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so that the total current is iT 100=12:5 8 A. And, at t 1, this divides equally between the two 5-

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.

5. The locus of points equidistant from two parallel planes is the plane parallel to them and midway between them.

and L 0:2 H, has a sinusoidal voltage v 150 sin 500t 0:785 V applied at t 0. Obtain the current for t > 0.

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

6. The locus of points equidistant from two concentric circles is the circle midway between the given circles and concentric to them.

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The solution is in two parts, the complementary function (ic ) and the particular solution ip , so that i ic ip . The complementary function is the general solution of (27) when the right-hand side is replaced by zero: ic ke 250t . The method of undetermined coe cients for obtaining ip consists in assuming that ip A cos 500t B sin 500t since the right-hand side of (27) can also be expressed as a linear combination of these two functions. Then

6. The locus of points equidistant from two concentric spheres is the sphere midway between the given spheres and concentric to them.

dip 500A sin 500t 500B cos 500t dt Substituting these expressions for ip and dip =dt into (27) and expanding the right-hand side, 500A sin 500t 500B cos 500t 250A cos 500t 250B sin 500t 530:3 cos 500t 530:3 sin 500t Now equating the coe cients of like terms, 500A 250B 530:3 and 500B 250A 530:3

Show that the p transconductance of a JFET varies as the square root of the drain current. p Ans: gm 2 IDSS =Vp0 iD

Locus in a Plane 7. The locus of the vertex of a right triangle having a given hypotenuse is the circle having the hypotenuse as its diameter.

Solving these simultaneous equations, A 0:4243 A, B 1:273 A. ip 0:4243 cos 500t 1:273 sin 500t 1:342 sin 500t 0:322 A and At t 0, i 0. i ic ip ke 250t 1:342 sin 500t 0:322 Applying this condition, k 0:425 A, and, nally, i 0:425e 250t 1:342 sin 500t 0:322 A A

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.)

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