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Eastern Mediterranean University Department of Electrical and Electronic Engineering EENG 341 ELECTRONICS I – INFE 242 ELECTRONICS FINAL EXAM Date Duration : 25 Jan 2010 : 120 min. ANSWER ALL 4 QUESTIONS (1) Figure 1 In the MOSFET circuit shown in Figure 1, the transistors have widths W1 = 10 μm, W2 = 40 μm, and their lengths are equal, i.e. L1 = L2 = 2 μm. Both transistors have kn 0.2 mA/V2 , and Vt = 1 V. Find the voltage Vo and the current ID. (25 pts) +5 V Q1 ID Vo Q2 +15 V (2) Figure 2 5 kΩ The MOSFET in the common-source amplifier shown in Figure 2 has the small signal parameters gm = 2 mA/V and 0 . The coupling and bypass capacitors are short circuits at the signal frequencies. (a) Draw the small-signal equivalent circuit of the amplifier. (10 pts) (b) Find the voltage gain Av vo / vi . (5 pts) (c) Find the overall voltage gain Gv vo / vsig . Rsig = 100 kΩ vo CC2 CC1 20 kΩ + vsig Rout + _ vi Rin 2 MΩ 1 kΩ CS (5 pts) (d) Find the input (Rin) and output (Rout) resistances. (5 pts) (3) -15 V Figure 3 The BJT in the circuit of Figure 3 has 1 and VCEsat = 0.2 V. At the edge of conduction VBE = 0.5 V. For active operation and saturation VBE = 0.7 V. (a) For VB = 0 V, find the collector and emitter voltages VC, VE. 6 mA VC 2 k (10 pts) (b) Find the value of VB at which the transistor enters cut-off. VB (7 pts) (c) Find the value of VB at which the transistor enters saturation. VE (8 pts) 2 mA 2 k (4) The BJT in the emitter-follower amplifier in Figure 4 has β = 100. The coupling capacitors may be taken as short circuits at the signal frequencies. Neglect the output resistance of the transistor. 200 kΩ (b) Calculate the small-signal parameters of the transistor. CC1 15 kΩ CC2 (a) Calculate the bias value of the emitter voltage VE. (5 pts) +10 V Figure 4 vsig 2 kΩ (5 pts) 2 kΩ iout (c) Draw the small-signal equivalent circuit of the i amplifier, and find its current gain Ai out . (15 pts) iin MOSFET: vo iin + _ BJT: Small-signal model: (a) Saturation-region current-voltage equation: 1 W 2 iD kn vGS Vt 2 L B + rπ Triode region: iD kn E gmvπ = βib gm (b) Small-signal model: + gmvgs vgs S 2I D , VGS Vt ro ro Small-signal parameters: VDS VGS Vt . gm vπ W 1 2 vGS Vt vDS vDS L 2 Note: for operation in the saturation region G C gmvπ 1 ID D ro IC VT VT 25 mV r gm ro VA IC SOLUTION EENG 341 FINAL EXAM Fall 2009-2010 1-) Because VDG = 0, both transistors operate in saturation. 1 W 1 W I D kn 1 (VGS 1 Vt )2 kn 2 (VGS 2 Vt )2 2 L1 2 L2 By KVL, VDS1 VDS 2 5 Also, VDS1 VGS 1 , VDS 2 VGS 2 3VGS1 10 Vt 9 V, VGS1 3 V ID L1 W2 . .(VGS 2 Vt ) 2(VGS 2 Vt ) L2 W1 (VGS 1 Vt ) VGS 2 5 VGS 1 Vo VGS 2 2 V 1 10 0.2 (3 1) 2 2 mA 2 2 2-) (a) Small-signal equivalent circuit G Rsig = 100 kΩ D + vsig + _ 2 MΩ Rin vo gmvgs vgs Rout 5 kΩ 20 kΩ S 1 kΩ vo v , vi vgs (1k ) gmvgs 3vgs vgs i vi 3 8 8 vo gmvgs (5k / /20k ) 2 4 vgs 8vgs vi Av V/V 3 3 vo vo vi 2 8 Overall voltage gain: Gv 2.54 V/V vsig vi vsig 2.1 3 Rin 2 M, Rout 5 k (b) Av Voltage gain: (c) (d) 3-) (a) Figure 3 VC 2(6 I C ) 12 2 I C 6 mA VE 2( I E 2) 2 I E 4 Assume that the transistor is active VB 0 V VBE 0.7 V IC VE 0.7 V I E 1.65 mA IC I E I E Let I C I E 0 (in cutoff) active. VC 12 V, VE 4 V VBE 0.5 V VB VE 0.5 3.5 V (c) At the edge of saturation I C I E and VCE (12 2 I C ) (2 I C 4) 16 4 I C VC 4.1 V, VE 3.9 V VC 2 k VB VC 8.7 V VCE 8.7 ( 0.7) 9.4 V (b) 6-IC VCE 0.2 V I C 3.95 mA VB VE 0.7 4.6 V IE IE -2 2 mA VE 2 k 4-) (a) KVL equation for the dc circuit +10 V 200k (1 )(2k ) I B 10 0.7 9.3 V 200 kΩ 9.3 V IB 23.13 A 402k VE (2k ) I E (2k )(1 ) I B 4.67 V IB + VBE _ 2 kΩ (b) I C I B 2.313 mA r (c) gm gm 2.313 mA 92.52 mA/V 25 mV 100 1.08 k 92.52 Small-signal equivalent circuit 15 kΩ ib B C iin vsig + _ 200 kΩ Rib βib rπ E iout 2 kΩ 2 kΩ 1 iout (1 )ib 50.5ib 2 vi r ib (2k / /2k )(1 )ib (102.08 k )ib ib 200 iin 0.6621iin 200 102.8 Ai 33.44 A/A Rib iout 50.5 0.6621iin vi 102.08 k ib