Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

Which way is better to simulate ADC power: RMS or Integrate?

Status
Not open for further replies.
E

eggzhang

Newbie level 4
Joined
Sep 6, 2008
Messages
6
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,325
adc power rms

I'm designing a ADC, and want to get the power consumption from Cadence.
I'm confused by two sayings:
one is based on Vdd*Irms, as supply voltage is constant, while current is various, so making use of the RMS function in Cadence to get the Irms, then get the power;
another is based on Vdd*Iintegrate/T, make use of the INTEG command in Cadence, get the inregrated value of current in a period, then get the power.

From simulation, the second way gives the less power. But I wonder which way is more correct?
 

Re: Which way is better to simulate ADC power: RMS or Integr

The integration method (which is effectively averaging) is the right one.
The immediate current demands are going to be served by the close-in decaps., so the current drawn from the supply is going to be the mean of the transient current.
 
Hi, saro_k_82,
I can't quite understand the second setence of your reply. What do you mean about the "close-in decaps."?
In my case, it is still in schematic simulation step.
Thanks.
 

Re: Which way is better to simulate ADC power: RMS or Integr

The load current can be decomposed in to the dc component, the fundamental switching frequency component, it's harmonics and noise. The job of the decoupling capacitors is to supply the immediate charge requirements of the device and they charge themselves at a slow rate from the supply.
Effectively the current taken from the supply will be the average of the current consumed by chip.
Only the mean current matters irrespective of whether the decaps are present or not., but the presence of it makes it easy to believe that way.
 
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top