However, at this stage, the reader may prefer to look upon the inductor as a device that tends to maintain the current reasonably constant during the switching action. (That is, it stores energy when the power device is “on”
and transfers this energy to the output when the power device is “off.”) I prefer the termchokefor the power inductor, because in this application it must support an element of DC current as well as the applied AC voltage stress. It will be shown later (Chapter 7) that the design of pure inductors (with zero DC current component) is quite different from the design of chokes, with their relatively large DC current component.
In the following section Mr. Pressman outlines the parameters that control the design and selection of this critical part.∼K.B.
The current waveform of the output inductor (choke) is shown in Figure 1.4f, and its characteristic “dual ramp” shape is defined in Section 1.3.2. Notice that the current amplitude at the center of the ramp is the mean value equal to the DC output currentIo.
We have seen that as the DC output load current decreases, the slope of the ramp remains constant (because the voltage acrossLoremains constant). But as the mean load current decreases, the ripple current waveform moves down toward zero.
At a load current of half the peak-to-peak magnitude of the ramp, Io = (I2 −I1)/2dI, the lower point of the ramp just touches zero.
At this point, the current in the inductor is zero and its stored energy is zero. (The inductor is said to have “run dry.”) If the load current is further reduced, there will be a period when the inductor current remains at zero for a longer period and the buck regulator enters into the “discontinuous current” operating mode. This is an important transition because a drastic change occurs in the current and voltage waveforms and in the closed loop transfer function.
This transition to the discontinuous mode can be seen in the real- time oscilloscope picture of Figure 1.6a.This shows the power switch current waveforms for a buck regulator operating at 25 kHz with an input voltage of 20 V and an output of 5 V as the load current is reduced from a nominal current of 5 A down to about 0.2 A.
The top two waveforms have the characteristic ramp-on-a-step waveshape with the step size reducing as the load current is reduced.
The current amplitude at the center of the ramp indicates the effective DC output current.
In the third waveform, whereIo = 0.95 A, the step has gone and the front end of the ramp starts at zero current. This is thecritical load currentindicating the start of the discontinuous current mode (or run- dry mode) for the inductor. Notice that in the first three waveforms, theQ1 “on” time is constant, but decreases drastically as the current is further reduced, moving deeper into the discontinuous mode.
In this example, the control loop has been able to maintain the out- put voltage constant at 5 V throughout the full range of load currents, even after the inductor has gone discontinuous. Hence it would be easy to assume that there is no problem in permitting the inductor to go discontinuous. In fact there are changes in the transfer function (discussed next) that the control loop must be able to accommodate.
Further, the transition can become a major problem in the boost-type topologies discussed later.
For the buck regulator, however, the discontinuous mode is not considered a major problem. For load currents above the onset of the discontinuous made, the DC output voltage is given byVo=V1Ton/T.
Notice the load current is not a parameter in this equation, so the voltage remains constant with load current changes without the need to change the duty ratio. (The effective output resistance of the buck regulator is very low in this region.) In practice the “on” time changes slightly as the current changes, because the forward drop acrossQ1 and the inductor resistance change slightly with current, requiring a small change inTon.
If the load is further reduced so as to enter discontinuous mode, the transfer function changes drastically and the previous equation for output voltage (Vo=V1Ton/T) no longer applies. This can be seen in the bottom two waveforms of Figure 1.6a. Notice the “on” time ofQ1 has decreased and has become a function of the DC output current.
FIGURE1.6 A 25-kHz buck regulator, showing the transition from the continuous mode to the discontinuous mode at the critical load current, with the inductorLorunning dry. Note, in Figure 1.6a ,line three above, that the “on”
time remains constant only so long as the inductor is in the continuous mode.
TIP The ratioTon/Tis normally referred to as the duty ratio D.The voltage formula for continuous operation is simplyVo = V1.D.However, for dis- continuous operation, the duty ratio becomes a function of the load current, and the situation is much more complicated. In the discontinuous mode, the output voltageVois given by the formula
Vo= V1.2D
D+(D2+(8L/RT))1/2
Since the control loop will maintain the output voltage constant, the effec- tive value of the load resistanceRwill be inversely proportional to the load current. Hence by holdingVo, V1,L,andTconstant, to maintain the voltage constant, requires that the remaining variable (the duty ratioD) must change with load current.
At the critical transition current, the transfer function will change from continuous mode in which the duty ratio remained constant with load change (zero output impedance) to the discontinuous mode in which the duty ratio must change with reducing load current (a finite output impedance). Hence in the discontinuous mode, the control loop must work much harder, and the transient performance will be degraded.∼K.B.
Dynamically, at load currents above the onset of the discontinu- ous mode, the output L/C filter automatically accommodated out- put current changers by changing the amplitude of the step part of the ramp-on-step waveforms shown in theQ1 andD1 waveforms of Figures 1.4dand 1.4e.To the first order, it could do this without chang- ing theQ1 “on” time.
The DC output current is the time average of theQ1 andD1 ramp current. Notice that in Figure 1.6a ,line three and line four, that at lower currents where the inductor has gone discontinuous and the step part of the latter waveforms has gone to zero, the only way the current can decrease further is to decrease the Q1 “on” time. The negative- feedback loop automatically adjusts the duty ratio to achieve this.
The dramatic change in the waveforms can be seen very clearly between Figure 1.7a(for the critical current condition) and Figure 1.7b (for the discontinuous condition). Figure 1.7b(2) shows theD1 current going to zero just beforeQ1 turns “on” (the inductor has dried out and gone discontinuous). With zero current inLo, the output voltage will seek to appear at the emitter ofQ1. However, the sudden transition results in a decaying voltage “ring,” at a frequency determined byLo and the distributed capacitance looking into theD1 cathode andQ1 emitter junction at pointV1. This is shown in Figure 1.7b(1).
TIP Although the voltage ring is not damaging, in the interest of RFI reduction, it should be suppressed by a small R/C snubber acrossD1.∼K.B.
FIGURE1.7 A 25-kHz buck regulator with typical waveforms.Q1 emitter voltage waveforms andD1 current waveforms for continuous conduction at the critical current (a) and in the discontinuous mode (b).