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L10Part2&amp;L11Part1 7.3_Small_Signal_BJT_1

L10Part2&amp;L11Part1 7.3_Small_Signal_BJT_1 - Lecture...

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Lecture 10 Part 2 Monday Feb. 1, 2010 Section 7.3 Small-Signal Operation Of BJT Differential Amplifier Zubair Rehman

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2 We set V B1 - V B2 = V d i C1 = α i E1 & i C2 = α i E2 Small Signal Operation (V d / V T ) < 1 V T = 25 mV For X < 1 e X = 1 + X + X 2 /2! + X 3 /3! + … = 1 + X e -X = 1 - X + (-X) 2 /2! + (-X) 3 /3! + … = 1 - X For X < 1 1 / 1+X = 1 - X + X 2 - X 3 + … = 1 - X 1 / 1-X = 1 + X + X 2 + X 3 + … = 1 + X Figure 7.12 The basic BJT differential pair configuration & T B B V V V E e I i 2 1 1 2 - + = T B B V V V E e I i 1 2 1 1 - + = We get & T d V V C e I i / 1 1 - + = α T d V V C e I i / 2 1 + + = α
3 Interpretation For Identical Q1 & Q2 The signal voltage V d appears to divide equally between the two transistors. i.e. V be1 = V d / 2 & V be2 = -V d / 2 because i c1 = g m1 V be1 & i c2 = g m2 V be2 - = + 2[1 Ι = + 1 + 1 Ι = + = - 2[1 Ι = - 1 + 1 Ι = T d t d T d C T d t d T d C V V I V V V V i V V I V V V V i 2 1 * 2 ] 2 / / 2 1 * 2 ] 2 / / 2 1 α α α α α α & & 2 1 1 1 1 I I i I i C c C C α = + = 2 2 2 2 2 1 d T c d T c V V I i V V I i α α - = = 2 2 2 2 2 I I i I i C c C C α = + = T T C m T T C m V V I g V V I g /2 Ι = = /2 Ι = = α α 2 2 1 1 2 2 2 2 1 1 d m c d m c V g i V g i - = = Figure 7.12 The basic BJT differential pair configuration

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Microelectronic Circuits - Fifth Edition Sedra/Smith 4 Figure 7.19 Equivalence of the BJT differential amplifier in (a) to the two common-emitter amplifiers in (b). This equivalence applies only
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