Finding the Maximum Fan-Out of a distribution network

In summary: One half cycle is 1/666 MHz which is 1.5 ns.In this time, the gate can charge to 0.63 of the final voltage: V(t)= V_final(1-e^(-t/RC)).So if you want 0.63*V_oh to be greater than V_ol, you need a fanout of 0.63 V_oh/V_ol.In summary, the allowed fanout for signals up to 333 MHz is 0.63 V_oh/V_ol.
  • #1
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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  • #2
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.
 

Related to Finding the Maximum Fan-Out of a distribution network

1. What is the maximum fan-out of a distribution network?

The maximum fan-out of a distribution network refers to the maximum number of downstream nodes that can be connected to a single upstream node. It is also known as the degree of a node, and it is an important measure for understanding the capacity and efficiency of a network.

2. How is the maximum fan-out of a distribution network calculated?

The maximum fan-out of a distribution network is calculated by counting the number of downstream nodes connected to a single upstream node. This can also be expressed as the ratio of the number of downstream nodes to the number of upstream nodes.

3. What factors affect the maximum fan-out of a distribution network?

The maximum fan-out of a distribution network is affected by various factors such as the capacity of the upstream node, the bandwidth of the network, the distance between nodes, and the type of network topology used. These factors can impact the overall performance and efficiency of the network.

4. Why is it important to find the maximum fan-out of a distribution network?

Finding the maximum fan-out of a distribution network is important for optimizing the network's performance and efficiency. It helps in understanding the network's capacity and identifying potential bottlenecks. By knowing the maximum fan-out, network engineers can make informed decisions about network design and upgrades.

5. How can the maximum fan-out of a distribution network be increased?

The maximum fan-out of a distribution network can be increased by adding more upstream nodes, increasing the capacity of existing nodes, or implementing a more efficient network topology. Network engineers can also use load balancing techniques to distribute the traffic evenly and increase the overall capacity of the network.

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