High frequency behavior to study the response of the

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High-Frequency Behavior To study the response of the amplifier at high frequencies, we insert the transistor capacitances, noting that and play no role because the source terminals of and are at ac ground. We thus arrive at the simplified topology shown in Fig. 11.58(c), where the overall transfer function is given by . How do we compute in the presence of the loading of the second stage? The two capacitances and are in parallel, but how about the effect of and ? We apply Miller’s approximation to both components so as to convert them to grounded elements. The Miller effect of was calculated above to be equivalent to k . The Miller multiplication of is given by fF. The first stage can now be drawn as illustrated in Fig. 11.58(d), lending itself to the CS analysis performed in Section 11.4. The zero is given by . The two poles can be calculated from Eqs. (11.70), (11.71), and (11.72): (11.176) (11.177) The second stage contributes a pole at its output node. The Miller effect of at the output is expressed as fF. Adding to this value yields the output pole as (11.178) (11.179) We observe that dominates the high-frequency response. Figure 11.59 plots the overall re- sponse. The midband gain is about 26 dB , around 20% lower than the calculated result. This is primarily due to the use of Miller approximation for . Also, the “useful” bandwidth can be defined from the lower -dB cut-off ( MHz) to the upper -dB cut-off ( MHz) and is almost one decade wide. The gain falls to unity at about 2.3 GHz. If not, then the circuit must be solved using a complete small-signal equivalent.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 588 (1) 588 Chap. 11 Frequency Response 10 6 10 7 10 8 10 9 10 10 30 20 10 0 10 20 30 Frequency (Hz) Magnitude of Frequency Response (dB) Figure 11.59 11.10 Chapter Summary The speed of circuits is limited by various capacitances that the transistors and other com- ponents contribute to each node. The speed can be studied in the time domain (e.g., by applying a step) or in the frequency domain (e.g., by applying a sinusoid). The frequency response of a circuit corresponds to the latter test. As the frequency of operation increases, capacitances exhibit a lower impedance, reducing the gain. The gain thus rolls off at high signal frequencies. To obtain the frequency response, we must derive the transfer function of the circuit. The magnitude of the transfer function indicates how the gain varies with frequency. Bode’s rules approximate the frequency response if the poles and zeros are known. A capacitance tied between the input and output of an inverting amplifier appears at the input with a factor equal to one minus the gain of the amplifier. This is called Miller effect. In many circuits, it is possible to associate a pole with each node, i.e., calculate the pole frequency as the inverse of the product of the capacitance and resistance seen between the node and ac ground.
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