Power consumption, rms current or avg current values?

Hi,

I am still confused about rms and avg power calculation, which is the right one to use to find the Power consumption in a circuit? I want to compare two comparators, one is opamp-based comparator that shows 3uW of rms and avg power values (they are the same here), but for a clocked comparator I got 2uW for the rms power and 50nW for the avg power, so it is huge difference?!

the rms current represent DC value, right? if yes, should it be always that Power consumption is measured by rms current and voltage values?

Hi,

comparators and RMS current? I´m confused. Please post datasheets or give links.

RMS or average.
There is a clear answer: It depends.

Some examples:
A resistor is an ohmic load. Double the voltage then automatically the current doubles. and thus the power becomes squared. --> this calls for RMS, because the "S" in RMS means "square".
An LED has about constant voltage. Double the current...but the voltage stays about the same. The power isn´t squared it is about linear to the current. --> use average for this.

Now assume you have a 12V supply, a PWM circuit and a 6 Ohms resistor.
With 100% duty cycle everything is clear. 12V at the 6 Ohms resistor causes 2A and thus you have 24V of power. at the resistor and at the power supply.
Now let´s say you have 25% duty cycle. Then at 25% of the time you have 12V and 2A at the resistor causing 24W, but 75% of time there is no voltage and no current and thus 0W.
25% of the time you have 24W ... or in other words you have a mean value of 6W.

Now what DC voltage do you need to connect to a 6Ohms resistor to get 6W of power? P = U x U / R --> U = sqrt (P x R) = sqrt ( 6W x 6Ohms) = 6V. This is what the RMS value says.
RMS value of an AC voltage generates the same power at an ohmic resistior as the same value as DC voltage.
So with 25% duty cycle of 12V you get 6V RMS: General calculation: U(RMS) = U(DC) x sqrt(duty cycle)

Now the same current that runs through the resistor runs through the power supply.
But the voltage at the power supply stays constant (in opposite to the resistor voltage). Therefore you don´t calculate the power supply current with "average". 25% the time 2A means an average current of 0.5A.
As a test we calculate the power that the power supply has to deliver. It is constant 12V times 0.5A average = 12v x 0.5A = 6W.
--> the power supply delivers the same power that the resistor consumes.

Crazy? but physically and mathematically correct:
Same circuit: But at the resistor you have to calculate with RMS and at the power supply you have to calculate with average.

Klaus

The fact that a resistor power squares vs current has absolutely nothing to do with rms. It squares because the power in a resistor is I^2×R.

but Pavg=Irms*Vrms (always, right?)

Is it even better to plot P(t)=I(t)*V(t) and then take the average? if this is true, the confusing results I got is:

1) Irms=3uA, Vrms=VDD=1V ==> Pavg=3uW
2) I plot P(t) and then take the average and I got -50nW, !! it should not be the same results?

- - - Updated - - -

KlausST said:

Hi,

comparators and RMS current? I´m confused. Please post datasheets or give links.

Klaus

Click to expand...

It's a comparator I designed (transistor-level). It's just an example to inquire the right way to find Power consumption in the circuit.

Hi,

Sorry, just saw that I used the German "U" instead of the international "V" as the sign for volts.
--> now using V"
****

The fact that a resistor power squares vs current has absolutely nothing to do with rms.

Click to expand...

Oh, yes it surely has to do with that.

As explained above.
When you change V, but I keeps constant then the power raises linear --> use average
When you change I, but V keeps constant then the power raises linear --> use average
When you change V and the same ratio I changes, than the power raises squared to V --> use RMS
When you change I and the same ratio V changes, than the power raises squared to I --> use RMS

P = V x I is true for all power calculations.
Now the direct relationship: V = R x I makes you to calculate P in three ways: but only for a ohmic resistor.
P = V x I ; now replayce V with the formula according ohm´s law: V = R x I
the result is: P = R x I x I = I^x R

The same you could do to get P = V^2 / R

Klaus

- - - Updated - - -

Hi,

for your comparator you have to differentiate between standby power and switching power.
And maybe even: with load and without load.

Maybe there are even more current values like bias current, pullup current.
I again recommend that you show us the document you refer to.

Klaus

For supply power delivered by a DC voltage, you are solely looking for Iavg.

but Pavg=Irms*Vrms (always, right?)

Click to expand...

Not always. For DC, yes.

For AC, (not necessarily sinusoidal), P = V(inst)*I(inst) and P(avg)=Integral(V.I) ---> you may need to use the phase factor.

P(avg)=Irms*Vrms will be true only for sine wave and a resistive load.

2) I plot P(t) and then take the average and I got -50nW

Click to expand...

If you get a negative result, it means I and V are of opposite sign (negative resistance) and your load is simply not resistive.

bio_man said:

but Pavg=Irms*Vrms (always, right?)

Is it even better to plot P(t)=I(t)*V(t) and then take the average? if this is true, the confusing results I got is:

1) Irms=3uA, Vrms=VDD=1V ==> Pavg=3uW
2) I plot P(t) and then take the average and I got -50nW, !! it should not be the same results?

Click to expand...

I think I found the problem that makes my confusion, in 1) Pavg=Irms*Vrms (I'm missing the Power Factor term cos(phi))

So based on the discussion, P_consumption=P_avg= average[i(t)*v(t)] is always right and for DC supply voltage P_consumption=P_avg=V*Iavg

I think now it is much clear, I appreciate your input ! thanks alot

If supply voltage is fixed then I like to use integ() on the
supply current, times voltage, divided by period. This is
time-averaged supply current.

RMS is good for signals that have a negative and positive
value over time. Supply current is unidirectional so an
averaging approach is good enough.

I have also on occasion used a cccs onto an integrating
capacitor and used the endpoint voltage, scaled by time
and C value.

I think now it is much clear, I appreciate your input !

Click to expand...

Glad that it is clearer now. For any given function f(x), we can define moments like:

mean= average (f(x)) over some interval;

second moment = average (f(x)*f(x)) over some interval;

third moment = average (f(x)*f(x)*f(x)) over some interval;

and all higher moments.

RMS is a simple construct: average (f(x)*f(x)) - average(f(x))*average(f(x))

The second moment is also commonly called "the sum of squares"- a very important quantity in many analysis.

Third and fourth moments are less widely used.