(b) Figure 3.1 shows the four input Or-gate is connected to...

(b) Figure 3.1 shows the four input Or-gate is connected to the D-Flip flops. Each of the two input OR gate has a propagation delay of 100ps and the contamination delay of 50ps. Each flip flop has a setup time of 60ps, a hold time of 20ps, a clock –to Q maximum delay of 70ps, and a clock to Q minimum delay of 50ps. Figure 3.1 Registered four-input Or-Gate Circuit I. Calculate the maximum frequency of the circuit, neglecting the Clock Skew. II. How much the clock skew can the circuit tolerate if it must operate at 2GHz? III. How much clock skew can the circuit tolerate before it might experience a hold time violation?

Transcript

00:01 Hello students, here we have four input or gate which is connected to deep fliprocks.

00:06 And we have the datas are here.

00:09 Propagation delay, contamination delay, set up time and whole time.

00:13 And the queue maximum delay and minimum delay are given.

00:17 So we have to find out the maximum frequency of circuit.

00:22 And also we have to find how much clocks queue the circuit can tolerate if it must operate at two gigahertz.

00:30 And also how much the clocks queue the circuit can tolerate before it might experience a whole time violation.

00:36 Okay? we have to find answers to these questions.

00:41 So we have the q maximum delay, it is ordered by t, p .cq, which is equal to 70.

00:52 Okay.

00:53 And we have minimum, which is cq, it will be 50.

01:03 Okay, and also we have for the organ, the propagation delay, which is 100, and the contamination delay, it will be that's equal to 50.

01:19 Okay, now, first we have to find out the time period of the circuit, okay? time period of circuit so we have so this is the equation for the time period now we can substitute the value which is greater than in equal to we have tpcq which is 70 plus we have the thought maximum propagation delay that will be 3 multiplied by a tpd plus t set up region 60.

02:10 So it is great to then equal to 70 plus 300 plus 60, which is greater than or equal to 430.

02:22 Now we have the equation for frequency which is 1 divided by t.

02:26 So frequency less than or equal to 1 divided by 430, alta -platabyte and means to minus 12...

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