Mathematical Analysis

An expression for the current through the load can be obtained as shown below. It can be assumed that the load current flows all the time. In other words, the load current is continuous. When diode D₁ conducts, the driving function for the differential equation is the sinusoidal function defining the source voltage. During the period defined by p < wt < 2p, diode D₁ blocks current and acts as an open switch. On the other hand, diode D₂ conducts during this period, the driving function can be set to be zero volts. For 0 < wt < p , the equation (1) shown below applies.

For the negative half-cycle of the source, equation (2) applies. As in the previous case, the solution is obtained in two parts. The expressions for the complementary integral and the particular integral are the same. The expression for the complementary integral is presented by equation (3). The particular solution to the equation (1) is the steady-state response and is presented as equation (4). The total solution is the sum of both the complimentary and the particular solution. For 0 < q < p, where wt = q, the total solution is presented as equation (5).

The difference in solution is in how the constant A in complementary integral is evaluated. In the case of the circuit without free-wheeling diode, i(0) = 0, since the current starts building up from zero at the start of every positive half-cycle. On the other hand, the current-flow is continuous when the circuit contains a free-wheeling diode also. Since the input to the RL circuit is a periodic half-sinusoid function, we expect that the response of the circuit should also be periodic. That means, the current through the load is periodic. It means that i(0) = i(2p).

Since the current through the load free-wheels during p < q < 2p , we get equation (6). We use ( q - p ) for the elapsed period in radians instead of q itself, since the free-wheeling action starts at q = p . From the total solution, we can get i(p) from equation (7) by substituing q = p. To obtain A, the following steps are necessary. From the total solution, obtain an expression for i(0) by substituting 0 for q. Also obtain an expression for i(p) by substituting p for q in equation (7). Using this expression for i(p) in equation (6), obtain i(2p) by letting q = 2p . Since i(0) = i(2p), we can obtain A from equation (8). In equation (8), the terms containing constant A are grouped on the left-hand side of equation and the other terms on the right-hand side.

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