barcode fonts for ssrs (a) Since IEQ we have, from (3.6), VBB ICQ RB 1 ICQ RE VBEQ ICQ 1 ICQ in Software

Generator Code 39 Full ASCII in Software (a) Since IEQ we have, from (3.6), VBB ICQ RB 1 ICQ RE VBEQ ICQ 1 ICQ

(a) Since IEQ we have, from (3.6), VBB ICQ RB 1 ICQ RE VBEQ ICQ 1 ICQ
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TRANSISTOR BIAS CONSIDERATIONS
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[CHAP. 5
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Rearranging gives ICQ VBB VBEQ VBB VBEQ RB 1 RB 1 RE RE @ICQ RB RE VBB VBEQ @ RB 1 RE 2 1
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and, from (5.10), S (b) Note in (2) that lim S 0.
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Now if ! 1 in (1), then ICQ % VBB VBEQ =RE constant.
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Temperature variations can shift the quiescent point by a ecting leakage current and base-toemitter voltage. In the circuit of Fig. 5-1, VBB 6 V; RB 50 k; RE 1 k; RC 3 k, 75; VCC 15 V, and the transistor is a Si device. Initially, ICBO 0:5 A and VBEQ 0:7 V, but the temperature of the device increases by 208C. (a) Find the exact change in ICQ . (b) Predict the new value of ICQ using stability-factor analysis.
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(a) Let the subscript 1 denote quantities at the original temperature T1 , and 2 denote quantities at T1 208C T2 . By (5.7), ICQ1 VBB VBEQ1 ICBO1 RB RE 6 0:7 0:5 10 6 51 103 3:1953 mA RB = RE 50 103 =75 1 103 ICBO2 ICBO1 2 T=10 0:5 10 6 220=10 2 A so Again by (5.7), ICQ2 Thus, (b) By (5.16) and (5.17), RB RE 50 1 30:6 RB = RE 50=75 1 75 0:6 10 3 SV RB RE 50 103 75 1 103 SI Then, according to (5.13), ICQ % SI ICBO SV VBEQ 30:6 1:5 10 6 0:6 10 3 0:04 0:0699 mA and ICQ2 ICQ1 ICQ 3:1953 0:0699 3:2652 mA VBB VBEQ2 ICBO2 RB RE 6 0:66 2 10 6 51 103 3:2652 mA RB = RE 50 103 =75 1 103 ICQ ICQ2 ICQ1 3:2652 3:1953 0:0699 mA VBEQ 2 10 3 T 2 10 3 20 0:04 V VBEQ2 VBEQ1 VBEQ 0:7 0:04 0:66 V
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Now, according to Section 5.2,
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In Problem 5.11, assume that the given values of ICBO and VBEQ are valid at 258C (that is, that T1 258C . (a) Use stability-factor analysis to nd an expression for the change in collector current resulting from a change to any temperature T2 . (b) Use that expression to nd ICQ when T2 1258C. (c) What percentage of the change in ICQ is attributable to a change in leakage current
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(a) Recalling that leakage current ICBO doubles for each 108C rise in temperature, we have ICBO ICBO jT2 ICBO jT1 ICBO j258C 2 T2 25 =10 1
CHAP. 5]
TRANSISTOR BIAS CONSIDERATIONS
Since VBEQ for a Si device decreases by 2 mV/8C, we have VBEQ 0:002 T2 25 Now, substituting SI and SV as determined in Problem 5.11 into (5.13), we obtain ICQ SI ICBO SV VBEQ RB RE I j 2 T2 25 =10 1 0:002 T2 25 RB RE CBO 258C RB RE (b) At T2 1258C, with the values of Problem 5.11, this expression for ICQ gives ICQ 30:6 0:5 10 6 2 125 25 =10 1 0:0006 0:002 125 25 15:65 mA 0:12 mA 15:77 mA (c) From part b, the percentage of ICQ due to ICBO is 15:65=15:77 100 99:24 percent.
In the constant-base-current-bias circuit arrangement of Fig. 5-7, the leakage current is explicitly modeled as a current source ICBO . (a) Find ICQ as a function of ICBO , VBEQ , and . (b) Determine the stability factors that should be used in (5.13) to express the in uence of ICBO , VBEQ , and on ICQ .
(a) By KVL, VCC IBQ Rb IEQ RE Substitution of (5.5) and (5.6) into (1) and rearrangement give ICQ % VCC VBEQ ICBO Rb RE Rb = RE 2 1
(b) Based on the symmetry between (2) and (5.7) we have, from Example 5.6, SI
+ VCC
Rb RE Rb = RE
SV
Rb RE
S
Rb VCC VBEQ ICBO Rb RE Rb RE 2
+ VCC RC RF
Rb ICBO IBQ RE
ICQ ICBO
IBQ RE
Fig. 5-7
Fig. 5-8
In the shunt-feedback bias arrangement of Fig. 5-8, the leakage current is explicitly shown as a current source ICBO . (a) Find ICQ as a function of ICBO ; VBEQ ; and . (b) Determine the stability factors that should be used in (5.13) to express the in uence of ICBO ; VBEQ ; and on ICQ .
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