POWER PLANNING ASSIGNMENT

• How power values are defined in .lib files

internal_power() {related_pin : "B";rise_power(energy_template_7x7) {index_1 ("0.008, 0.04, 0.08, 0.12, 0.16, 0.224, 0.28");index_2 ("0.01, 0.06, 0.1, 0.15, 0.2, 0.25, 0.3");values ( \"0.002557, 0.002618, 0.002627, 0.002631, 0.002634, 0.002636, 0.002636", \"0.00255, 0.002611, 0.00262, 0.002625, 0.002627, 0.002629, 0.002629", \"0.00254, 0.002602, 0.002611, 0.002615, 0.002618, 0.00262, 0.00262", \"0.002531, 0.002594, 0.002602, 0.002607, 0.002609, 0.002611, 0.002612", \"0.002524, 0.002586, 0.002595, 0.0026, 0.002603, 0.002604, 0.002605", \"0.002514, 0.002576, 0.002586, 0.002591, 0.002594, 0.002596, 0.002596", \"0.002507, 0.00257, 0.002581, 0.002586, 0.002588, 0.00259, 0.002591");}

◦ What are ccs models.

▪ Accurate delay calculation is critical for timing closure of complex digitaldesigns.

▪ At 90nm and below, physical effects present new challenges for delaycalculation.

▪ Top-level interconnect is becoming more resistive with narrower metalwidths, resulting in cases where the interconnect impedance is much greater than the drive resistance of the driving cell.

▪ In addition, second order physical effects such as the Miller effect are now becoming first order and must be accounted for in the timing analysis.

▪ The concern over power in today’s smaller technologies is also presentingnew requirements for operation of all or part of the design at lower voltages, increasing the need for analysis over a range of voltages without aunique characterization for each operating point.

• At 90nm, max allowed slot width = 12un , process requirements is 36un to

address EM/IR. What would be the right approach to address this?

◦ You need 3 straps of 12 un width to approach the process requirements of 36un

DMOHANTY

 

POWER PLANNING ASSIGNMENT

• For a 10K gate design with 500 MHz frequency and typical voltage of 1.2V,

die size of 5mmx5mm, tr of 0.3ns, R of 30Ω, having probability of transitions

as 50% and leakage power of 20mW. Calculate Avg. power & Total power.

◦ E(sw)=50% probability of transition in a given time period◦ therefore E(sw)=0.5◦ t=rc;◦ 3ns=30*c◦ therefore c=0.3ns/30◦ c=10pF◦ Total Power is given as:▪ cv2fE(sw) + v.imax(tr+tf/2)f+leakage power ▪ 10pf*(1.2)2*500MHz*0.5+1.2*40mA*0.3ns*500MHz+20mW▪ Total Power = 31mW

▪ Integrating the instantaneous power over the period of interest, the energyEVDD taken from the supply during the transition is given by

▪ EVDD= 0->∞∫I. VDD(t).VDD.dt▪ =VDD. 0->∞∫ CL.(dvout/dt).dt▪ = CL.VDD. 0->VDD∫.dvout▪ = CL.VDD2

◦ Average power = 10pf*(1.2)2=14.5pW

• What is IR Drop ?

IR Drop is the problem of voltage drop of the power and ground due to highcurrent flowing through the power-ground resistive network.

DMOHANTY