Bit error rate (BER) calculator
Measure mode divides the bit errors you counted by the bits sent. Test mode answers the planning question: how many bits must pass, with at most a given number of errors, before you can claim the link meets a BER target at a confidence level, and how long that takes at your line rate.
Runs in your browser: nothing you enter is sent anywhere.
How to use it
- Measure: enter the bit errors from your tester or counters and the bits sent in the same period.
- Plan a test: enter the target as an exponent (12 for 10⁻¹²), the confidence level, how many errors you will tolerate and the line rate in Gbit/s.
- Read how many bits to send and the test time.
- BER = errors ÷ bits
- Test plan: find λ such that P(X ≤ E | λ) = 1 − CL for a Poisson count X with E allowed errors
- For E = 0: λ = −ln(1 − CL)
- Bits needed = λ ÷ target BER; test time = bits ÷ line rate
Worked example
Target BER 10⁻¹², 95 % confidence, no errors allowed, 10 Gbit/s link:
- λ = −ln 0.05 = 2.9957.
- Bits needed: 2.9957 × 10¹² = 2,995,732,273,554.
- Test time: 2.9957 × 10¹² ÷ 10¹⁰ = 299.57 s, that is 4.99 min.
Tips
- An error during the test does not end it. Look up the new bit count for one or two errors and keep running.
- Measure mode gives a point estimate. A few errors in a short run say little; for a reliable figure collect at least tens of errors.
- Count bits, not bytes or frames. Frame error counters need converting: each lost frame can hide one or many bit errors.
Frequently asked questions
How is bit error rate calculated?
BER = bit errors ÷ total bits sent. 3 errors in 5 × 10⁹ bits give 6 × 10⁻¹⁰, which is one error per about 1.67 billion bits.
How long must a BER test run to prove 10⁻¹²?
With no errors allowed and 95 % confidence, you need −ln(0.05) ÷ BER ≈ 3 ÷ 10⁻¹² = 3 × 10¹² bits. At 10 Gbit/s that is about 300 seconds, 5 minutes. Allowing one error raises the factor to about 4.74, two errors to about 6.30.
Why is the rule of thumb 3 divided by the BER?
Errors on a link are rare and independent, so their count follows a Poisson distribution. With an expected λ errors, the chance of seeing none is e^−λ. For this to be only 5 %, λ = ln 20 ≈ 3. So if 3 ÷ BER bits pass without error, a worse BER would very likely have shown an error.
What BER do networks require?
Many Ethernet standards specify 10⁻¹² for the link, and links with forward error correction often quote a pre-FEC BER of around 10⁻⁴ to 10⁻⁵ that FEC then corrects. Check the standard for your interface.