Imagine a transmitter + transmitter antenna with the following characteristics:

  • 433Mhz transmitter transmitting 10dBm
  • Into an impedance matched (50 ohms) antenna, omni-directional, 3dBi gain

And a receiver:

  • 50 ohm impedance antenna, receiver circuitry presents 50 ohms
  • Receiving antenna is an identical 3dBi omni-directional antenna.

How would I go about calculating the voltage the receiver circuitry would see? I'm trying to learn RF electronics by building a detector for a constant 433.92Mhz carrier, and I can't predict how my diode will behave without knowing the voltage applied across it.

My attempt from first principles:

10 dBm is 10mW of power, 3dBi of gain means 10mW X 2 = 20mW radiated from the antenna under ideal conditions.

Based on the inverse square law, we get:

$$ P = { 0.02 \over 4 \Pi r ^2 } $$

$$ P = { 0.02 \over 4 \Pi 40 ^2 } $$

$$ P = { 0.02 \over 4 \Pi 40 ^2 } $$

$$ P = 9.94718394^{-7} $$

Yipes that seems small.

This is the bit where I get stuck. Can I just multiply it by the receiving gain (3dBi, basically 2x) and sub it into ohms law (with the impedance of the antenna, 50 ohms) combined with the power formula (P = IV)?

$$ P = ({V \over R}) V $$

$$ 2 \times 9.94718394^{-7} = ({V \over 50}) V $$

After doubling the power (because the receiving antenna gives me 3dBi gain) and plugging this formula into Wolfram Alpha to solve, I get ~10mV.

This seems low and I'm not confident with the logic of my derivation. Is this correct?

  • 1
    \$\begingroup\$ 10 dBm is 10 mW of power and not 10 mV. \$\endgroup\$
    – Andy aka
    Commented Sep 16, 2018 at 9:54
  • \$\begingroup\$ 10mW is what I meant, edited question to have correct units. Thanks. \$\endgroup\$ Commented Sep 16, 2018 at 10:02

2 Answers 2


You should consider the Friis transmission loss (or link loss) equation for free space: -

Loss (dB) = 32.45 + 20\$log_{10}\$(f) + 20\$log_{10}\$(d)

Where f is in MHz and d is in kilometres. This equation tells you how many dB of power loss you can expect at a given distance with a given carrier frequency.

It is based on free-space and accounts for the reduction in received signal as frequency rises due to the received antenna becoming smaller. This equation is also for antennas that transmit equally in all directions (so called isotropic antennas). With real antennas there is always a gain because real antennas don't transmit equally in all directions hence there is always one direction where there is a higher power density of transmission.

The equation also assumes that your antennas are not so close that a proper EM wave hasn't formed. Translated, this means that your antennas must be at least one wavelength apart and, as you appear to be using 40 metres in your question and, your frequency is 433 MHz, there is no problem.

Putting numbers in gets you a link loss of 32.45 dB + 52.73 dB - 27.96 dB or 57.22 dB. Given your antennas have a gain of 3 dB each the link loss reduces to about 51.2 dB. Sanity check using on-line calculator: -

enter image description here

Given your transmit power is +10 dBm, the received signal power will be 52.2 dB down on +10 dBm or -41.2 dBm or about 0.076 uW.

Related Q and A.

Another related Q and A

  • \$\begingroup\$ Awesome that theres an equation for loss-over-distance, thanks! In terms of working from the received power (3.09uW) to voltage however, can I just substitute it into ohms law & P=IR (as I did above) to calculate the voltage? \$\endgroup\$ Commented Sep 16, 2018 at 10:37
  • \$\begingroup\$ Yes you can be do be aware that a dipole antenna will have an effective source resistance (aka radiation resistance) of about 73 ohms and the power delivered in the equations above assume that the correct resistance is matched to the antenna resistance. Be also aware that an antennas impedance varies dramatically when the antenna length doesn't precisely match the wavelength of the carrrier. \$\endgroup\$
    – Andy aka
    Commented Sep 16, 2018 at 10:41

Using the formula Power = VoltsRMS ^2 / Resistance, and then changing Resistance to Impedance, we have

Power = VoltsRMS^2 / Impedance

and we solve for VoltsRMS

VoltsRMS = squareroot (Power * Impedance)

Now at one milliWatt power (0.001 watt) and 50 ohms, the math produces

VoltsRMS = squareroot (0.001 * 50) = squareroot (1 / 20) = 1 / squareroot(20)

VoltsRMS = 1 / 4.47 = 0.223 volts rms

VoltsPP = 0.223 * 2 * 1.414 = 0.632 voltsPP

Are your numbers consistent?


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