Finding the Maximum Fan-Out of a distribution network

AI Thread Summary
The discussion focuses on determining the maximum fan-out (h) for a distribution network driven by a 5V source with a 1kΩ Thevenin resistance and 100fF load per gate. The key parameters include V_OL at 1V and V_OH at 4V, with a target signal frequency of 333 MHz. The time required to charge the parallel gates from V_OL to V_OH is critical, as it affects the ability to maintain signal integrity at high frequencies. The relationship between the source's current capacity and the number of gates is emphasized as crucial for calculating the allowable fan-out. Understanding these dynamics is essential for ensuring satisfactory signal propagation.
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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