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AC + v = Vm sin t supply S
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Vm RL vo Vm + 2 t
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Figure 113 AC-AC converter circuit and waveform
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Finally, DC-AC converters, or inverters, are used to convert a xed DC supply to a variable AC supply; they nd application in AC motor control The operation of these circuits is rather complex; it is illustrated conceptually in the wave-
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VBE 1 + 05 Duty cycle = = t1/T VS DC supply Load <vo > 0 t1 T t + VBE 0 vo VS t1 T t
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Figure 114 DC-DC converter circuit and waveform
forms of Figure 115, where it is shown that by appropriately switching two pairs of transistors it is possible to generate an alternating current waveform (square wave)
VG1, VG2 1 + M1 VG1 VS DC supply Load 05 0 VG2 vo VS + 0 VS 0 T/2 05 T 1 t vZ V 0 iZ 1 RZ iZ T/2 T t M3 VG3 VG3, VG4 1 0 T/2 T t 05
M4 VG4
Figure 115 DC-AC converter circuit and waveform
Each of the circuits of Table 111 will be analyzed in greater detail later in this chapter
Voltage regulator
VOLTAGE REGULATORS
You will recall the discussion of the Zener diode as a voltage regulator in 8, where we introduced a voltage regulator as a three-terminal device that acts nearly as an ideal battery Figure 116 depicts the appearance of a Zener diode i-v characteristic and shows a block diagram of a three-terminal regulator A simple Zener diode is often inadequate for practical voltage regulation In some cases, the Zener resistance alone might cause excessive power dissipation
Unregulated supply
Load
Figure 116 Zener diode characteristic and voltage regulator circuit
11
Power Electronics
in the Zener diode (especially when little current is required by the load) A more practical and often-used circuit is a regulator that includes a series pass transistor, shown in Figure 117 The operation of this voltage regulator is as follows If the unregulated supply voltage, vS , exceeds the Zener voltage, vZ , by an amount suf cient to maintain the BJT in the active region, then vBE V and vZ VZ , the Zener voltage Thus, the load voltage is equal to vL = VZ V = Constant (111) and is relatively independent of uctuations in the unregulated source voltage, or in the required load current The difference between the unregulated source voltage and the load voltage will appear across the CE junction Thus, the required power rating of the BJT may be determined by considering the largest unregulated voltage, VSmax : PBJT = (VSmax VL )iC (VSmax VL )iL (112)
The operation of a practical voltage regulator is discussed in more detail in Example 111
EXAMPLE 111 Analysis of Voltage Regulator
Problem
iC + VCE _ _ + VBE
+ V _ S
Determine the maximum allowable load current and the required Zener diode rating for the Zener regulator of Figure 117
+ IL
RB iZ
+ VZ _
Solution
VL RL
Known Quantities: Transistor parameters; Zener voltage; unregulated source voltage; BJT base and load resistors Find: IL max and PZ
Figure 117 Practical voltage regulator
RB = 47
Schematics, Diagrams, Circuits, and Given Data: VS = 20 V; VZ = 127 V;
; RL = 10
Transistor data: TIP31 (see Table 112)
Assumptions: Use the large-signal model of the BJT Assume that the BJT is in the active region and the Zener diode is on and therefore regulating to the nominal voltage Analysis: Figure 118 depicts the equivalent load circuit Applying KVL we obtain:
+ RB + 127 V I VL RL
VZ = VBE + RL I From which we can compute the load current:
VZ V 127 13 = 114 A = RL 10
We then note that I is also the BJT emitter current, IE
Part II
Electronics
Applying KVL to the base circuit, shown in Figure 119, we compute the current through the base resistor: IRB = VS VZ 20 127 = 0155 A = RB 47
+ _ 20 V
Knowing the base and emitter currents, we can determine whether the transistor is indeed is the active region, as assumed With reference to Figure 1110, we do so by computing VCB and VBE to determine the value of VCE : The base voltage is xed by the presence of the Zener diode: VB = VZ = 127 V VCB = IRB RB = 0155 47 = 73 V VE = IL RL = 114 10 = 114 V Thus, VCE = VCB + VBE = VCB + (VB VE ) = 73 + (127 114) = 86 V This value of the collector-emitter voltage indicates that the BJT is in the active region Thus, we can use the large-signal model and compute the base current and subsequently the Zener current: 114 IE = = 1036 mA IB = +1 10 + 1 Applying KVL at the base junction (see Figure 1111), we nd IRB IB IZ = 0 IZ = IRB IB = 0155 01036 = 514 mA and the power dissipated by the Zener diode is:
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