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Unformatted text preview: Prelab 7: Frequency Response Name: Lab Section: v 1 ( t ) A C i ( t ) v 2 ( t ) Figure 1: Amplifier with a “Miller” capacitor The Miller effect plays an important role in determining the poles of an amplifier. If there is gain, A , across the capacitor, C , as shown in Figure 1, then the current across C can be written as: i ( t ) = C d dt ( v 1 ( t )- v 2 ( t )) = C d dt ( v 1 ( t )- Av 1 ( t )) By distributing the derivative, this simplifies to: i ( t ) = C (1- A ) d dt v 1 ( t ) Therefore, the equivalent capacitance looking into v 1 ( t ) is the capacitance C multiplied by (1- A ). If the gain, A , is large enough, this Miller effect can make the capacitor dominate over other capacitances and thus, contribute to the dominant pole of the amplifier. Similarly, the equivalent capacitance looking into v 2 ( t ) can be derived as C ( 1- 1 A ) .- + V BIAS- v in + R S V CC = 5 V R C . 25 mA v OUT Figure 2: Common emitter amplifier. Note that this amplifier has negative gain....
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This note was uploaded on 04/06/2009 for the course ECSE 2050 taught by Professor Monahella during the Spring '08 term at Rensselaer Polytechnic Institute.
- Spring '08