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The DC Load Line Method

Published in: Electronics | Physics
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Ohm's law is applicable to resistors only. In case of devices such as diodes, transistors, FETs, etc., the way they behave on application of a voltage across them is vastly different from how resistors behave (V = IR)' moreover, their behavior is often described by much more complicated equations than V = IR. The DC load line method is one of the ways in electronics to solve problems graphically and, therefore, avoid dealing with seemingly nasty and intractable equations.

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    The DC Load Line Method Kishan Sinha Department of Physics and Astronomy University of Nebraska-Lincoln Given an electronic circuit, our first job is to find out the operating point of that circuit. Operating point refers to a set of parameters which includes voltages across all electrical components in the circuit, and the current flowing through the circuit. DC load line technique is one of many techniques to figure out the operating point of an electronic circuit. In case a circuit has only resistive components (resistor), figuring out its operating points is a straightforward task. All one needs to do is apply Kirchhoffs voltage and current laws, and Ohm 's law (V = IR for resistors). What makes resistive devices easy to deal with is their linear behavior. The resistance (or impedance) of these devices do not change regardless of the voltage across them. Thus, given the voltage across these devices, the current can be obtained simply by applying Ohm's law (I = V/R). Depending on the complexity of the circuit one may need to resort to techniques such as Thevenin's and Norton's theorems. Application of these theorems render a complex circuit mathematically more tractable. What if we are given a circuit where one or more components are non-resistive, hence non-linear, meaning they do not obey Ohm's law V= IR? Unlike resistors, impedance of a non-linear device changes with the voltage across it and, also, with the frequency of the applied voltage. Thus, we can no longer say I = V/R or (i = v/z) for R (or z) is no longer a constant. Application of Kirchhoffs voltage and current laws in such cases leads us nowhere. How, then, do we figure out operating point of circuit? Let's consider a simple circuit where a resistor R and a non-resistive device Z are in series as shown in figure 1. Our job here is to find the operating point (I, v R, Vz). Let's begin by listing everything we know about this circuit. V cc cc z Figure 1 and 1=1 R = Iz (required by law of charge conservation) . V = IR (Ohm's law for the resistor R) .... (2)
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    We do not have an expression for voltage and current in Z. At this point we can 't say anything about voltage across and current through Z except that whatever they might be they must be consistent with equations (1). So, we will need to solve this puzzle using only (1) and (2). Let's draw two axes: x being the voltage across the mysterious Z, and y being the current through the circuit. It is important here to keep in mind that current through Z is equal to the current through R and the current through the entire circuit (I = IR = Iz ) Since we don't know how Z will behave, let's first plot the extreme possibilities: when Z is infinite and zero. A) When Z T, entire voltage will appear across Z. Hence, Vz = Vcc. B) When Z = 0, the voltage across is 0 but the current through the circuit now is simply Vcc/R. We plot these two extremes on our I-V plot (figure 2). We have, thus, obtained a range of voltages (0 Vcc) that can appear across Z 1 This is the maximum possible current in our circuit given the power supply Vcc and resistor R. This indicates the value of current when the impedance of Z is zero or negligible. cc Voltage across the component B can never exceed this limit. cc This would happen when impedance Z is too large (practically infinite) compared to R. (voltage across the component Z) If you think about it, this is addition of voltages across the resistor and the component B; and, this can't ever change. Figure 2 Now, let's pick a point arbitrarily in the range [0, Vcc], say Vz, (figure 3) to represent voltage across Z. Then, the voltage across R MUST be Vcc-Vz. and hence, the current through R must be IR = (Vcc-Vz)/R (Ohm 's law for resistor). But, I = IR = Iz implying (Vcc-Vz)/R is also the current through Z: Iz = (Vcc-Vz)/R = I (the current through the entire circuit). We plot this current I and voltage Vz on our I-V plot.
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    1 IR = Vcc/R IR = ycE_yz cc Let's say this is the Then, this MUST be voltage across Z the voltage across R VT = Vz+ VR always Figure 3 We do this for all points in the range [0, Vcc], and we end up with a line (figure 4). It shouldn 't be surprising that this line represents the equation I = (Vcc-Vz)/R. This is what we call as the DC load line. Note that we haven't yet considered the nature of Z. So far, we have figured out only a range for current (I) and voltage (Vz) possible in the given circuit that would be consistent with equations (1) and (2). 1 1 = vcc/R The DC load line for the given circuit Figure 4 It's time to pin down the exact voltages and current through the circuit. We can't apply Ohm's law to our mysterious non-linear device but we can certainly figure out its characteristics by applying a range of voltages across it and measuring corresponding current through it. Manufacturers do this for us, and on Data Sheet for a device they always provide the I-V characteristics of that device. It may look like figure 5.
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    1 Figure 5 Figure 5 tells us how voltage and current across Z are related point-by-point. For a resistor it would simply be a straight line (I = V/R) with a slope I/R. All we now need to do is find a point on figure 5 which would be consistent with equations (1) and (2). We do this superimposing figures 4 and 5: This is how the voltage across the component Z and the current through are related. 1 1 = v cc/R Operating point of the circuit. This is the only point where the current through the resistor and component Z are equal; the law conservation of charge is satisfied only at this point Figure 6
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    1 cc Figure 7 We find a common point to the DC load line and I-V of our mysterious device. At this point, the current through the resistor R and the device Z is the same. This is required by the law of conservation of charge. Since nothing "normal" can by-pass this law, the entire circuit must work at this operating point. IMPORTANT: Note that we didn 't need to consider any characteristic of Z until we had the DC load line completely figured out. All we really used to construct our DC load line was a power supply V cc and a resistor R. I could have replaced V Z with a resistor, or a diode, or a transistor, or anything, and the DC load line would have remained unaltered.

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