# What is needed link budget? [closed]

Assuming I want to send a signal 1 km away from a 2.4 GHz antenna with 5 DBI gain and recieving antenna of 2.15 DBI gain. What is the minimum needed power at the receiver (not the minimum transmitter power) to get a 1:1 SNR ratio.

• No it's not a homework – ObsessionWithElectricity Sep 20 '18 at 2:12
• Depends on your noise power... – DSWG Sep 20 '18 at 2:31
• My noise power is around -100dbm – ObsessionWithElectricity Sep 20 '18 at 9:32

Here is an example of link budget analysis, with some assumptions of certain values:

-174dBm/Hertz is Boltzmann/Nyquist/Johnson/thermal noise floor

+70 dB for 10^+7 Hertz bandwidth

+0.5dB for losses from Antenna to front-end-matching

+0.5dB for front-end-matching losses (a PI network with low Q)

0dB for Intermodulation Distortion Products that fall in-channel and tho deterministic in nature will degrade the SNR. Taylor Series is the useful math.

-6dB for the benefit of that PI network

+3dB for Noise Figure of the Low Noise Amplifier

+6dB for SNR needed to achieve 1% data packet re-send rate [lots of room to play with this]

-174 + 70 +0.5 +0.5 +0 -6 +3 + 6 = -174 + 74 = -100dBm

With 0dBm (50 ohm) being 0.223 volt RMS, the level of -100dBm is 100,000:1 smaller, or 2.23 microVolts RMS.

This is the power required OUT OF THE Receiver antenna.

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Given Watts Law; { Power = Voltage ^2 / Resistance }, then for the 1 milliWatt ( that 0 dBm baseline is 1mW) power, the RMS voltage is sqrt( 0.001 * 50) = 0.223 volts RMS at the 0dBm level. The -120 dBm level is 0.223 microVolts RMS. The -180 dBm level is 0.223 nanoVolts RMS. And yes, some radio systems can perform quite well at the -180 dBm level (NASA deep space ground stations).

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The "+6dB for SNR" has lots of room to play. Turbo Coding should let you easily work at +6dB SNR, with excellent (low) bit error rates (BER). At 6dB, you are well above Shannon Limit.

Without any error correction, as in classic NRZ/biphase 2-level signaling, you need to be near +18dB SNR and with low jitter in where you sample the data-eye.

For analog FM, you'll need +18dB SNR or more; for analog FM you can reduce the bandwidth from 10MHz to 200KHz.

I included no margin for pathlosses nor for rain fading or nor for multi-path null / peak. But I did tell you this was AT output of the antenna, right?

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Path Loss will be

PL = +22dB + 10*log10( [distance/wavelength]^2 )

or

PL = +22dB + 20*log10( distance/wavelength )

At 1,0000 wavelengths or about 120 meters, the baseline energy spreading between TX antenna and RX antenna (its a range^2 spreading) will cause 22+60 = 82 dB reduction in energy density.

Assuming ZERO dB antenna gains, you'll need this power out of your transmitter:

-100dBm + 82 = -18 dBm

Given 0 dBm is 0.223 volts RMS, and with the 3 factors of 6dB in the required power, the TX voltage must be 0.223 * ( 0.5 * 0.5 * 0.5 ) = 0.223/8 or about 0.028 volts RMS at 50 ohm impedance.

In all this, I'm assuming ZERO dB antenna gains. NASA (and perhaps NSA) have dishes with +70dB gains. These are great for collecting energy from the Pluto probes, out at 8GHz and 8 Billion miles.

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Now discuss the "0dB" contribution of Intermodulation Products. This next section is mostly from a book by Wambacq and Sansen "Distortion of analog integrated circuits". From page 355, having googled "ip2 distortion of bipolar junction", we have the 2 results

IP2h = 4 * n * Vt

IP3h = 2*sqrt(6) * n * Vt

where Vt = 0.026 volts at room temperature. These results are using a weak-inversion FET model (often called sub-threshold) where the FET exhibits an exponential response like the bipolar junction devices exhibit; Wambacq&Sansen compute "n", the nonideality factor, caused by non-abrupt PN junctions (from what I've read) as 1.275x for the 0.5 micron FET technology (e.g. the gradient of the doping). All this makes IP2h = 131 milliVolts at room temp, and IP3h = 161 milliVolts at room temp. You may find other, more suitable, distortion IP2 and IP3 values in other references. But what do these mean? How to interpret?

First off, round both of these values to 0.2 volts, and proclaim InterceptPoint2 = InterceptPoint3 = 0.2 volts. And memorize that number. This is -14 dBv, or 1/5 of a volt.

Consider IP2. Input 1/5 volt into such an amplifer, and the output 2nd harmonic distortion will be the same amplitude as the amplified signal. Thus -14 dBv input to a 10 dB amplifier will produce +4dBv output of signal and +4dBv of the 2nd harmonic.

Consider IP2 again. Reduce the input to -100dBv, and have gain of 10dB. The output signal will be -90dBv. The output distortion (the 2nd harmonic) will be 2dB per dB weaker than the IP2 intercept. Given the delta is -14dBv - 100dBv, or 86dB, we double that to -172 dB; the output distortion is 172 db below the Intercept of -14dBv, thus the 2nd harmonic distortion is -14 - 172 = -186dBv.

Given the output signal is -90dBv, the SINAD based on distortion is 96dB, and other deterministic and random behaviors will limit the BitErrorRate.

What if you have signal carrier at 2.4Ghz and -100dBv, and the front end with poor selectivity lets in 1.2GHz at -50dBv (which is not dBm)? What is the SNR or the SINAD? [warning: the diagram below does not address this]

simulate this circuit – Schematic created using CircuitLab

• First of all power is denoted in watts not volts. – ObsessionWithElectricity Sep 20 '18 at 8:29
• Second, I want to know the minimum power I need to be received at the recieving end (which will be typically in nanowatts or picowatts) not the power the receiver takes from my circuit but the raw power it gets from air. – ObsessionWithElectricity Sep 20 '18 at 8:31
• He has given you those numbers, -100dBm out of the RX antenna (Assuming no feedline losses) for a 10MHz channel bandwidth. -100dBm is 0.1pW via the obvious maths. – Dan Mills Sep 20 '18 at 9:42
• Seriously, go and get a communications systems book, we cannot teach you fundamental radio engineering on this site, the topic is just too huge. Pozars "Microwave and RF Design of Wireless Systems" might work, his "Microwave Engineering" book is probably a little too low level. – Dan Mills Sep 20 '18 at 9:52
• Actually I am not an engineer or engineering student so I have just learnt these topics from here and there. – ObsessionWithElectricity Sep 20 '18 at 10:33