Analysis of the Ideal Circuit

Analysis of the circuit is carried out based on the following assumptions. The circuit is ideal. It means when the switch is ON, the drop across it is zero and the current through it is zero when it is open. The diode has zero voltages drop in the conducting state and zero current in the reverse-bias mode. The time delays in switching on and off the switch and the diode are assumed to be negligible. The inductor and the capacitor are assumed to be lossless.

1. The responses in the circuit are periodic. It means especially that the inductor current is periodic. Its value at the start and end of a switching cycle is the same. The net increase in inductor current over a cycle is zero. If it is non-zero, it would mean that the average inductor current should either be gradually increasing or decreasing and then the inductor current is in a transient state and has not become periodic.
2. It is assumed that the switch is made ON and OFF at a fixed frequency and let the period corresponding to the switching frequency be T. Given that the duty cycle is D, the switch is on for a period equal to DT, and the switch is off for a time interval equal to (1 - D)T.
3. The inductor current is continuous and is greater than zero.
4. The capacitor is relatively large. The RC time constant is so large, that the changes in capacitor voltage when the switch is ON or OFF can be neglected for calculating the change in inductor current and the average output voltage. The average output voltage is assumed to remain steady, excepting when the change in output voltage is calculated.
5. The source voltage V_S remains constant.

Inductor Current with Switch Closed

When the switch is closed, the equivalent circuit that is applicable is shown in Fig. 2. The source voltage is applied across the inductor and the rate of rise of inductor current is dependent on the source voltage V_S and inductance L. The differential equation describing this condition is:

If the source voltage remains constant, the rate of rise of inductor current is positive and remains fixed, so long as the inductor is not saturated. Then equation (1) can be expressed as :

The switch remains ON for a time interval of DT in one switching cycle and hence DT can be used for )t. The net increase in inductor current when the switch is ON can be obtained from equation (2) to be:

Inductor Current with Switch Open

When the switch is open, the circuit that is applicable is shown in Fig. 3. Now the voltage across the inductor is:

Given that the output voltage is larger than the source voltage, the voltage across the inductor is negative and the rate of rise of inductor current, described by equation (5), is negative. Hence if the switch is held OFF for a time interval equal to (1 - D)T, the change in inductor current can be computed as shown in equation (6)

.

The change in inductor current reflected by equation (6) is a negative value, since V_o > V_S. Since the net change in inductor current over a cycle period is zero when the response i_L(t), the sum of net changes in inductor current expressed by (4) and (6) should be zero. That is,

On simplifying equation (7), we get that

It has been stated that when i_L(t) is periodic, the net change in inductor current over a cycle is zero. Since change in inductor current is related to its volt-seconds, the net volt-seconds of the inductor has to be zero. The expression for the net volt-seconds can be obtained from equation (6) and it can be seen that the numerator of equation (7) should be zero. That is,

The value of D varies such that 0 < D < 1 and it can be seen from equation (8) that output voltage is greater than the source voltage, and hence this circuit is called the boost converter. The output voltage has its lowest value when D = 0 and then the output voltage equals the source voltage. When D approaches unity, output voltage tends to infinity. Usually D is varied in the range 0.1 # D # 0.9 .

The waveforms of inductor voltage and inductor current are shown in Fig. 4. These waveforms are drawn assuming that both the output and the source voltage remain steady. These waveforms illustrate how the inductor voltage is related to its current.

Output Voltage Ripple with Switch Closed

In this sub-section, the change in output voltage is calculated. It needs to be emphasized that the peak-to-peak ripple in output voltage is quite small for a well-designed circuit. For the inductor, the net change in inductor current over a cycle is zero when i_L(t) is periodic. For the capacitor, the net change in capacitor voltage over a cycle is zero when it is periodic. When the switch is closed, the equivalent circuit in Fig. 2 shows that the boost converter is split into two sub-circuits, with the loop currents decoupled from each other. When the switch is closed, the output voltage is sustained by the capacitor. During this period, the capacitor discharges part of its stored energy and it re-acquires this energy when the switch is open. When the switch is open, part of the inductor current charges the capacitor since the inductor current usually remains larger than the current through the load resistor. From Fig. 2,

When current through a capacitor charges it up, its rate of rise of capacitor voltage is positive since the capacitor voltage is increasing. When the switch is open, the capacitor is discharging its energy with its voltage falling and the current through the capacitor is then a negative value. The output voltage remains positive and hence the output current is positive and it is the negative of the capacitor current, as can be seen from Fig. 2. Since the change in output voltage is quite small, it can be assumed that the load current remains constant at its average value and equation (10) can be now expressed as:

When the capacitor current is constant, its voltage changes linearly with time. Here the period for which the switch is closed is DT and the DT can be used in place of )t and the peak-to-peak ripple in output voltage expressed as )v_o is then:

Equation (12) yields the value of the peak-to-peak ripple in output voltage. In equation (12), 1/f replaces T since T is the reciprocal of switching frequency.

Figure 5 shows how the capacitor current and voltage vary over a cycle. The ripple in output voltage is exaggerated in Fig. 5, whereas in practice it would be much smaller. If the output voltage is drawn to scale, the ripple in output voltage would not be noticeable.

Expression for Average Inductor Current

The average inductor current can be found out by equating the power drawn from the source to the power delivered to the load resistor. Again the ripple in output voltage is ignored and it is assumed justifiably that the output voltage remains steady at its average value. Power P_o absorbed by load resistor is then:

It can be seen from the circuit in Fig. 1 that the current drawn from the source flows through the inductor. Hence the average value of inductor current is also the average value of source current. Let the average inductor current be I_L. Then power P_S supplied by the source is then:

After equating equations (13) and (14), we get the average inductor current as:

Since load current I_o is:

Using equations (8) and (16), equation (15) can be re-presented as:

Since 0 < D < 1, it can be seen from equation (17) that I_L > I_o.

 
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