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Download MicroCMOS design by Bang-Sup Song PDF

By Bang-Sup Song

"Starting on the transistor point, this publication covers simple system-level CMOS layout thoughts acceptable to trendy SoCs. The textual content makes use of sensible layout examples to demonstrate circuit layout in order that readers can enhance an intuitive instead of traditional analytic knowing. System-level wisdom is equipped upon figuring out primary innovations of noise, jitter, and frequency and section noise. This material Read more...

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31) Then, the input resistance can be obtained as Ri = vx R2 = R1 + ≈ R1 . 32) Due to this shunt feedback, the opamp input node becomes the virtual ground with the resistance level of R 2/(1 + ao) ~ 0. However, the input resistance of the noninverting ­amplifier is still very high due to the high input resistance of the opamp. 17, the output resistance of the inverting amplifier can be derived. Ignore the current through the feedback network when looking into the output port, ix = vx + ao fvx .

3 CS-CD This combination is the CS amplifier followed by a CD buffer amplifier, which is a source follower. 23). The low output resistance can drive the low-impedance load, while the high gain of the CS amplifier is buffered from the load. 4 CG-CS, CG-CG, CG-CD These three combinations with the CG amplifier as the first stage are the same as the previous three combinations CS-CS, CS-CG, and CS-CD, respectively. 23). 5 CD-CS This combination is not useful in the MOS transistor design. In the BJT design, the emitter follower followed by the common-emitter amplifier is the most useful and is called the Darlington pair.

4 Driving-point small-signal terminal resistances of transistors. || ro ~ 1 ~ gm − gmb 1 gm 33 Amplifier Design Rx = ri || 1 1 ||ro ≈ . 8) In the MOS case, there is no input resistance, but one factor to consider is the body effect because the substrate voltage is not constant. 9) . 9) The parallel combination of 1/(gm − gmb) and ro can be approximated as 1/(gm − gmb), which is about 10% ~ 20% lower than 1/gm. That is, if the source is floating, the effective transconductance of a MOS device increases to be (gm − gmb) due to the body effect.

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