Finding the Maximum Fan-Out of a distribution network

[KasraMohammad]

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

 

[rude man]

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.