模拟集成电路设计英文课件：frequency_response_of_amp

2. For a pole (zero) frequency ωm of at LHP , the phase begins to fall (rise) at approximately 0.1 ωm, experiences at change of -45°(+45) at ωm, and approaches a change of -90 ° (+90°) at 10 ωm. That of RHP, change inversely.
– Miller’s Theorem: If the circuit of Fig. (a) can be converted to that of Fig. (b), the Z1=Z/(1-Av) and Z2=Z/(1-Av-1), where Av=VY/VX。
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概述-密勒效应
LHP: left half-plane at complex frequency plane;
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6.1 General Considerations
• 6.1.1 Miller Effect
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– Importance note of Miller’s theorem (cont.)
• Strictly speaking, the value of Av=VY/VX must be calculated at the frequency of interest. However, in many cases we use the low-frequency value of Av to gain insight into the behaviro of the circuit.
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• If the impedance Z forms only signal path between X and Y, then the conversion is often invalid
• Nevertheless, Miller’s theorem proves useful in cases where the impedance Z appears in parallel with the main signal.
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6.1 General Considerations
• 6.1.0 Bode Plots (波特图)
1. The slope of the magnitude changes by ＋20 dB/dec at every zero
frequency and by -20 dB/dec at every pole frequency.
Design of Analog Integrated Circuits
Frequency Response of Amplifier
概述-密勒效应
Outline
• 6.1 General Considerations • 6.2 Frequency Response of Common-Source
Stage • 6.3 Frequency Response of Common-Drain
Stage (optional) • 6.4 Frequency Response of Common-Gate
Stage • 6.5 Frequency Response of Cascode Stage • 6.6 Frequency Response of Differential Pair
• If applied to obtain the input-output transfer function, Miller’s theorem can not be used simultaneously to calculate the output impedance.
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The charge drawn by CF from Vin
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– Importance note of Miller’s theorem
• Miller’s theorem does not stipulate the conditions under which this conversion is valid. If the conversion is invalid, the results of miller’s theorem are incorrect.
That is,
Similarly,
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– Example 1
Consider the circuit shown in Fig. (a), where the voltage amplifier has a negative gain equal to –A and is otherwise ideal. Calculate the input capacitance of the circuit. Solution: Using Miller’s theorem to convert the circuit to that shown in Fig.(b), we have Z=1/(CFs) and Z1=[1/(CFs)]/(1+A). That is, the input capacitance is equal to CF(1+A)。
– Proof: The current flowing through Z from X to
Y is equal to (VX-VY)/Z. பைடு நூலகம்or the two circuits to be equivalent, the same current must flow through Z1. Thus,