Hi all,

Consider a simple inverter feed with a clock at a certain frequency. Which is the correct way to calculate the power dissipation? and why?

a. rms(VDD(t)) * rms(IDD(t))
b. average(VDD(t) * IDD(t))

where VDD is the supply voltage, IDD is the supply current.

thanks,
chase

Hi,

Thanks aw, but I have further question.

Assume a square wave voltage source with 50% duty cycle, connected to a resistor in series. Then I will agree that the mean power will be the product of rms voltage and rms current.

However for an inverter, assumed clocked at 1/T and at supply voltage V, therefore the charge Q, being dumped into the ground is C*V every T seconds. Therefore energy being dissipated per cycle is Q*V, power, P = Q*V/T. Since Q = ∫I dt, then power, P = 1/T *∫I dt * V. Isn't it the 1/T * ∫I dt = average current instead of RMS current?

Can anyone clarify that?

Thanks,
chase

Short answer: do not use Vrms*Irms.

The correct answer to "how do I measure the average power correctly" is (b), average(V*I) for pin V_DD, assuming that:
(i) the inverter has one supply V_DD
(ii) the negative supply is tied to ground
(iii) there is no net power in/out of any other pin

More generally, the average power dissipated in a block with multiple pins will be the average of the sum of V*I on all pins:

P_AVE= average( V₁*I₁ + V₂*I₂ ... V_N*I_N)

P=Vrms*Irms is true only for purely resistive (real) loads.  As a trivial counter-example, imagine that your block contains only an ideal capacitor and you drive it with a sine wave.

The paper at www.eznec.com incorrectly states that P=Vrms*Irms is always true, because it only considers a resistive load.  It makes a slightly different point, which is that the RMS value of (I*V) has no particular significance.  I would recommend this article instead:

http://en.wikipedia.org/wiki/AC_power