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Fig. 7-24
Substituting (2) and (3) into (1) and rearranging give     V Yin s Cgk 1 o Cpg VS Now, from the result of Problem 7.21, Vo RL VS RL rp so (4) becomes     RL Yin s Cgk 1 C RL rp pg
(b) The output admittance is Yo and I2 Ip Ipk IL Vo Vo Ipk sCpk Vo 7 (8)
Let Yo0 be the output admittance that would exist if the capacitances were negligible; then  so that (c) Yo s Ip Yo0 Vo   R L rp 1 Cpg Cpk Yo0 RL 9 (10)
From (6) and (10) we see that high-frequency triode operation can be modeled by Fig. 7-9(b) with a capacitor Cin Cgk 1 RL = RL rp Cpg connected from the grid to the cathode, and a capacitor Co 1 RL rp =RL Cpg Cpk connected from the plate to the cathode.
Supplementary Problems
7.24 7.25 Find the input impedance as seen by the source vi of Example 4.2 if CC is large. Ans: 940 k
Show that the p transconductance of a JFET varies as the square root of the drain current. p Ans: gm 2 IDSS =Vp0 iD
SMALL-SIGNAL MIDFREQUENCY FET AND TRIODE AMPLIFIERS
[CHAP. 7
In the ampli er of Fig. 4-15, R1 20 k; R2 100 k; R3 1 M; rds 30 k;  150 (see Problem 7.2), and RS 1 k. Find (a) Av vo =vi ; b Ai id =ii , and (c) Zo . Ans: a 0:829; b 843; c 198:7  Find the voltage gain of the CG ampli er of Fig. 7-13(a). Ans: Av vo =vi  1 RD = RD rds  1 RS Find the voltage gain Av2 v2 =vi for the circuit of Fig. 7-25(a). Figure 7-25(b) is a small-signal equivalent circuit in which impedance re ection has been used for simpli cation. Ans: Av2 RD = RD rds  1 RS Let RL1 RL2 ! 1 for the ampli er of Fig. 7-25(a). If RD RS , the circuit is commonly called a phase splitter, since v2 v1 (the outputs are equal in magnitude but 1808 out of phase). Find Av1 v1 =vi and, by comparison with Av2 of Problem 7.28, verify that the circuit actually is a phase splitter. Ans: Av1 RS = RD rds  1 RS For the circuit of Fig. 7-25(a), model the MOSFET by NMOSG of Example 4.4 except use Vto 4 V. Let VGG 2 V, VDD 15 V; RD RS 1:5 k; RL1 RL2 10 k, and CC1 CC2 100 F. Use SPICE analysis to show that v1 v2 , thus substantiating the claim of Problem 7.29 that the circuit is a phase splitter. (Netlist code available from author website.)
4 RD CC2 2 D + RL2 1 +
L2 _
+ VDD
( m + 1)RSL
S 3 _ + _ 0 (a) CC1 6 + RS RL1
L1 _ Li
rds + _ mLgs _ S (b) + RDL
L2 _
Fig. 7-25 For the ampli er circuit of Example 7.4, reduce the value of the bypass capacitor CS to 0:01 F so that RS no longer appears shorted to ac signals and assess the impact on voltage gain. (Netlist code available from author website.) Ans: Av 1:22 1398 The series-connected JFETs of Fig. 4-23 are identical, with  70; rds 30 k; RG 100 k; and RD RL 4 k. Find (a) the voltage-gain ratio Av vL =vi ; b the current-gain ratio Ai iL =ii , and (c) the output impedance Ro . Ans: a Av 9:32; b Ai 233; c Ro 2:16 M The JFET ampli er of Fig. 4-33 has RG 1 M; RS1 800 ; RS2 1:2 k; and RL 1 k. The JFET obeys (4.2) and is characterized by IDSS 10 mA, Vp0 4 V; VGSQ 2 V; and  60. Determine (a) gm by use of (7.3), (b) rds , and (c) the voltage-gain ratio Av vL =vi . Ans: a 2:5 mS; b 24 k; c 0:52
CHAP. 7]
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