I have a dBW value and dBW I understand to normally be given as representing a RMS Power (however this is not specificically ever stated and I am told its just assumed?)

from dBW to Watts = \$10^{\frac{dBW}{10}}\$. This then gives RMS power in Watts

At what point do I convert this from a RMS to a peak value?

  1. Multiply RMS Power in Watts by \$\sqrt{2}\$ before using \$V = \sqrt{P\cdot 50}\$
  2. Or multiply the result from \$V = \sqrt{P\cdot 50}\$ by \$\sqrt{2}\$

Because they give two different values...

  • \$\begingroup\$ Please confirm: 1) sqrt((sqrt(2)*power)*50); 2) sqrt(power*50)*sqrt(2); \$\endgroup\$ – CapnJJ Jun 7 '18 at 18:49

In terms of watts, the relevant quantity is an average value. RMS watts make no sense. RMS volts make sense and RMS volts x RMS current equals average power and nothing else. If you want to know the power peak level and the voltage and current are sinwaves that are in phase, the peak power is twice the average power. If volts and amps are not in phase then you can’t estimate the peak power from the average power without knowing the phase angle.

Please read this for confirmation that RMS power is a misnomer. And here are a couple of examples of power waveforms when V and I are in phase and when they are not in phase: -

enter image description here

Note that the power waveform shape and size remains the same but its average level falls when V and I are not in phase. Picture source


Power is proportional to voltage squared. RMS power is proportional to the square of RMS voltage. Peak power is related to the square of peak voltage. Peak voltage is related to RMS by sqrt(2), so peak power is 2 x RMS power. Peak-to-peak is the same (for a resistive load), because the power will always be positive.

  • \$\begingroup\$ Rms power is a misnomer. There is no meaningful physical quantity as RMS power. Please don’t propagate myths. \$\endgroup\$ – Andy aka Jun 7 '18 at 19:03
  • \$\begingroup\$ So when do i multiply by sqrt of 2 \$\endgroup\$ – Natalie Johnson Jun 8 '18 at 6:25

I am assuming that you are calculating the power through a \$50\Omega\$ load, as such

$$\begin{align}P_{AVG} &= V_{RMS}\cdot I_{RMS} = \frac{V_{RMS}^2}{R}\\ &\Downarrow \\ V_{RMS} &= \sqrt{P_{AVG}\cdot R} \end{align}$$

The right way to multiply by \$\sqrt{2}\$ is when using \$V_{RMS}\$. So you can write that:

$$V_{peak} = \sqrt{2}\cdot V_{RMS}$$

We also have a similar equation for the current.

$$I_{peak} = \sqrt{2}\cdot I_{RMS}$$

So in order to calculate the peak power, you'll have to use:

$$\begin{align}P_{peak} &= V_{peak}\cdot I_{peak} \\ &= \sqrt{2}V_{RMS}\cdot \sqrt{2}I_{RMS} \\ &= 2V_{RMS}I_{RMS} \\ &= 2P_{AVG} \end{align}$$

I hope that helps.

  • \$\begingroup\$ Rms power is a misnomer. No such thing exists as a meaningful quantity. \$\endgroup\$ – Andy aka Jun 7 '18 at 19:01
  • \$\begingroup\$ I edited my post to account for you remarks. \$\endgroup\$ – Sven B Jun 7 '18 at 19:28
  • \$\begingroup\$ @Andyaka Check now to see if that changes your mind. Sven, you have to @ someone to give them a notification in their mailbox. \$\endgroup\$ – KingDuken Jun 7 '18 at 19:37
  • \$\begingroup\$ Thanks for the hint! I kind of assumed it didn't work as it doesn't turn it into a link (to the user). \$\endgroup\$ – Sven B Jun 7 '18 at 19:46
  • \$\begingroup\$ @KingDuken why should it change my mind? Are you somehow expecting me to congratulate the answerer on his learning experience. I know this sounds rude but what do you realistically expect me to do or say? \$\endgroup\$ – Andy aka Jun 7 '18 at 20:27

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