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.
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The sounds that the pen makes is so satisfying 🙈😝
ASMR by Razavi
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.
Waiting eagerly for circuit theory I & II and the rest of the series.
@41:50 i think it should be ro1 not ro2 at the mosfet M1 . Brilliant video !
how do we find ro1?
@@fettahyldz460 1/(lambda*Id)
What about decomposing ro1 into input and output side according to Miller's Effect? Will the results for ω1, in and ω1, out be any different?
Thank You
Can't we break ro2 using miller and then have a more accurate representation?
Miller itself is an approximation!
nice
Why didnt you add rpi in parallel with RE and 1/gm?
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.
rpi is 1/gm
@@avirathi9209 wtf bro? rpi is are no 1/gm
@@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.
Do you guys know how to find ro1?
1/(chanel length modulation* drain current)
@@bappadas9547 I already graduated