How Many µA of Common-Mode Current Fail CISPR 32 Class B?

Engineers often hear that a few microamps of common-mode current on a cable are enough to fail a radiated-emission test. This post puts real numbers on that claim using the short-cable (Hertzian dipole) model and the CISPR 32 Class B limits, then converts the result into a reading you can check with a current probe before going to the lab.

The model

For a cable much shorter than the wavelength (L < λ/10) carrying a common-mode current I, the maximum field over a ground plane at distance r is approximately E = 1.257 × 10⁻⁶ · f · L · I / r, with f in Hz, L in metres, I in amperes and r in metres. The factor already includes the ×2 ground reflection of an open-area or semi-anechoic site. Note that E rises with frequency and cable length, so the same current is far more harmful at 200 MHz than at 30 MHz until the cable stops being electrically short.

Worked example: 1 m cable at 30 MHz

Take a 1 m USB or power cable and the CISPR 32 Class B limit at 10 m, which is 30 dBµV/m (quasi-peak) from 30 to 230 MHz. 30 dBµV/m is 31.6 µV/m. At 30 MHz the wavelength is 10 m, so a 1 m cable is exactly λ/10 and the model still applies. Solving for current: I = E · r / (1.257 × 10⁻⁶ · f · L) = 31.6 × 10⁻⁶ × 10 / (1.257 × 10⁻⁶ × 30 × 10⁶ × 1) ≈ 8.4 µA.

So roughly 8.4 µA, or 18.5 dBµA, of common-mode current on one metre of cable reaches the Class B limit. If you want the usual 6 dB of margin, the target is about 4.2 µA (12.5 dBµA).

How the allowed current changes with frequency and length

Because the model is only valid for electrically short cables, the useful length shrinks as frequency rises. The table keeps each case at or below λ/10.

Values from E = 1.257 × 10⁻⁶ · f · L · I / r with ground reflection included. Above λ/10 the real cable behaves like a resonant antenna and the model is no longer reliable.

The 3 m row gives almost the same current as the 10 m row because the 3 m limit is 10 dB higher while the distance ratio of 10/3 is about 10.5 dB. That is why the 10 dB rule of thumb for converting between 3 m and 10 m limits works for far-field sources.

Turning it into a current-probe check

A current probe clamped around the whole cable reads the common-mode current directly: I (dBµA) = receiver reading (dBµV) − transfer impedance Z_T (dBΩ). With a probe whose calibration sheet gives Z_T = 15 dBΩ at 30 MHz, the 8.4 µA limit (18.5 dBµA) corresponds to a receiver reading of 33.5 dBµV. With 6 dB margin, aim for 27.5 dBµV or lower.

• Clamp around the entire cable (all conductors and the shield) so differential currents cancel and only common-mode current is measured.
• Slide the probe along the cable and keep the maximum reading; standing waves make the current position-dependent.
• Use the Z_T value at each frequency from the probe's own calibration data, not a single nominal value.
• Use the same detector and bandwidth as the final test (120 kHz RBW, peak for a quick scan) so the comparison is meaningful.

What this tells you about fixes

Microamp-level currents are tiny compared with the milliamps flowing in the circuit, so they come from small imbalances: a ground connection with a few nanohenries of inductance, a shield bonded by a pigtail, or a missing filter at the connector. A ferrite clamp that adds about 100 Ω of common-mode impedance at 30 MHz can cut the current by several times when the source impedance is low, which is often the cheapest way to win the 6 dB back. Confirm the result with the probe, then recheck the margin against the limit including your measurement uncertainty.