Finding the Maximum Fan-Out of a distribution network

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SUMMARY

The discussion focuses on calculating the maximum fan-out (h) of a distribution network driven by a 5V source with a Thevenin resistance of 1kΩ, where each gate has a load capacitance of 100fF. The objective is to ensure that signals can propagate satisfactorily at a frequency of 333 MHz. The key parameters include a minimum output voltage (V_OH) of 4V and a maximum output voltage (V_OL) of 1V. The fan-out capacity is determined by the time required for the source to charge the parallel gates from V_OL to V_OH and vice versa.

PREREQUISITES
  • Understanding of Thevenin's theorem and equivalent circuits
  • Knowledge of capacitive reactance (Xc = 1/(Cjw))
  • Familiarity with voltage levels in digital logic (V_OH and V_OL)
  • Basic principles of signal propagation and frequency analysis
NEXT STEPS
  • Calculate the time constant for the RC circuit formed by the Thevenin resistance and the total load capacitance.
  • Learn about the impact of frequency on signal integrity in digital circuits.
  • Explore methods to optimize fan-out in digital design.
  • Investigate the effects of load capacitance on signal propagation delay.
USEFUL FOR

Electrical engineers, digital circuit designers, and students studying electronic circuit design who are interested in optimizing fan-out and understanding signal propagation in distribution networks.

KasraMohammad
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Homework Statement


A V_s=5V source, with Thevenin resistance R_s=1kΩ, drives a certain number h(fanout) gates, each of which is modeled as a C_g=100fF load. The gates are driven simultaneously (i.e., in parallel). You are given: V_OL= 1V, V_OH = 4V. Ignore the wire resistance. What is the allowed fanout (h, as a numerical value) such that signals up to 333 MHz can propagate satisfactorily?


Homework Equations



Xc = 1/(Cjw) , V=IR , C(total parallel) = C1 + C2 + C3 +...etc



The Attempt at a Solution



My understanding is that the capacity of the fan-out has to do with the current being driven out of the source. Given V_OH which I believe is the minimum output needed to reach state '1' on the gate, the voltage node at the Capacitors in parallel must be equal or greater than V_OH. This is as far as I got, but I am not sure my line of thinking is correct
 
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KasraMohammad said:

Homework Statement


A V_s=5V source, with Thevenin resistance R_s=1kΩ, drives a certain number h(fanout) gates, each of which is modeled as a C_g=100fF load. The gates are driven simultaneously (i.e., in parallel). You are given: V_OL= 1V, V_OH = 4V. Ignore the wire resistance. What is the allowed fanout (h, as a numerical value) such that signals up to 333 MHz can propagate satisfactorily?


Homework Equations



Xc = 1/(Cjw) , V=IR , C(total parallel) = C1 + C2 + C3 +...etc



The Attempt at a Solution



My understanding is that the capacity of the fan-out has to do with the current being driven out of the source. Given V_OH which I believe is the minimum output needed to reach state '1' on the gate, the voltage node at the Capacitors in parallel must be equal or greater than V_OH. This is as far as I got, but I am not sure my line of thinking is correct

That's quite right.

The idea is the 1K source takes time to charge the parallel-wired input gates from V_ol to V_oh or the reverse.

Assume the 333 MHz is a square wave. Then your concern is the time to get from V_ol to V_oh and from V_oh to V_ol.
 

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