of impedance as the derivative:

(1.6)

Note from Equation 1.6 that the impedance of a resistor is equal to its resistance.

Now consider an ideal DC power supply. Its function is to provide a fixed voltage regardless of the load current. Since its voltage does not change, Equation 1.6 shows that its impedance is zero.2 Therefore, when computing the impedance of the circuit in Figure 1.3, we treat the power supply, VS, as an AC ground.

Similarly the attenuation provided by the circuit of Figure 1.3 refers to the ratio of the AC voltage at the output of IC1 to the AC voltage at A. When computing attenuation, the power supply, VS, is also an AC ground.

The preceding discussion shows that for computing input impedance and gain, 3 the DC supply, VS, is an AC ground which places R1 in parallel with R2 in Figure 1.3. We therefore simplify Figure 1.3 by representing R1, R2, and VS as a DC Thevenin source as shown on the right side of Figure 1.4a. Since the capacitor appears as a short circuit to the desired signal and since the Thevenin voltage source is an AC ground, the circuit of Figure 1.4b can be used to compute the input impedance and gain.

Figure 1.4 Simplifications of the circuit of Figure 1.3 are used to compute input impedance and circuit gain. A Thevenin equivalent allows R1, R2, and VS to be shown as a Thevenin equivalent (a). Since the capacitor is an AC short and since voltage source VTH is an AC ground, the circuit can be further simplified (b).

From Figure 1.4b the input impedance is

(1.7)

And the gain is computed using the voltage divider theorem

(1.8)

The analysis above can be used to design the coupler based on a design specification as shown in the following example.

Example 1.2. The circuit shown in Figure 1.5 is a coupling circuit for an AC signal....