As I stated, for a steady DC output, such as from a linear regulator, the average current is the same as the RMS current (for DC the RMS calculation is trivial).
Do you actually expect 200 MHz current components loading the power supply? Did you forget to install bypass capacitors in your circuit?
LDO pass transistor power dissipation is (Vin-Vout)*avg(Iout). Additional considerations for SOA. Irms isn't present in the calculations.
Here I again have to disagree.if you have the current waveform for one digital block, since it isn't DC, use RMS to find a measure of the current you need,
then multiply by 10 for all 10 digital blocks, and add about 20% margin for variations block to block and over temperature, etc
when measuring the load current I realise that the current goes both ways (+ and -),
You ought
to apply some margining to "DC" (time averaged) load
current, to switching impulses' magnitude and rep rate
to make sure your filter bank is making the intra-cycle
supply current acceptably stable and correct.
OK got it.
Thanks for the explanation, now I have a better understanding on AVG/RMS.
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