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Here has 2 memory device A and B:
The active current of A is 10 mA, B is 20 mA;
and leakage current of A is 20 uA, B is 10 uA.
For a hand on SOC design, like cellphone processor chip, which memory device is better choice? A or B?
Please help me.
I have searched the sense amplifier related information of SOC embedded memory.
Could anyone told me the power of sense amplifier is what percentage of SOC embedded memory read power?
Thanks a lot.
I find out the data following:
180nm process : the capacitance of M1 : 0.2fF/um.
32nm process : the capacitance of M1 : 0.2fF/um.
But the metal 1 line area of 32nm process should smaller much than 180nm process.
Why their capacitance is almost the same...
On metal line has 700um long, 0.25um width,
for 0.2fF/1um, the capacitance of the metal line is 0.2*700=140fF.
If the metal line need run up to 1GHz.
The peak current is V/R, R=t/5C=1ns/5*140fF=1.4k.
But the metal width is 0.25um=> only can run 250uA.
current carrying capacity in different layer at 45 nm process
I estimate one plan now.
Ｉ need current carrying capacity in M1 M2 M3 M4 with 140nm width metal line at 45 nm process.
Could any one help me ?
I have experience with block simulation in digital design.
I wish to learn large system verification including hardware/software co-sim.
Could someone give the completed example or book of it ?
Thanks a lot.
I need the 0.18um SRAM's active power(mA) and area(mm^2) as following :
(1) 4096X32 bits run at 50Mhz and multiplexer sets to 16.
(2) 32768X32 bits run at 50Mhz and multiplexer sets to 16.
Please help me to get the related information. Any Feb's data is good.
As we know, due to large RC time constant of SRAM bit line, so it need sense amplifier for read mechanics.
Because the read bit line and write bit line is the same line of SRAM, so their RC time constant is the same.
How could write timing can fast as read timing?
The standand cell information is following:
.subckt BUFX1 Y A
X_g0 Y net7 inv pl=0.18u pw=1.92u nl=0.18u nw=1.3u
X_g1 net7 A inv pl=0.18u pw=0.71u nl=0.18u nw=0.47u
The cell size is 0.18*(1.92+1.3+0.71+0.47)=0.18um*4.4um
But the data sheet shows the standand cell size...