barcode fonts for ssrs SMALL-SIGNAL MIDFREQUENCY FET AND TRIODE AMPLIFIERS in Software

Making USS Code 39 in Software SMALL-SIGNAL MIDFREQUENCY FET AND TRIODE AMPLIFIERS

Fig. 7-20

the marked spectra magnitudes with the negative sign accounting for the 1808 phase shift observed from inspection of the instantaneous waveforms. Av vL 0:621 2:48 0:250 vS

In the cascaded MOSFET ampli er of Fig. 7-21, CC ! 1. Av vL =vi and (b) the current-gain ratio Ai iL =ii .

gm2 rds2 RL + LL _

R12 _

Fig. 7-21

(a) The small-signal equivalent circuit is given in Fig. 7-22. Using the result of Example 7.1, but replacing RD with RD1 kRG2 where RG2 R21 kR22 , we have Av1 gm1 rds1 RD1 kRG2 rds1 RD1 kRG2

iL gm2Lgs2 RD2 rds2 + LL _

Fig. 7-22 gm2 rds2 RD2 kRL rds2 RD2 kRL gm1 gm2 rds1 rds2 RD1 kRG2 RD2 kRL rds1 RD1 kRG2 rds2 RD2 kRL

(2) (3)

(b) Realizing that RG1 R11 kR12 , we have Ai where Av is given by (3). iL v =R R o L Av G1 ii vi =RG1 RL

For the JFET-BJT Darlington ampli er of Fig. 7-23(a), nd (a) the voltage-gain ratio Av ve =vi and (b) the output impedance Ro . Assume hre hoe 0 and that RG ) R1 ; R2 .

rds m+1 + m L m + 1 gd _ D (b)

_ Ro (a)

Fig. 7-23 (a) The small-signal equivalent circuit is given in Fig. 7-23(b), where the CD model of the JFET (see Problem 7.5) has been used. Since ib id and vgd vi , KVL yields    rds vi id hie hfe 1 id R1 R2  1  1

CHAP. 7]

By Ohm s law, ve hfe 1 id R1 R2 Solving (1) for id , substituting the result into (2), and rearranging give Av  hfe 1 R1 R2 ve vi rds  1 hie hfe 1 R1 R2 2

(b) We replace R1 R2 with a driving-point source oriented such that vdp ve . With vi deactivated (short circuited), vgd 0. Then, by Ohm s law, ib and by KCL, idp hfe 1 ib Substituting (3) into (4) and rearranging give Ro vdp rds  1 hie idp  1 hfe 1 4 vdp hie rds =  1 3

For a triode with plate characteristics given by Fig. 7-8, nd (a) the perveance  and (b) the ampli cation factor .

(a) The perveance can be evaluated at any point on the vG 0 curve. Choosing the point with coordinates iP 15 mA and vP 100 V, we have, from (4.9),  iP

(b) The ampli cation factor is most easily evaluated along the vP axis. From (4.9), for the point iP 0, vP 100 V; vG 4 V, we obtain  vP 100 25 4 vG

Use the current-source small-signal triode model of Fig. 7-9(a) to derive the voltage-source model of Fig. 7-9(b).

We need to nd the Thevenin equivalent for the circuit to the left of the output terminals in Fig. 7-9(a). If the independent source is deactivated, then vg 0; thus, gm vg 0, and the dependent current source acts as an open circuit. The Thevenin resistance is then RTh rp . The open-circuit voltage appearing at the output terminals is vTh gm vg rp  vg where   gm rp is the ampli cation factor. Proper series arrangement of vTh and RTh gives the circuit of Fig. 7-9(b).

For the ampli er of Example 7.6, (a) use (7.9) to evaluate the plate resistance and (7.10) to nd the transconductance.

a rp %  vP  218 152  10 k iP vG 4 14:7 8:1 10 3

gm %

Find an expression for the voltage gain Av vp =vg of the basic triode ampli er of Fig. 4-12, using an ac equivalent circuit.

The equivalent circuit of Fig. 7-9(b) is applicable if RL is connected from P to K. division in the plate circuit, vp RL vg R L rp so Av vp RL vg RL rp Then, by voltage

In the ampli er of Problem 4.27, let vS 2 cos !t V. (a) Draw the ac load line on Fig. 4-31. (b) Graphically determine the voltage gain. (c) Calculate the voltage gain using small-signal analysis.

(a) If capacitor CK appears as a short circuit to ac signals, then application of KVL around the plate circuit of Fig. 4-30 gives, as the equation of the ac load line, VPP VGQ iP RL vP . Thus, the ac load line has vertical and horizontal intercepts VPP VGQ 300 4 25:5 mA RL 11:6 103 as shown on Fig. 4-31. (b) We have vg vS ; thus, as vg swings 2 V along the ac load line from the Q point in Fig. 4-31, vp swings a total of 2Vpm 213 145 68 V as shown. The voltage gain is then Av 2Vpm 68 17 2Vgm 4 and VPP VGQ 296 V

where the minus sign is included to account for the phase reversal between vp and vg . (c) Applying (7.9) and (7.10) at the Q point of Fig. 4-31 yields  vP  202 168  4:86 k rp iP vG 4 15 8 10 3  i  15:5 6:5 10 3 4:5 mS gm P   vG vP 180 3 5 Then,   gm rp 21:87, and Problem 7.21, yields Av RL 21:87 11:6 103 15:41 R L rp 11:6 4:86 103

The input admittance to a triode modeled by the small-signal equivalent circuit of Fig. 7-9(b) is obviously zero; however, there are interelectrode capacitances that must be considered for highfrequency operation. Add these interelectrode capacitances (grid-cathode capacitance Cgk ; plategrid, Cpg ; and plate-cathode, Cpk ) to the small-signal equivalent circuit of Fig. 7-9(b). Then (a) nd the input admittance Yin , (b) nd the output admittance Yo , and (c) develop a highfrequency model for the triode.

(a) With the interelectrode capacitances in position, the small-signal equivalent circuit is given by Fig. 7-24. The input admittance is Yin IS I I2 1 VS VS VS sCgk VS 1=sCgk VS Vo sCpg VS Vo 1=sCpg 1