Say we have a circuit with an inductive load in it. From the instant that electrical power is supplied to the circuit (by the closing of a switch say), the inductive load will accumulate stored energy. If an attempt is made to open the switch this energy will arc across the contacts of the switch, and could cause damage to the circuit components. Freewheeling diodes are placed across inductive loads to provide a path for the release of energy stored in the load when the load voltage drops to zero.

Figure 3.1 shows a diode circuit with a freewheeling diode. Diode D_m is the freewheeling diode. The circuit operation is divided into two modes. Mode 1, which begins when the switch is closed, and Mode 2, which begins when the switch is opened.

Mode 1 begins when the switch is closed at time t = 0. The equivalent circuit for this mode is shown in figure 3.1b.

Examination of the equivalent circuit for Mode 1, using voltage simple analysis, reveals the following:

Now we know that for an inductor

Therefore

            ..............(3.1)

We need to obtain an equation for the current in the circuit. This can be obtained by taking the Laplace transform of the equation, making I(s) the subject of the formulae, and then taking the inverse Laplace.

Recall

The Laplace Transform of a function F(t) is denoted by L{F(t)} and is defined by        ......(L.1) The Integral above is a function of the parameter s; therefore we call that function F(s) such that F(s) = L{F(t)} An important property of the Laplace function is its linearity, similar to the Differential function. If F1(t) and F2(t) have Laplace transforms and if c1 and c2 are arbitrary constants,        ......(L.2) An important Theorem (that will not be proven here) in working in the Laplace plane is as follows: If F(t) is continuous for t³  0 and is also of exponential order as t ® ¥ , and if F´(t) is of "Class A"*, then L{F´(t)} = sL{F(t)} - F(0)       ......(L.3) Where a function of Class A is one that is Sectionally Continuous over every finite interval in the range t ³  0 Of exponential order as t ® ¥ We can also say, from L.3, that L{F¢¢(t)} = sL{F¢(t)} - F¢(0) L{F¢¢(t)} = s[sL{F(t)} - F(0)] - F¢(0) F¢¢(s) = s²F(s) - sF(0) - F¢(0)       ......(L.4) and the process can be repeated as many times as we wish for more derivatives.

Let us assume that the initial conditions are zero, i.e i(t=0)=0. The Laplace transform can be applied to equation 3.1, using L.4 yielding

            ..............(3.2)

Solving for I₁(s) yields

            ..............(3.3)

where

            ..............(3.4)

The inverse transform of equation (3.3) in the time domain yields,

            ..............(3.5)

When the switch is opened at time t = t₁ at the end of this mode, the current becomes

            ..............(3.6)

If t₁ is sufficiently long such that the exponential term in equation 3.6 becomes negligible, then the steady-state current is given by

            ..............(3.7)

This mode begins when the switch is opened and the load current starts to flow through the freewheeling diode D_m. The voltage equation for this mode is given by

            ..............(3.8)

with initial conditions

Applying the Laplace transform to equation 3.8 yields

and solving for I₂ yields

            ..............(3.9)

where

Applying the inverse Laplace transform to equation 3.9 yields

            ..............(3.10)

Equation 3.10 is the freewheeling current which decays exponentially to zero at time t = t₂.

Last updated June 27th 2001
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The Department of Electrical & Computer Engineering,
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University of the West Indies,
St. Augustine