The resistor, capacitor, and inductor are linear circuit elements in that their v-i relations are linear in the mathematical sense. Voltage and current sources are (technically) nonlinear devices: stated simply, doubling the current through a voltage source does not double the voltage. A more blatant, and very useful, nonlinear circuit element is the diode (learn more). Its input-output relation has an exponential form.
Because of the diode's nonlinear nature, we cannot use impedances nor series/parallel combination rules to analyze circuits containing them. The reliable node method can always be used; it only relies on KVL for its application, and KVL is a statement about voltage drops around a closed path regardless of whether the elements are linear or not. Thus, for this simple circuit we have
We need to detail the exponential nonlinearity to determine how the circuit distorts the input voltage waveform. We can of course numerically solve [link] to determine the output voltage when the input is a sinusoid. To learn more, let's express this equation graphically. We plot each term as a function of for various values of the input voltage ; where they intersect gives us the output voltage. The left side, the current through the output resistor, does not vary itself with , and thus we have a fixed straight line. As for the right side, which expresses the diode's v-i relation, the point at which the curve crosses the axis gives us the value of . Clearly, the two curves will always intersect just once for any value of , and for positive the intersection occurs at a value for smaller than . This reduction is smaller if the straight line has a shallower slope, which corresponds to using a bigger output resistor. For negative , the diode is reverse-biased and the output voltage equals .
What utility might this simple circuit have? The diode's nonlinearity cannot be escaped here, and the clearly evident distortion must have some practical application if the circuit were to be useful. This circuit, known as a half-wave rectifier, is present in virtually every AM radio twice and each serves very different functions! We'll learn what functions later.
Here is a circuit involving a diode that is actually simpler to analyze than the previous one. We know that the current through the resistor must equal that through the diode. Thus, the diode's current is proportional to the input voltage. As the voltage across the diode is related to the logarithm of its current, we see that the input-output relation is