Convert Power to a Peak Voltage, when to multiply by sqrt(2) - Not so simple?

Question

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...

Answer 1

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: -

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

Answer 2

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.

Answer 3

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.