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iD RS +1 VP
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We can expand the above equation to obtain: 2IDSS RS IDSS 2 2 R i 1+ 2 VP VP S D iD + IDSS = 0
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and solve for the given values to obtain iD = 6 mA and iD = 15 mA These lead to values of vGS = 6 V and vGS = 15 V Since the condition vGS = VP is satis ed only for the second solution, we conclude that the circuit of Figure 944(a) indeed operates as a current source in the saturation region, as indicated in Figure 944(b) The strength of the current source is iD = 15 mA Note that this value is different from the parameter IDSS , which is the nominal current source strength when RS = 0
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Comments: The source resistor can be adjusted to yield the desired source current in
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spite of variations in IDSS
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914 What is RDS for the circuit of Figure 944(b) 915 What is the drain current for the circuit of Figure 944(b) 916 Determine the actual operating point of the JFET of Example 912, given the
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choice of standard resistor values
917 Repeat the design of Example 912 (ie, calculate RS and RD ) if the required drain current is halved
CONCLUSION
Transistors are three-terminal electronic semiconductor devices that can serve as linear ampli ers or switches The bipolar junction transistor (BJT) acts as a current-controlled current source, amplifying a small base current by a factor ranging from 20 to 200 The operation of the BJT can be explained in terms of the device base-emitter and collector i-v characteristics Large-signal linear circuit models for the BJT can be obtained by treating the transistor as a controlled current source Field-effect transistors (FETs) can be grouped into three major families: enhancement MOSFETs, depletion MOSFETs, and JFETs All FETs behave like voltage-controlled current sources FET i-v characteristics are intrinsically nonlinear, characterized by a quadratic dependence of the drain current on gate voltage The nonlinear equations that describe FET drain characteristics can be summarized in a set of universal curves for each family
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CYU 91 CYU 92 CYU 93 CYU 95 CYU 96 CYU 97 CYU 98 CYU 99 CYU 910 CYU 911 CYU 913 CYU 914 CYU 915 CYU 916 CYU 917
1 A = (ri +RS ) (roRL L ) ; ri 0, ro 0 +R
ro R L 1 (ri +RS ) (ro +RL )
ri r RL A = (ri +RS ) (roo+RL )
Saturation VBB = 5 V; VCE = 644 V 374%; because RE provides a negative feedback action that will keep IC and IE almost constant Saturation; 85 mA 159 mW The MOSFET is in the ohmic region Choosing the smaller value of vGS , RS = 207 k , vGS = 286 V, iD = 0586 mA The answer is not unique: Selecting the larger gate voltage, we nd RS = 115 k Approximately 400 200 Approximately 25 mA iD = 389 mA; vDS = 1057 V; vGS = 058 V RS = 634 ; RD = 6366 k
Part II
Electronics
HOMEWORK PROBLEMS
Section 1: Bipolar Transistors 91 For each transistor shown in Figure P91, determine
whether the BE and BC junctions are forward- or reverse-biased, and determine the operating region
+ 07 V + _ 4V _ (b) _ 06 V + _ 54 V 10 V 15 V 02 V
magnitudes of the voltages across the emitter-base and collector-base junctions are 065 and 73 V Find a VCE b IC c The total power dissipated in the transistor, de ned here as P = VCE IC + VBE IB
_ _ 06 V + + (a) + + 03 V
95 Given the circuit of Figure P95, determine the
emitter current and the collector-base voltage Assume the BJT has V = 06 V
15 k 30 k
07 V _ _ (c)
Figure P95
+ (d)
96 Given the circuit of Figure P96, determine the
operating point of the transistor Assume a 06-V offset voltage and = 150 In what region is the transistor
Figure P91
92 Determine the region of operation for the following
transistors: a npn, VBE = 08 V, VCE = 04 V b npn, VCB = 14 V, VCE = 21 V c pnp, VCB = 09 V, VCE = 04 V d npn, VBE = 12 V, VCB = 06 V
62 k 33 k
18 V
93 Given the circuit of Figure P93, determine the
operating point of the transistor Assume the BJT is a silicon device with = 100 In what region is the transistor
15 k 12 k
Figure P96
97 Given the circuit of Figure P97, determine the
820 k 22 k
emitter current and the collector-base voltage Assume the BJT has a 06-V offset voltage at the BE junction
20 k 12 V 20 V 20 V 39 k
910
Figure P97
98 If the emitter resistor in Problem 97 (Figure P97) is
Figure P93
94 The magnitudes of a pnp transistor s emitter and
base currents are 6 mA and 01 mA, respectively The
changed to 22 k , how does the operating point of the BJT change 99 The collector characteristics for a certain transistor are shown in Figure P99
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