I really don't know much about radio signals.

But I want to know that is there any mathematical relation between antenna dB and the output of an RF circuit (I have heard they say this circuit is 200mW for example)?

I want to improve the range of an RF circuit. but I don't know whether it is possible to do it by a different antenna (with higher dB) or I have to make the output of my signal more strength?

  • \$\begingroup\$ Is it just dBs are log scale? audioholics.com/frequent-questions/… \$\endgroup\$
    – kenny
    Commented Sep 17, 2013 at 20:06
  • \$\begingroup\$ @kenny that is all directly applied to audio, they are using the scale for how it is used in audio equipment, some of that is general, but most of that is audio specific. \$\endgroup\$
    – Kortuk
    Commented Sep 17, 2013 at 20:47

2 Answers 2


You can do it either way. First you need to get all your units together. In small RF circuits you might see the output noted in dBm, which is the absolute power referenced to one milliwatt. To convert from dBm to W or mW and in reverse, these equations from Wikipedia provide the answer,

$$ P = 1mW \times 10^{{dBm}\over{10}}\\ P = 1W \times 10^{{(dBm - 30)}\over 1}\\ dBm = 10log_{10} {P\over1mW}\\ dBm = 10log_{10} {P\over1W} + 30 $$

So, if your system is rated at an output 200mW, you'll calculate an output of 23dBm. This will be used in the equation at the bottom of this answer.

The gain of an antenna is not a literal boost in power output, it's a perceived gain in one direction or on one axis from what one would expect from either an isotropic radiator (a perfect sphere) or a dipole radiator (there are other less common reference points). That is, if you were standing a certain distance from a perfect isotropic radiator and measured a power of 0 dB it would read 0 dB at every point with the same distance from the antenna, but, a real antenna with a gain of 2dB will read 2dB at some point with the same distance from the antenna, but certainly not every point at that distance. If the antenna is listed with gain units of dB, it's likely with respect to an isotropic radiator, but it could be with respect to a dipole. Some manufacturers use the \$ dB_i\$ or \$dB_d\$ to denote this difference.

As an illustration between the two, in the image below all points on the line would read the same output power. enter image description here

If only the non-isotropic antenna were on, and one measured the power output at the crossing point of the axis and the isotropic radiator the power would be higher than 0dB, say 3dB, then that antenna would be said to have a gain of 3dB. But, if the measurement was made on the opposite side the gain would be lower than 0dB, perhaps -5dB.

If you want to calculate how much power you will receive you can use the Friis transmission equation to get a rough estimate.

$$ P_r = P_t + G_t + G_r + 10log_{10}({\lambda \over {4\pi R}}) $$

The \$P_r\$ is the power received the \$P_t\$ is the power transmitted that you calculated from above, the \$G\$s are gains of the transmitting and receiving antennas respectively (in dB) and the final bit is the isotropic antenna equation, which will tell you how much power was 'lost' because it was not directed at your receiving antenna. So, as you can see, increasing either the gain of the antenna or the power transmitted will increase the power received.

So in summary, you can either get an antenna with a higher gain and point it correctly to increase signal strength or you can increase the power delivered to the antenna. Either one, done correctly, will increase the power output of your system (or at least the power received by the other end of the system).

  • 3
    \$\begingroup\$ Gain for an antenna is not in units of dBm, it is in units of dB. If you are measuring power it is dBm, but if you are talking about your antenna radiating 3dBm instead of the reference of 0dBm then you have a gain of 3dB on your antenna. \$\endgroup\$
    – Kortuk
    Commented Sep 17, 2013 at 20:50
  • \$\begingroup\$ @Kortuk They are both units of absolute power, so either one is correct. However, you're right that they are usually listed in dB, so I've edited the example. \$\endgroup\$
    – Samuel
    Commented Sep 17, 2013 at 20:57
  • 3
    \$\begingroup\$ dB is not a unit of absolute power, it is a ratio. When rating an antenna you rate it for much relative gain you will see at your maximum relative to input, not for how much power it radiates. Your answer seems spot on, you even explain it is relative to an isotropic antenna or a dipole. \$\endgroup\$
    – Kortuk
    Commented Sep 17, 2013 at 21:01
  • 2
    \$\begingroup\$ -1, No, a dB is NOT "the absolute power referenced to one watt" - that would be a dBW. A dB is a ratio - it doesn't have a reference until you state one. \$\endgroup\$ Commented Sep 17, 2013 at 21:03
  • \$\begingroup\$ Kortuk and Chris you are both correct. I was thinking one thing and saying another. Thanks for the corrections. \$\endgroup\$
    – Samuel
    Commented Sep 17, 2013 at 21:09

I want to improve the range of an RF circuit

The link-loss formula tells you how many decibels are lost between transmitter and receiver in free-space: -

Loss (dB) = \$ 32.45 + 20log_{10}[\$frequency(MHz)\$] + 20log_{10}[\$distance(km)\$]\$

For instance, if you transmit at 433MHz and your distance is 10km, the link loss is 32.4dB + 52.7dB + 20dB = 105dB. If you are only transmitting 1km then the loss is 85dB. It's worth mentioning at this point that the formula only applies to far-field waves and this means any distance greater than 10 x wavelength. At 433MHz far field is approximately 7m but this would be for simple antennas like dipoles. Parabolic dish antennas, due to the focusing of signals have a greater far field.

It allows you to calculate the dB loss between isotropic antennas in free space. Real antennas have a gain compared to isotropic antenna so the loss is mitigated by the antenna gains. If dipole antennas are used you get 2 x 1.76dB knocked off the loss and it becomes about 102dB.

If transmit output is 20dBm, using the example above you'll get a receive power of -82dBm.

But how much bandwidth can you put down and successfully receive. This document entitled "the essentials of radio wave propagation" (by Christopher Haslett) is a mine of information on this subject and it relates bandwidth and required power at a receiver with the formula: -

\$P_{required}(dBm) = -154 + 10log_{10}\$(bit rate)

If you are to receive 512kbps the minimum power necessary is -154dBm + 57dBm = 97dBm

The link loss formula is for free space i.e. no obstacles to get in the way and no strange effects caused by our planet. The document linked to above also goes into some of the formulas used to calculate link loss when "Earth" is taken into account. The Okumura–Hata model (page 35) gives the following: -

\$loss (dB) = 146.8 - 13.82log_{10} h + (44.9-6.55log_{10} h)log_{10} d \$

This is specifically for 900MHz (although it links to the original document that covers other frequencies) and assumes a mobile phone antenna height of 1.5m. In the formula above h is transmitting station height and d is distance.

The document linked to is a source of great information and is probably worth a good read.

  • \$\begingroup\$ I really appreciate if you find your books and fill the details as you said. but exactly I want to know if I have a 433MHz transmitter and receiver with let say 10mW output power, can I receive the data from that at 10 kilometer range?? which dB should my antenna be? Imaging my bandwidth needs to be 512Kbps. \$\endgroup\$ Commented Sep 18, 2013 at 5:03

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