Shannon Channel Capacity Calculator
Enter a channel bandwidth and signal-to-noise ratio to get the theoretical maximum error-free data rate and the spectral efficiency in bits per second per hertz.
The ceiling every link works under
Shannon capacity is the honest answer to “how fast could this channel possibly go?” It shows why widening a channel buys throughput directly while cranking up power buys ever less, and it gives you a target to measure a real modem against. When a link falls far short of it, the gap points at the modulation, coding, or interference — not at physics.
The signal-to-noise ratio that drives it comes out of a link budget: build received power with the RF Link Budget Calculator, convert powers with the dBm to Watt Converter, and estimate the path loss behind the SNR with the Free-Space Path Loss Calculator.
Frequently Asked Questions
What does the Shannon–Hartley theorem say?
It sets the absolute ceiling on error-free data rate over a noisy channel: C = B · log2(1 + SNR), where B is the bandwidth in hertz and SNR is the linear signal-to-noise ratio. No modulation or coding scheme, however clever, can beat it — it is a law of information theory, not a limit of today's hardware. Capacity grows linearly with bandwidth but only logarithmically with signal power.
How do I enter the SNR — in dB or as a ratio?
Enter it in decibels, the way it is usually measured. The tool converts it to the linear ratio the formula needs with SNR = 10^(SNR_dB/10). So 0 dB is a ratio of 1, 10 dB is 10, 20 dB is 100, and 30 dB is 1000. Because capacity depends on the log of (1 + SNR), each extra 10 dB of signal adds only a few more bits per second per hertz — diminishing returns from raw power.
What is spectral efficiency?
Spectral efficiency is capacity divided by bandwidth, in bits per second per hertz — how much data each hertz of spectrum can theoretically carry at the given SNR. It equals log2(1 + SNR) and is the number to compare channels or modulation schemes on, independent of how wide they are. A 20 dB channel tops out around 6.7 bits/s/Hz in theory.
Will a real link actually reach this rate?
No — it is an upper bound. Practical systems reach a fraction of the Shannon limit because real modulation and coding, protocol overhead, imperfect synchronisation, and interference all cost throughput; implementation gaps of several dB from capacity are normal. Use the figure to judge whether a channel could support a target rate and to compare designs, not as a throughput you should expect to measure. Educational estimate.
A theoretical upper bound only — real links reach a fraction of it after modulation, coding, and overhead. Educational estimate, not a throughput guarantee.