BJT3.pdf

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BJT Amplifier Circuits As we have developed different models for DC signals (simple large-signal model) and AC signals (small-signal model), analysis of BJT circuits follows these steps: DC biasing analysis: Assume all capacitors are open circuit. Analyze the transistor circuit using the simple large signal mode as described in pages 77-78. AC analysis: 1) Kill all DC sources 2) Assume coupling capacitors are short circuit. The effect of these capacitors is to set a lower cut-off frequency for the circuit. This is analyzed in the last step. 3) Inspect the circuit. If you identify the circuit as a prototype circuit, you can directly use the formulas for that circuit. Otherwise go to step 4. 4) Replace the BJT with its small signal model. 5) Solve for voltage and current transfer functions and input and output impedances (node- voltage method is the best). 6) Compute the cut-off frequency of the amplifier circuit. Several standard BJT amplifier configurations are discussed below and are analyzed. For completeness, circuits include standard bias resistors R 1 and R 2 . For bias configurations that do not utilize these resistors ( e.g., current mirror), simply set R B = R 1 k R 2 → ∞ . Common Collector Amplifier (Emitter Follower) R E R 2 V CC v i v o R 1 c C DC analysis: With the capacitors open circuit, this circuit is the same as our good biasing circuit of page 110 with R C = 0. The bias point currents and voltages can be found using procedure of pages 110-112. AC analysis: To start the analysis, we kill all DC sources: R E v o R 1 R 2 v i R E R 2 v i v o R 1 CC V = 0 c C C E c C B ECE65 Lecture Notes (F. Najmabadi), Winter 2006 123
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We can combine R 1 and R 2 into R B (same resistance that we encountered in the biasing analysis) and replace the BJT with its small signal model: v i R B R E i Δ B o v v i i Δ C i Δ B v o R E C c Δ BE v C c r π r π r o r o β Δ B β Δ B B C E R B C + _ B E i i The figure above shows why this is a common collector configuration: collector is shared between input and output AC signals. We can now proceed with the analysis. Node voltage method is usually the best approach to solve these circuits. For example, the above circuit will have only one node equation for node at point E with a voltage v o : v o - v i r π + v o - 0 r o - β Δ i B + v o - 0 R E = 0 Because of the controlled source, we need to write an “auxiliary” equation relating the control current (Δ i B ) to node voltages: Δ i B = v i - v o r π Substituting the expression for Δ i B in our node equation, multiplying both sides by r π , and collecting terms, we get: v i (1 + β ) = v o 1 + β + r π 1 r o + 1 R E = v o " 1 + β + r π r o k R E # Amplifier Gain can now be directly calculated: A v v o v i = 1 1 + r π (1 + β )( r o k R E ) Unless R E is very small (tens of Ω), the fraction in the denominator is quite small compared to 1 and A v 1.
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