Voltage Swing for NMOS: Digital ICs by Jan M. Rabaey

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SUMMARY

The discussion focuses on the voltage swing in NMOS transistors as described in "Digital Integrated Circuits" by Jan M. Rabaey. Key points include the understanding that the voltage at node X charges to Vdd - Vtn1 due to the behavior of source followers M1 and M2, where M1 turns off when X exceeds Vdd - Vth. Additionally, the voltage drop across M2 can be negligible, leading to the conclusion that the swing is accurately represented as Vdd - Vtn1 rather than Vdd - Vtn1 - Vtn2, as the overdrive voltage does not cascade in this configuration.

PREREQUISITES
  • Understanding of NMOS transistor operation
  • Familiarity with voltage thresholds (Vtn and Vth)
  • Knowledge of source follower configurations
  • Basic concepts of overdrive voltage (Vov) and current flow in MOSFETs
NEXT STEPS
  • Study NMOS transistor characteristics and their impact on digital circuits
  • Learn about source follower configurations and their applications in IC design
  • Research the implications of overdrive voltage on MOSFET operation
  • Explore the concept of voltage swing in digital integrated circuits
USEFUL FOR

Electrical engineers, digital IC designers, and students studying semiconductor devices who seek to deepen their understanding of NMOS transistor behavior and voltage swing in digital circuits.

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TL;DR
Voltage swing for nmos is Vdd - Vth? How?
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Ref: Digital Integrated Circuits by Jan M. Rabaey

I have trouble understanding two things:
1) Why will x charge to Vdd - Vtn1 and why not Vdd?
2) If x in left charges to Vdd - Vtn1, then in right also it would charge to Vdd - Vtn1... so Vy would charge to Vdd - Vtn1 - Vtn2... so why have they written the swing as vdd - Vtn1 and why not Vdd - Vtn1 - Vtn2?
 
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1. When A and B are connected to Vdd, M1 is a source follower, if the source X then rises above Vdd–Vth, M1 will turn off and so X will stop rising

2. Both M1 and M2 are source followers with their source voltages set between common and Vdd–Vth. The voltage dropped across M2 can be zero so Vth is not cascaded.
 
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Thanks. I understood the 1 part. But not the second part. How will drop across M2 be zero. Drop across M2 is overdrive voltage, right? I.e Vds.

Vx can be Vdd - Vth. If Vy also becomes Vdd-Vth, then there will be no current flow as Vds = 0. No potential difference so no current flow.
 
Think of an NMOS FET as a variable resistor, with Ron set by Vgs. The voltage drop across M2 = Vsd, will depend on output load current multiplied by Ron of M2. For low currents it will be only a few mV, so close to zero.
 
Ok, but if Vds is less than Vov, then the mosfet would go in linear mode, right? That's not a favorable mode (I don't know why though. Most switch mode should be cut off or saturation)
 
244748
 
The voltage drop increases when the drain is used to drive the gate of another Mosfet as shown in left part of above pic. Thanks for this :)
 

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