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Here is the design procedure for the midterm 2: 1) calculate td(max) -- either “low->high”; “high->low” First calculate the worst-case delay of the FA, from either low->high, or high->low. This now is annotated as the FA delay. 2) t(cycle) = 2*td(max) Set the clock cycle time to be equal to 2*td(max). Hence, simulate the FA assuming that the clock frequency is 1/t(cycle). 3) measure Iavg*Vdd*2td(max) (static leakage) Don’t switch anything with the FA – keep all inputs stable. Then, measure the Iavg current in Vdd. Hence, the energy dissipated during a single cycle of 1/t(cycle) => Ecycle (static) = Iavg * Vdd * t(cycle). This is your static leakage energy consumed in a single t(cycle) period. 4) measure Iint*Vdd*2*td(max) (BOTH dynamic + static energy together) Now, set your clock period of your FA to t(cycle) = 2*td(max). Use Iint, integrating ALL the current from the beginning of the clock to the end of t(cycle). Ecycle (total) = Iint(for entire t(cycle)) * Vdd. This is your TOTAL energy/computation, including both your static leakage energy and your dynamic energy together. 5) rise-time issue: use different inverter to drive D-FF clock One problem with the simulations is that the rise/fall time of the input and the clocks can greatly affect the

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Page 1: eecs.oregonstate.edueecs.oregonstate.edu/research/vlsi/teaching/ECE471_WIN13/... · Web viewHere is the design procedure for the midterm 2: calculate td(max) -- either “low->high”;

Here is the design procedure for the midterm 2:

1) calculate td(max) -- either “low->high”; “high->low”

First calculate the worst-case delay of the FA, from either low->high, or high->low. This now is annotated as the FA delay.

2) t(cycle) = 2*td(max)

Set the clock cycle time to be equal to 2*td(max). Hence, simulate the FA assuming that the clock frequency is 1/t(cycle).

3) measure Iavg*Vdd*2td(max) (static leakage)

Don’t switch anything with the FA – keep all inputs stable. Then, measure the Iavg current in Vdd. Hence, the energy dissipated during a single cycle of 1/t(cycle) => Ecycle (static) = Iavg * Vdd * t(cycle). This is your static leakage energy consumed in a single t(cycle) period.

4) measure Iint*Vdd*2*td(max) (BOTH dynamic + static energy together)

Now, set your clock period of your FA to t(cycle) = 2*td(max). Use Iint, integrating ALL the current from the beginning of the clock to the end of t(cycle). Ecycle (total) = Iint(for entire t(cycle)) * Vdd. This is your TOTAL energy/computation, including both your static leakage energy and your dynamic energy together.

5) rise-time issue: use different inverter to drive D-FF clock

One problem with the simulations is that the rise/fall time of the input and the clocks can greatly affect the dynamic current through Vdd. i.e. fast rise times result in large capacitive switching. I suggest using another inverter as a ‘buffer’ to isolate your fast clock edge from the real input(s) to your D-FF. Make sure to put this separate inverter not on the same VDD as the VDD you are measuring, however. For example, a previous student Maggie Watkins suggested making the rise/fall time of the clock/data equal to 1/10 of the clock period.

6) Vt process variability issue

The second part suggests to change ‘vt’ of the FA, to emulate process variability. The ‘real’ way people do this in industry is to run Monte Carlo, which randomizes across a Gaussian distribution the possible effects that might occur, and run 100’s/1000’s of iterations in the hopes of finding how sensitive the design is to failure across a 3-sigma event.

Page 2: eecs.oregonstate.edueecs.oregonstate.edu/research/vlsi/teaching/ECE471_WIN13/... · Web viewHere is the design procedure for the midterm 2: calculate td(max) -- either “low->high”;

For this simple assignment, the easiest way to do this is to change the ‘vt’ of the devices. This will be a ‘worst-case’ simulation, but simplifies things because you do not need to run Monte Carlo.

Put two voltage sources in series with the transistors in the FA, both for the NMOS and the PMOS. Make this 1-sigma for each, and ‘effectively’ makes each PMOS/NMOS seem much slower (for example, a 50mV higher Vt for each transistor). Note that 1-sigma for 25um technology will be substantially different than 1-sigma for 65nm-CMOS. Assume AVT=4mV/um.

You can ignore the gate length variation for this portion.