display barcode in ssrs report Applying the quadratic formula to (5) and disregarding the extraneous root yields IDQ1 1:27 mA. in Software

Print ANSI/AIM Code 39 in Software Applying the quadratic formula to (5) and disregarding the extraneous root yields IDQ1 1:27 mA.

Applying the quadratic formula to (5) and disregarding the extraneous root yields IDQ1 1:27 mA.
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CHARACTERISTICS OF FIELD-EFFECT TRANSISTORS AND TRIODES
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Now the use of KVL around the left drain-source loop gives VDD VSS VDSQ1 IDQ1 RD ISQ RS Substituting (1) into (6) and solving the result for RD leads to the desired result: RD VDD VSS VDSQ1 2IDQ1 RS 15 5 7 2 1:27 10 3 3 103 4:20 k IDQ1 1:27 10 3 6
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For the series-connected identical JFETs of Fig. 4-23, IDSS 8 mA and Vp0 4 V. If VDD 15 V, RD 5 k; RS 2 k, and RG 1 M, nd (a) VDSQ1 ; b IDQ1 ; c VGSQ1 , (d) VGSQ2 , and (e) VDSQ2 .
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Fig. 4-23 (a) By KVL, VGSQ1 VGSQ2 VDSQ1 (1) But, since IDQ1  IDQ2 , (4.2) leads to     VGSQ1 2 VGSQ2 2 IDSS 1 IDSS 1 Vp0 Vp0 or, VGSQ1 VGSQ2 (2)
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Substitution of (2) into (1) yields VDSQ1 0. (b) With negligible gate current, KVL applied around the lower gate-source loop requires that VGSQ1 IDQ1 RS . Substituting into (4.2) and rearranging now give a quadratic in IDQ1 :  2    2 Vp0 Vp0 1 2R 2 IDQ1 S IDQ1 0 3 RS RS IDSS Vp0 Substitution of known values gives
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2 IDQ1 4:5 10 3 IDQ1 4 10 6 0
from which we obtain IDQ1 3:28 mA and 1.22 mA. The value IDQ1 3:28 mA would result in VGSQ1 < Vp0 , so that value is extraneous. Hence, IDQ1 1:22 mA. c VGSQ1 IDQ1 RS 1:22 10 3 2 103 2:44 V
CHAP. 4]
CHARACTERISTICS OF FIELD-EFFECT TRANSISTORS AND TRIODES
(d) From (1) with VDSQ1 0, we have VGSQ2 VGSQ1 2:44 V. (e) By KVL, V DSQ2 VDD VDSQ1 IDQ1 RS RD 15 0 1:22 10 3 2 103 5 103 6:46 V
Identical JFETs characterized by iG 0; IDSS 10 mA, and Vp0 4 V are connected as shown in Fig. 4-24. Let RD 1 k; RS 2 k, and VDD 15 V, and nd (a) VGSQ1 ; b IDQ2 , (c) VGSQ2 ; d VDSQ1 , and (e) VDSQ2 .
VDD RD
Q2 RG RS
Fig. 4-24 (a) With negligible gate current, (4.2) gives   VGSQ1 2 IG2 IDQ1 0 IDSS 1 Vp0 so VGSQ1 Vp0 4 V
(b) With negligible gate current, KVL applied around the lower left-hand loop yields VGSQ2 VGSQ1 IDQ2 RS Substituting (1) into (4.2) and rearranging give
2 IDQ2
 2       Vp0 VGSQ1 RS Vp0 VGSQ1 2 1 2 1 0 IDQ2 RS Vp0 Vp0 RS IDSS
which becomes, with known values substituted,
2 IDQ2 8:4 10 3 IDQ2 1:6 10 5 0
The quadratic formula may be used to nd the relevant root IDQ2 2:92 mA. (c) With negligible gate current, KVL leads to VGSQ2 VGSQ1 IDQ2 RS 4 2:92 10 3 2 103 1:84 V (d) By KVL, VDSQ1 VDD IDQ1 IDQ2 RD IDQ2 RS VGSQ2 15 0 2:92 10 3 1 103 2:92 10 3 2 103 1:84 8:08 V
CHARACTERISTICS OF FIELD-EFFECT TRANSISTORS AND TRIODES
[CHAP. 4
By KVL, VDSQ2 VDD IDQ1 IDQ2 RD IDQ2 RS 15 0 2:92 10 3 1 103 2:92 10 3 2 103 6:24 V
Fixed bias can also be utilized for the enhancement-mode MOSFET, as is illustrated by the circuit of Fig. 4-25. The MOSFET is described by the drain characteristic of Fig. 4-9. Let R1 60 k; R2 40 k; RD 3 k; RL 1 k; VDD 15 V, and CC ! 1. (a) Find VGSQ . (b) Graphically determine VDSQ and IDQ .
RD R1 CC +
CC +
R2 _ _
Fig. 4-25
(a) Assume iG 0.
Then, by (4.3), VGSQ VGG R2 40 103 VDD 15 6 V R2 R1 40 103 60 103
(b) The dc load line is constructed on Fig. 4-9 with vDS intercept VDD 15 V and iD intercept VDD =RL 5 mA. The Q-point quantities can be read directly from projections back to the iD and vDS axes; they are VDSQ % 11:3 V and IDQ % 1:4 mA.
For the enhancement-mode MOSFET ampli er of Problem 4.20, let vi sin !t and graphically determine vo .
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