Chapter09

Microelectronic Circuit Design

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CHAPTER 9 9.1 Since V REF =− 1.25 V , and v I 1.6 V , Q 1 is off and Q 2 is conducting. v C 1 = 0 V and v C 2 α F I EE R C ≅− I EE R C 2 mA () 350 0.700 V 9.2 I C 2 I C 1 = exp V BE V T ⇒∆ V BE = 0.025ln 0.995 F I EE 0.005 F I EE = 0.132 V (a) v I = V REF +∆ V BE 1.25 + 0.132 1.12 V v I = V REF V BE 1.25 0.132 1.38 V (b) v I = V REF V BE 2.00 + 0.132 1.87 V v I = V REF V BE 2.00 0.132 2.13 V 9.3 Since V REF 2 V v I 1.6 V Q 2 is off and Q 1 is conducting. v C 2 = V and v C 1 F I EE R C I EE R C 2.5 mA 700 1.75 V Note that Q 1 is beginning to enter the saturation region of operation, but V BC = +0.15 V is not really enough to turn on the collector-base diode. (See Problems 9.5 or 5.61.) 9.4 v I = V REF + 0.3 V Q 1 on; Q 2 off. I C 1 = F I EE I EE = 0.3 mA | I C 2 = 0 v C 1 = 0 I C 1 R 1 + R C 0.3 mA 3.33 k Ω+ 2 k 1.60 V v C 2 = 0 I C 1 R 1 0.3 mA 3.33 k 0.999 V 9-1
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9.5 With V BE = 0.7 and V BC = 0.3, the transistor is technically in the saturation region, but calculating the currents using the transport model in Eq. (5.13) yields β F = α F 1 F = 0.98 1 0.98 = 49 | R = R 1 R = 0.2 1 0.2 = 0.25 i C = 10 15 exp 0.7 0.025 ⎟ − exp 0.3 0.025 10 15 0.25 exp 0.3 0.025 ⎟ − 1 = 1.446 mA i E = 10 15 exp 0.7 0.025 ⎟ − exp 0.3 0.025 + 10 15 49 exp 0.7 0.025 ⎟ − 1 = 1.476 mA i B = 10 15 49 exp 0.7 0.025 ⎟ − 1 + 10 15 0.25 exp 0.3 0.025 ⎟ − 1 = 29.52 µ A At 0.3 V, the collector-base junction is not heavily forward-biased compared to the base- emitter junction, and I C = 48.99 I B F I B . The transistor still acts as if it is operating in the forward-active region. 9.6 (a) For Q 2 off, V H = 0 V. For Q 2 on, I C I E and I E = 0.2 0.7 −− 2 () 1.1 x 10 4 = 100 A V L ≅− 4000 I E =− 0.400 V (b) Yes, these voltages are symmetrically positioned above and below V REF , i. e. V REF ± 0.2 V, and the current will be fully switched. See Parts (d) and (e). (c) For v I = 0, I C I E = 0 0.7 2 1.1 x 10 4 = 118 A R = 0.4 V 118 A = 3.39 k (d) Q 2 is cutoff. Q 1 is saturated with V BC = +0.4 V. (e) Q 1 is cutoff. Q 2 is saturated with V BC = +0.2 V. (f) 0.2 V and 0.4 V are not large enough to heavily saturate Q 1 or Q 2 . Although the transistors are technically operating in the saturation region, the transistors still behave as if they are in the forward-active region. (See problem 10.5). 9-2
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9.7 a () For v I = V L , I E = 0.2 0.7 −− 2 1.1 x 10 4 = 100 µ A | V L ≅− 4000 I E =− 0.400 V For v I = V H = 0 V I C I E = 0 0.7 2 1.1 x 10 4 = 118 A R = 0.4 V 118 A = 3.39 k P = 2 V 100 A + 118 A 2 = 218 W b R EE ' = R EE 5 = 11 k 5 = 2.20 k R C 1 ' = R C 1 5 = 4 k 5 = 800 R C 2 ' = R C 2 5 = 3.39 k 5 = 678 9.8 V H = 0 V BE 0.7 V V L 5 mA 200 ( ) 0.7 1.70 V V REF = V H + V L 2 1.2 V V =5 mA 200 = 1.00 V 9.9 V H = 0 V BE 0.7 V V L 1 mA 600 ( ) 0.7 1.30 V V REF = V H + V L 2 1.0 V V =1 mA 600 = 0.600 V 9.10 I EE = 40 .3 mA = 1.2 mA I 3 = I 4 = .1 mA = 0.4 mA R C = 2 k 4 = 500 9.11 a R C = V I EE = 0.8 V 0.3 mA = 2.67 k V H = 0 V BE 0.7 V V L 0.8 V BE 1.5 V V REF = V H + V L 2 1.10 V b NM H = NM L = V 2 V T 1 + ln V V T 1
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This note was uploaded on 03/05/2008 for the course EE 307 taught by Professor Braun during the Spring '08 term at Cal Poly.

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Chapter09 - CHAPTER 9 9.1 Since VREF = -1.25V , and v I =...

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