0195136047.pdf
374 SEMICONDUCTOR DEVICES vGS = VT = 4 V vGS = 11 V 6 V 8 V 9 V 10 V VT 25 20 Drain current iD , mA 15 10 (a) (b) Cutoff region ...
7.4 FIELD-EFFECT TRANSISTORS 375 Solution (a) ForvGS=1 V: Since it is less thanVT, the MOSFET is in the cutoff region so that iD ...
376 SEMICONDUCTOR DEVICES Figure 7.4.9p-channel enhancement MOSFET.(a)Cross-sectional structure.(b)Symbol. MOSFETs, like JFETs, ...
7.4 FIELD-EFFECT TRANSISTORS 377 iD = 0 S + − G vGS ≤ −VP n+ n+ p n+ n+ p (b) D Depletion regions S G vGS < 0 n+ n n+ p d (a) ...
378 SEMICONDUCTOR DEVICES Solution Applying Equation (7.4.11): vGS=VP [{ iD IDSS + ( vDS VP ) 2 }( VP 2 vDS ) − 1 ] = 4 [{ 1 7 + ...
7.6 LEARNING OBJECTIVES 379 fact that typical semiconductors do not exhibit the magnetic properties needed to realize practical ...
380 SEMICONDUCTOR DEVICES MOSFET small-signal equivalent circuit (for low frequencies) and its application for simple circuit c ...
PROBLEMS 381 that would yieldI=− 8 μA. 7.2.4A silicon diode is forward-biased withV= 0. 5 V at a temperature of 293 K. If the di ...
382 SEMICONDUCTOR DEVICES + − 1 mA 1 0.8 0.6 0.4 0.2 0.9 0.7 0.5 0.3 0.1 0 (b) (a) 0.5 12 1.5 2.5 3 id, mA 4 V 2 kΩ vd + − id 2 ...
PROBLEMS 383 + − 500 kΩ 300 kΩ i 2 i 1 10 V Figure P7.2.13 − + 8 mA 1 2 3 4 5 0 (b) (a) 0.2 0.4 0.6 0.8 1 id, mA 2 V 500 Ω vd + ...
384 SEMICONDUCTOR DEVICES + − 9 V 300 Ω 100 Ω iD 600 Ω Diode Figure P7.2.16 v(t) vo(t) v(t) t 600 Ω 6 V 0 10 mS 200 Ω 10 V Ideal ...
PROBLEMS 385 VS VZ RZ = 0 VL = VZ RL VZ/RL Source resistance Regulator resistor DC source Regulator Load + RS I R − + + − − Figu ...
386 SEMICONDUCTOR DEVICES D 1 vL RL D 3 D 4 D 2 vS(t) = VS sin ωt + +− − Figure P7.2.30 RL vL(t) V 2 D 2 − + V 1 RS Source Limit ...
PROBLEMS 387 +−vπ ∆vBE = 0.05 V ∆vL RL = 10 kΩ 0.03 vπ ro C B E rπ + − + − Figure P7.3.7 E −2.0 −1.5 −1.0 −0.5 0 (b) Load line − ...
388 SEMICONDUCTOR DEVICES Figure P7.3.9 Q 2 iE 2 iB 2 e iE 1 b Q 1 c iC 2 iCC iC 1 iB 1 Figure P7.3.10 RB RC −VCC Figure P7.3.11 ...
PROBLEMS 389 7.3.13If the circuit of Example 7.3.1 is to switch from cutoff to saturation, find the condition onvS, given that t ...
390 SEMICONDUCTOR DEVICES 7.4.5For ap-channel JFET in its active region, specify the polarities of voltages and the directions o ...
PROBLEMS 391 + + + + + − − G RD iG = 0 iD VDD VGG vg = √ 2 Vg sin ωt vG vD Figure P7.4.6 ID, mA VGS, V VDS, V VGS = 0 V −0.5 V − ...
392 SEMICONDUCTOR DEVICES VDS, V VGS ≤ 2.5 V 5.5 V VGS = 5.0 V 4.5 V 4.0 V 3.5 V 3.0 V ID, mA 10 8 6 4 2 (^0246810121416) Figure ...
8 Transistor Amplifiers 8.1 Biasing the BJT 8.2 Biasing the FET 8.3 BJT Amplifiers 8.4 FET Amplifiers 8.5 Frequency Response of ...
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