# nvSRAM average Vcc current Level 5
Level 5   Hello,

Regarding the current consumption of nvSRAM, the data sheet has the following description.

tRC = 25 ns, max 70 mA,

tRC = 45 ns, max 52 mA

Please send it if there is a calculation formula of current consumption.

Best Regards,

Naoaki Morimoto

1 Solution

# Re: nvSRAM average Vcc current Employee
Employee   Hi Naoaki,

There is a way to estimate the current for different frequencies from values available in the datasheet.

Usually active current (ICC) has two components - Static (due to circuits biased in active mode) + switching currents which vary linearly with frequency.

Let us say fixed component is x mA and the switching current component is y mA.

According to datasheet-

200-ns or 5 MHz current is 35 mA;    which equates to  x+5y=  35 ----eqn 1;

45- ns or ~22 MHz current is  52 mA; which equates to x+22y=52 --- eqn 2.

Solving eqn 1 and eqn2 gives x= 30 mA; y = 1 mA or 1 mA/MHz.

Therefore, 25 ns or 40 MHz becomes - 30 + 40*1= 70 mA.

If you want to calculate, say for 33-ns or 30 MHZ, it will equate to 60 mA.

I hope this helps. Please let me know if any questions.

Rgds,

Shivendra

4 Replies

# Re: nvSRAM average Vcc current Moderator Moderator   Hello,

Please also mention the part number.

Thanks,

# Re: nvSRAM average Vcc current Level 5
Level 5   The part number is CY14B104K-ZS45XIT.

Regards,

Naoaki Morimoto

# Re: nvSRAM average Vcc current Level 5
Level 5   Do you have any update on this?

Regards,

Naoaki Morimoto

# Re: nvSRAM average Vcc current Employee
Employee   Hi Naoaki,

There is a way to estimate the current for different frequencies from values available in the datasheet.

Usually active current (ICC) has two components - Static (due to circuits biased in active mode) + switching currents which vary linearly with frequency.

Let us say fixed component is x mA and the switching current component is y mA.

According to datasheet-

200-ns or 5 MHz current is 35 mA;    which equates to  x+5y=  35 ----eqn 1;

45- ns or ~22 MHz current is  52 mA; which equates to x+22y=52 --- eqn 2.

Solving eqn 1 and eqn2 gives x= 30 mA; y = 1 mA or 1 mA/MHz.

Therefore, 25 ns or 40 MHz becomes - 30 + 40*1= 70 mA.

If you want to calculate, say for 33-ns or 30 MHZ, it will equate to 60 mA.

I hope this helps. Please let me know if any questions.

Rgds,

Shivendra 