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From our parts list we will add an inductor and diode to our heater circuit and get the schematic shown Figure 3-4. 
Figure 3-4: Switched Inductor Converter
Unlike our capacitor example, this circuit is 100% efficient with ideal components. It's major problem is terrible load transient response. If the load is removed, the current must continue to flow in the inductor and the output voltage tends to go to infinity. If the load is increased by two, the output voltage is halved, since the inductor current remains the same until adjusted by the L/R time constant. Because of the terrible load transient response this circuit is rarely used as a voltage regulator. But it is interesting to note that in the type of dc-dc converter we are discussing in this section, an inductor is always needed, but a capacitor is not. It is also interesting to note that this is the equivalent circuit of a switching-mode control of current in a motor field winding, which is represented as an inductor and series resistor.
Note that the diode is necessary to provide a current path for the inductor current when the switch opens. Otherwise the voltage across the inductor would build up to several hundred volts until the energy stored in the inductor would dissipate as an arc in the switch, charge the parasitic capacitance of the inductor windings, or result in some other undesired phenomenon. A rule of thumb, which we will use later for selecting a value of inductance in an LC filter, is to start with an inductor value that results in a peak-to-peak inductor current that is 10% of the full load current. For our example of a 12 V input, 5 V output, 20 A load, and 20 kHz switching frequency, the inductor can be calculated from the formula V = L*(di/dt). This is a formula you will constantly use in switching-mode power supply design, hence worth memorizing. Example:
L = V*Ton/I
V = 12 V - 5 V = 7 V
Ton = D*T = 0.417*50E-6 = 20.85E-6
I = 20 A * 10% = 2 A
L = (7 V)*(20.85E-6 s)/(2 A) = 72.3 uH
=> 75 uH The resulting two amps (1.94 A) peak-to-peak ripple current gives about a half volt peak-to-peak ripple current in the 5 V, 20 A load. Note that you can decrease the ripple current to as small as you want by increasing L, or make L smaller by allowing more ripple current. Do not use this information for design without independent verification of the information. Editor: Jerrold Foutz |
