We are investigating the use of a GreenPak device for a low power application and want to estimate the current usage. The application will use digital only macrocells.
Does the ''Chip Quiescent" current in the 'Typical Current Estimated for Each Macrocell' tables cover the use of multiple digital macrocells (eg. DFFs, LUTs, Counters and I2C), regardless of how many are used? Or is that an average that we should multiply by the number of those digital macrocells we expect to use?
Thanks
Device:
Hello BenC,
The chip quiescent current is related to the internal POR circuitry. That is going to be constant, regardless of what else is running. The base die datasheet contains a table that explains the various blocks current consumption vs. VDD (1.8, 3.3, and 5.0V). You can use this table to get a very good estimate of total chip current consumption. It would not include current through VDD for various output configurations (such as an open drain NMOS with a 100 kohm pull up resistor). If low current is your goal, I would highly recommend looking at the SLG46855V base die. It's quiescent current is 80 nA at 3.3V. I should note that the current consumption also does not account for switching voltage levels (so if a MHz signal is piped into the GreenPAK, current will be higher than if a kHz frequency is the input). Let us know if you have any other questions,
Regards,
Chuck
Hi Chuck,
Is there a way to estimate power when using blocks not listed in the table?
For instance, using one of the application note (AN-CM-286) examples; '16-bit FSM with I2C output'. How would we estimate the current draw for that implementation of the example counter, as well as the change in current draw if a second counter was subsequently added to the design?
Cheers,
Ben
Hi Ben,
The short answer is that we would not have an easy way to "estimate". We would need to power that design up on a bench and test it. Given that the design file is readily available, as long as you have a development kit with the appropriate GreenPAK base die, you can perform the current consumption test pretty quickly. I wish I had a better answer for you.
Regards,
Chuck