01:25 - Intro and Review of Lecture 23 10:18 - Frequency Response of CB/CG Stages 27:10 - Example: Frequency Response of CG Stage with the P-MOS Current Source Load >> 35:33 - The Output Node - Taking Channel-Length Modulation into Account
01:29 Examples of High-Speed Circuits 04:08 Exploring high-speed circuits and their applications in real-life scenarios. 09:21 Challenges in transmitting high data rates 12:17 Understanding frequency response and resistor usage in common gate stage 18:29 Finding poles by inspection simplifies analysis of high-speed circuits 21:23 Understanding input and output poles in common base stage circuits. 27:46 Replacing resistor with current source for higher gain 31:09 Reduction of capacitances simplifies the circuit design 36:55 Understanding pole frequency in high-speed circuit analysis 39:38 Analysis of high-speed circuits with MOSFET source degeneration 45:37 Analysis of common-base/gate stages and finding pole frequencies. Crafted by Merlin AI.
If you write the small signal model and calculate the input impedance(Vx/Ix) you will find out the it will be 1/gm. The reason is that Vx=-Vpi in the small signal model. So there is no rpi. It will be cancled out.
@@tag1343 You are right about Vx = -Vpi. However, Rpi does not get eliminated due to this reason. For simplicity if we ignore early effect in BJT(no Ro in the small signal model) and write a KCL at the emitter node for the common base topology. We would indeed find the impedance looking into the emitter to be Rpi||(1/gm). The reason it gets eliminated is because Rpi||(1/gm) = Rpi/(gmRpi + 1). gmRpi is the beta of the BJT which is assumed much greater than 1. The impedance would then just become Rpi/gmRpi which is just 1/gm cancelling Rpi.
