Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic 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

[SOLVED] Average vs RMS load-current for LDO design

komax

Junior Member level 3
Joined
Dec 2, 2019
Messages
31
Helped
0
Reputation
0
Reaction score
0
Trophy points
6
Activity points
312
I am designing an LDO that will be used to power a small digital circuit. I have access to said digital circuit and have been simulating my LDO with it.

My question: when designing my pass device to support maximum load current, which numbers should I use, average or RMS current? And why?

Follow up question: when measuring the load current I realise that the current goes both ways (+ and -), so if I were to measure RMS the negative current will be squared and counted as well while in reality this negative current simply represents kick back current towards the LDO due to switching (I think), wouldn't counting these current be over-design? Should I make some adjustment to the waveform to get rid of negative parts before taking RMS then?

Thank you in advance for your kind help.
 

c_mitra

Advanced Member level 5
Joined
Nov 13, 2012
Messages
3,395
Helped
854
Reputation
1,708
Reaction score
807
Trophy points
1,393
Activity points
26,012
when measuring the load current I realise that the current goes both ways (+ and -),..
It is surprising; a small digital circuit as a load to a LDO regulator is not expected to act like this! Can you please provide more details?
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,460
Helped
3,943
Reputation
7,884
Reaction score
3,813
Trophy points
113
Activity points
115,755
Hi,

My question: when designing my pass device to support maximum load current, which numbers should I use, average or RMS current? And why?
* For devices where the voltage (across it´s pins) is directly proportional (like a resistor): use RMS
* for devices with about constant voltage (across it´s pins), independent of current (like diodes, LEDs): use AVG.

...in most cases. It depends what you want to calculate

Klaus
 

komax

Junior Member level 3
Joined
Dec 2, 2019
Messages
31
Helped
0
Reputation
0
Reaction score
0
Trophy points
6
Activity points
312
It is surprising; a small digital circuit as a load to a LDO regulator is not expected to act like this! Can you please provide more details?
It's actually 10x of the same digital circuit, so in total they are not so small I guess. The range of RMS current I'm getting is 2.5mA typical and 5mA worst case.
The circuit itself is a driver circuit that is switching at 200MHz, I didn't dig deeper on what else are inside.
 

komax

Junior Member level 3
Joined
Dec 2, 2019
Messages
31
Helped
0
Reputation
0
Reaction score
0
Trophy points
6
Activity points
312
Thanks for the reply.
How about if I want to calculate the current consumption of a switching digital blocks, do I use RMS or AVG?

Basically I want to make sure that I design the size of my pass device sufficiently to support the digital circuit while it's switching at its highest frequency and heaviest activities. I have the waveform of this current (kind look like noise with peaks here and there, e.g. 5mA peak, 10mA, etc), I'm wondering if I should calculate RMS or AVG for designing the LDO's pass device.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,460
Helped
3,943
Reputation
7,884
Reaction score
3,813
Trophy points
113
Activity points
115,755
Hi,

How about if I want to calculate the current consumption of a switching digital blocks, do I use RMS or AVG?
This s a vague question, thus I can´t answer it.

A digital block has several units inside.. Show us a setch of your "building block" and show what voltage and what current you want to use for power calculations.

If you have the sketch you should easily find out by yourself whether V and I are in a proportional relationship or not.

Klaus
 

c_mitra

Advanced Member level 5
Joined
Nov 13, 2012
Messages
3,395
Helped
854
Reputation
1,708
Reaction score
807
Trophy points
1,393
Activity points
26,012
For a DC current, the average and the RMS are the same.
Not correct. This is true only for a constant current.

I guess I need not elaborate.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,460
Helped
3,943
Reputation
7,884
Reaction score
3,813
Trophy points
113
Activity points
115,755
Hi,

Not correct. This is true only for a constant current.

I guess I need not elaborate.
When I think about DC current, then it´s a straight horizontal line. Thus in my eyes every DC current is constant.

****
But maybe you see "DC" as every signal that just does not go negative. Then you are correct.


Klaus
 

c_mitra

Advanced Member level 5
Joined
Nov 13, 2012
Messages
3,395
Helped
854
Reputation
1,708
Reaction score
807
Trophy points
1,393
Activity points
26,012
When I think about DC current, then it´s a straight horizontal line...
In that case, the rectified output of an AC is not DC; but I am not sure what is the more common understanding.
 

wwfeldman

Advanced Member level 2
Joined
Jan 25, 2019
Messages
693
Helped
169
Reputation
338
Reaction score
167
Trophy points
43
Activity points
5,022
I would size this LDO pass element depending on desired Vout, say 5V, drop out voltage, say 1V, so i need at least 6 V in
the current requirement has to be the maximum current the load will draw. plus a little margin
what is the actual input voltage?

your device is drawing, ??? per digital circuit at its highest frequency? (let's say 5 mA for fun)
multiply that by 10, unless you can guarantee the maximum number of your digital circuits that may be"ON" at the same time is less than 10
(or 50 mA)

and add about 10 to 20% margin. (60 mA)

since you now know current and voltage, find power and add a heat sink.

so this (hypothetical) LDO pass element will drop 1 V at 60 mA, or 60 mW
your device will drop the maximum voltage across the pass element times the maximum current (worst case - so design for it)

check the LDO specs to size capacitors on input and output, especially if you need to pick so it doesn't oscillate
 

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
46,978
Helped
13,934
Reputation
28,116
Reaction score
12,574
Trophy points
1,393
Location
Bochum, Germany
Activity points
273,927
For a DC current, the average and the RMS are the same.
Not for pulsed DC. But that's just theory and doesn't actually matter for the question.

The voltage regulator power dissipation only depends on the average DC current, the circuit must be additionally able to provide expectable peak current. No need to measure or calculate RMS numbers.

The circuit itself is a driver circuit that is switching at 200MHz.
Do you actually expect 200 MHz current components loading the power supply? Did you forget to install bypass capacitors in your circuit?
 

frankrose

Advanced Member level 3
Joined
Nov 27, 2013
Messages
772
Helped
216
Reputation
432
Reaction score
210
Trophy points
43
Activity points
5,369
How about if I want to calculate the current consumption of a switching digital blocks, do I use RMS or AVG?

Basically I want to make sure that I design the size of my pass device sufficiently to support the digital circuit while it's switching at its highest frequency and heaviest activities. I have the waveform of this current (kind look like noise with peaks here and there, e.g. 5mA peak, 10mA, etc), I'm wondering if I should calculate RMS or AVG for designing the LDO's pass device.
Hi,

Use RMS. It is the most talkative to estimate the load current with its all DC and AC components, and some overdesign is still necessary.
For example the voltage ripple on your LDO output should be low as possible, and probably design only for nominal current consumption is not enough to keep ripple low.
Bypass capacitor is not a solution always. Big load capacitance (even nF capacitors) can cause stability issues at certain LDOs, however to handle very fast peaks like in your case some capacitance (pF range) at the output is recommended.
If you design integrated circuit also necessary to know the AVG and PEAK values. Electromigration checker use different rule constants for those.
If it is not integrated I still suggest to calculate/measure AVG and PEAK values too, obviously they can tell you more info about your circuit.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,460
Helped
3,943
Reputation
7,884
Reaction score
3,813
Trophy points
113
Activity points
115,755
Hi,

I don't agree.
One should not mix RMS and AVG.
Each value makes sense, but one value can not replace the other.

One application where one can see the difference is the power dissipation calculation of SCRs.
(If interested, read some application notes about this topic.)
At SCRs there is a combination of two major power dissipation sources:
* one is the diode drop where you need to use AVG current
* and the other is the ohmic power dissipation, where you need to use RMS.
Both calculated values combine to the total power dissipation.
--> you can not use RMS only and you can not use AVG only.

Klaus
 

frankrose

Advanced Member level 3
Joined
Nov 27, 2013
Messages
772
Helped
216
Reputation
432
Reaction score
210
Trophy points
43
Activity points
5,369
Digital load current is like noise, as described in #5, and noise is characterized with its RMS value.
I agree with the difference, but SCR example is misleading now I think, or maybe it is answer for an other post.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,460
Helped
3,943
Reputation
7,884
Reaction score
3,813
Trophy points
113
Activity points
115,755
Hi,

I see your point.

But I also see that the OP asks about LDO dimensioning and load current.
And when I hear "power supply" I see a constant input voltage, whare a "squared value" makes no sense.

As already written in post #3v
...in most cases. It depends what you want to calculate
And the requested sketch ( in post#7) can clarify this.

Without exactly knowing what the OP really wants to calculate I can't give no detailed answer.

Klaus
 

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
46,978
Helped
13,934
Reputation
28,116
Reaction score
12,574
Trophy points
1,393
Location
Bochum, Germany
Activity points
273,927
LDO pass transistor power dissipation is (Vin-Vout)*avg(Iout). Additional considerations for SOA. Irms isn't present in the calculations.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,460
Helped
3,943
Reputation
7,884
Reaction score
3,813
Trophy points
113
Activity points
115,755
Hi,

This is exactly my idea, too (without more information from the OP)

Klaus
 

komax

Junior Member level 3
Joined
Dec 2, 2019
Messages
31
Helped
0
Reputation
0
Reaction score
0
Trophy points
6
Activity points
312
Hi,

This is exactly my idea, too (without more information from the OP)

Klaus
Sorry for the late reply, just hit morning here.

This illustrate what I want to do: Vout vs Iload

In the X-axis is a sweep of load DC current of the LDO, on the Y-axis is the LDO output. In this example I can see that 5mA is the maximum limit for my LDO and thus it can safely support any load up to that point.

My question is when I am given a digital blocks to design the LDO for, how do I approximate the equivalent DC current that I need to support based on the current waveform of said digital block. In other words, what is my "5mA" in that picture if I have the load current waveform, should it be based on average or rms?

The digital blocks themselves are a bunch of digital gates and inverters, so I don't think the V and I are proportional like resistor.
 

crutschow

Advanced Member level 5
Joined
Feb 22, 2012
Messages
3,737
Helped
891
Reputation
1,780
Reaction score
869
Trophy points
1,393
Location
L.A. USA Zulu -8
Activity points
20,992
LDO pass transistor power dissipation is (Vin-Vout)*avg(Iout). Additional considerations for SOA. Irms isn't present in the calculations.
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).
 

Toggle Sidebar

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top