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Linear 1D Barcode Creator In Java Using Barcode generator for Java Control to generate, create 1D image in Java applications. Print Barcode In ObjectiveC Using Barcode encoder for iPad Control to generate, create barcode image in iPad applications. where T is the period of vAC (t) Now, if we let vAC (t) = A sin t we can express the average (DC) value of the load voltage VL = 1 T T /2 (1112) A sin t dt = (1 + cos t) A 2 (1113) vtrigger
11
Power Electronics
in terms of the ring angle, , de ned as
t = (1114) Vm 0 Vm Vm 0
By evaluating the integral of equation 1013, we can see that the (DC) load voltage amplitude depends on the ring angle, : 2 t VL = (1 + cos ) A 2 (1115) The following examples illustrate applications of thyristor circuits
2 t Figure 1126 Halfwave controlled recti er waveforms
EXAMPLE 116 Thyristorbased Variable Voltage Supply
Problem
Analyze the thyristorbased variable voltage supply shown in Figure 1127 Determine: (1) the rms load voltage as a function of the ring angle and (2) the power supplied to the resistive load at zero ring angle and at ring angles equal to /2 and + vL(t) 120 V + ~ AC _ Control
2 0 Control circuitry iG(t) iAK(t) + vAK(t) Solution
Known Quantities: Load resistance
Find: VL , PL  =0 , PL  = /2 , PL  =
Schematics, Diagrams, Circuits, and Given Data: VAK on = 0 V; RL = 240 pulsed gate current, iG (t), is timed as shown in Figure 1128 Assumptions: The thyristor acts as an ideal diode when on (VAK > 0) The
Part II
Electronics
vS(t) 120 2 iG(t) I 0 T+ T
Analysis: 1 Load voltage calculation As explained in the preceding section, the load voltage will have the appearance shown in Figure 1129 The rms value of the load voltage as a function of the ring angle, , is therefore computed as follows: (120 2)2 2 VL ( ) = sin t d( t ) 2 (120 2) 1 (1 cos(2 t ))d( t ) = 2 1 (120 2) 1 + sin(2 ) = 2 2 2 Load power calculation We can now compute the load power for each of the three values of : PL = For = 0: V2 = RL V2 RL 120 2 2 240 PL =
= 30 W; vL(t) 120 2 2 t 11
Power Electronics
for = /2: V2 = RL 120 2 1 1 2 2 240 120 2 1 1 2 240
PL = for = : = 15 W PL =
V2 = RL
Comments: Note that no power is wasted when the ring angle is set for zero load
voltage This would not be the case if a resistive voltage divider were used to adjust the load voltage EXAMPLE 117 Automotive Battery Charger
Problem
Qualitatively explain the operation of the automotive battery charger shown in Figure 1130
D1 R1 R2 Rp + VR Z C T2 R3
T1 120 VAC D2 + 12 V
D3 R4
Figure 1130 Automotive battery charger
Solution
Analysis: The charging circuit is connected to a standard 110V singlephase supply
Diodes D1 and D2 form a fullwave recti er (see Figure 842); resistors R1 and R2 and thyristor T2 form a variable voltage divider Assume that thyristor T2 is not in the conducting state and that the anode voltage of D3 is such that D3 conducts Then T1 will be red near the beginning of the positive halfcycle of the AC source voltage, and its period of conduction will be long, providing a substantial current to the battery (resistors R4 and Rp are suf ciently large that most of the current owing through T1 will go to the battery) The potentiometer Rp is set so that when the battery voltage is low, the voltage VR is not suf cient to turn on the Zener diode, Z Thus, Z is effectively an open circuit, and T2 remains off (recall that we had initially assumed T2 to be off this con rms the correctness of the assumption) As the battery charges to a progressively higher value, Z Part II
Electronics
will eventually conduct; when Z conducts, a gate current is injected into T2 , which is then turned on When T2 conducts, the voltage across the R2 T2 series connection becomes signi cantly lower, because T2 is now nearly a short circuit Resistors R1 and R2 are selected so that when T2 conducts, D3 becomes reversebiased Once this condition occurs, T1 is turned off and charging stops You see that the circuit has builtin overcharging protection EXAMPLE 118 Thyristor Circuit
Problem
Determine the value of R on the circuit of Figure 1131 such that the average load current through the thyristor is 1 A + vS
iL RG
R vt
Solution
Known Quantities: Resistances and source voltage Find: Resistor R such that iL = 1 A
V; R1 = 75
Schematics, Diagrams, Circuits, and Given Data: vS = 200 V rms, 250 Hz; VAK
; RG = 1 k ; C = 1 F
Assumptions: The thyristor acts as an ideal diode when on (VAK > 0) Analysis: Figure 1132 depicts the relative timing of the source voltage, vS (t), thyristor current, iL (t), and triggering voltage, vt (t) The expression for the source voltage is: vS (t) = 2 200 sin (2 250t)

