# Average vs RMS load-current for LDO design

1. ## Re: Average vs RMS load-current for LDO design

Thanks for the answer. I was inclined for RMS because it's the bigger number between the 2, but it's actually much bigger, 3x, so if I am designing based on this my circuit is that much bigger which give me area concern. So I really don't want to be over designing much more than necessary, hence my question here.

2. ## Re: Average vs RMS load-current for LDO design

Originally Posted by crutschow
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).
But the current waveform that I'm measuring is from the power supply of digital blocks and are noise-like, for this waveform the RMS and average current value between the same time-range is different. Which should I choose as DC equivalent current for my LDO?

Basically this image, shows what I'm trying to achieve. The X-axis is the DC load current (swept) and Y-axis is LDO output. As can be seen, 5mA is the maximum the LDO can support in this example. My question is how should I approximate what my "5mA" is based on a certain current waveform obtained from digital block switching.

3. ## Re: Average vs RMS load-current for LDO design

Originally Posted by FvM

Do you actually expect 200 MHz current components loading the power supply? Did you forget to install bypass capacitors in your circuit?
Hmm.... I'm actually not sure what you mean here. The 200MHz current switching seems to load the LDO greatly since I see the output drops immediately when it starts switching.

4. ## Re: Average vs RMS load-current for LDO design

Originally Posted by FvM
LDO pass transistor power dissipation is (Vin-Vout)*avg(Iout). Additional considerations for SOA. Irms isn't present in the calculations.
So it looks like from this equation it is IAVG instead of IRMS, am I interpreting it correctly?

Just to clarify my initial question, please take a look at this image.
The X-axis is DC load current being swept, the Y-axis is my LDO output. In this example the maximum current the LDO can support is 5mA.
My question is, given a certain digital blocks which current I can measure (ie. put a probe between LDO output and the digital block and plot the current), what is my equivalent "5mA" that I need to design based on this current waveform? I can measure RMS or AVG and the 2 have very different numbers, about 3x different at worst-case.

Of course the safe thing to do here is just to pick RMS since it's the bigger number, but it requires a pass transistor that is sooo much bigger and it's causing area concerns in my design =(
So I really don't want to over design much more than necessary.

I tried talking to several of my senior engineer colleagues here but it seems opinions are split, so I can't make a decision, especially without understanding the "why" it's RMS/AVG.

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5. ## Re: Average vs RMS load-current for LDO design

in post 4, you said there are 10 identical digital blocks your LDO regulator is powering

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

however, if it is possible that all 10 blocks draw the maximum current at the same time, you need to design for that.

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6. ## Re: Average vs RMS load-current for LDO design

Hi,

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
Here I again have to disagree.
If one takes the signal as noise...which means a band of uncorrelated frequencies...then the amplitudes need to be added in "square and squareroot" manner. This is what RMS means.
If one has 10 uncorrelated but from the values identical RMS values, then the resuling RMS value is sqrt(10) of the single RMS value, and not 10x the single RMS value.

But you may directly multiply average values to get "10x the single value".

Klaus

7. ## Re: Average vs RMS load-current for LDO design

I actually have all 10 blocks simulated with the LDO so I don't need to extrapolate from 1 to 10, my confusion is rather on whether I should take RMS or AVG of this waveform.

The waveform looks something like this.

8. ## Re: Average vs RMS load-current for LDO design

Hi,

Is it accurate to treat the digital switching signal similarly to noise? Because the switching occurs periodically on a certain switching frequency unlike noise. It does looks like noise though. I am now more confused than ever about the notion of RMS.

The current waveform looks something like this.

Would you recommend RMS or AVG to get DC representation of such current waveform?

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9. ## Re: Average vs RMS load-current for LDO design

It's unlikely your LDO will respond to the digital switching
transients (orders of magnitude slower control loop. than
technology edge-rates). You will depend on some filter
bank to "soak up" the impulse energy.

Since the current is likely unidirectional (as seen by the
LDO pass transistor) "average" is as good as "RMS" as
far as I'm concerned. But what "average"? 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.

Beware the poor series resistance of MOS caps in this
application, optimize that by layout style and maybe
supplement with some MOM parallel plates for the high
frequency components.

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10. ## Re: Average vs RMS load-current for LDO design

when measuring the load current I realise that the current goes both ways (+ and -),
But this needs more details.

11. ## Re: Average vs RMS load-current for LDO design

Hi,

It still is unclear to me what path you really want to measure, but I don't want to ask a third time for a sketch...

If it's about power supply requirement, then in most cases AVG is the way to go. But I recommend to use bypass capacitors.
As already stated - they will already average the current, in a way that it is almost constant.
They won't modify (much) the overall consumption, thus it won't influence power requirement, and the average current should be (almost) the same. But RMS will be greatly reduced.

Klaus

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12. ## Re: Average vs RMS load-current for LDO design

A significant point missed is that RMS current only works as an indication of power with a resistive load.
For a constant voltage load, such as a switching power supply, the average current is the indicator of power.
Using RMS current for that will give an erroneously high power indication.

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13. ## Re: Average vs RMS load-current for LDO design

In my opinion RMS is a fancy average calculation. The integral symbol and the division with a time period means the average calculation in the equation. The squared signal only relates to convert signal to a unit which is quasi equvivalent with power. It better describes a waveform which has a big DC and more smaller AC components than usual average calculation in term of power dissipation.

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14. ## Re: Average vs RMS load-current for LDO design

Sorry for not providing a sketch yet, mainly I was unsure on what sketch you required. i did try to draw the vout vs load current to illustrate.

In any case, I think I am now clear that AVG is the way to go. I also found some simulation method to confirm this and it seems like it is indeed AVG instead of RMS.

Thank you for all the advice and explanation.

15. ## Re: Average vs RMS load-current for LDO design

OK got it.

Thanks for the explanation, now I have a better understanding on AVG/RMS.

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16. ## Re: Average vs RMS load-current for LDO design

Thanks for the input.
Could you elaborate more on this part:
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.
I don't quite understand what you mean here.

17. ## Re: Average vs RMS load-current for LDO design

Originally Posted by komax
OK got it.

Thanks for the explanation, now I have a better understanding on AVG/RMS.
A quick summary:

Avg value of noise = const DC offset + avg (noise) --> zero (second term) (Avg of noise is always zero except for the offset; or you can consider the offset as the average if you wish)

(noise refers to noise voltage)

RMS value of noise = transferred effectively + dissipated locally (this addition is valid for the total energy but need to convert to voltage)

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