Electromagnetic Compatibility Course Notes - Supplement
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Problem Solutions
Chapter 1
1. For each of the EMC problems below, identify the probable source,
coupling path and receptor.
a.) Source: Microwave Oven, Coupling Path: Radiated Field, Receptor: Computer Display
b.) Source: Starter Motor, Coupling Path: Conducted (through power wiring), Receptor: Car Radio
c.) Source: Lightning, Coupling Path: Conducted (Power or Communication Wiring), Receptor: Modem
d.) Source: Electric Blanket, Coupling Path: Radiated (likely) or Conducted through Power Wiring, Receptor: WiFi Router or Device
e.) Source: Airport Radar or Transmitter, Coupling Path: Radiated Field, Receptor: Tire Pressure Monitoring System
2. What are alternative names for conducted, inductive, and capacitive coupling?
Conducted Coupling ⇒ Common Impedance Coupling
Inductive Coupling ⇒ Magnetic-Field Coupling
Capacitive Coupling ⇒ Electric-Field Coupling
3.
4. 150 mV = 43.5 dB(mV) = 104 dB(μV) = –3.5 dBm
5. 6 dB (same as ratio of rms amplitudes)
6. +12 dB - 6 dB = +6 dB (or a factor of 2 in voltage), so output amplitude is 2 volts.
7. The term 50% gain is ambiguous. Does it mean 50% of the original amplitude (factor of 0.5) or 50% more (factor of 1.5)?
By the first interpretation, answers are 0.5 V, 1 V, 2V and 5V. By the second interpretation, answers are 1.5 V, 2 V, 3 V and 6V. These answers would all be different, if we interpreted the 50% gain to be a power gain instead of a voltage gain. Note that a 100% gain is generally interpreted as a doubling of the amplitude. However, a 200% gain is also generally considered a doubling of the amplitude. Signal gains, losses, or error margins should usually be expressed in dB in order to avoid misinterpretation.
8. 46 dB(μV/m) - 40 dB(μV/m) = 6 dB. Note that it is NOT 6 dB(μV/m), which is a signal amplitude totally unrelated to the correct answer.
9. 200 μV/m = 46 dB(μV/m). 100 μV/m = 40 dB(μV/m). 46 dB(μV/m) - 40 dB(μV/m) = 6 dB. Note that we could also say that the measured field is 200 μV/m - 100 μV/m = 100 μV/m over the limit. In most cases, however, it will be more helpful to know the ratio in dB rather than the difference in the field strengths.
10. Bulk Current Injection is an immunity test.
11. Radiated Immunity and Bulk Current Injection testing should be done in a shielded room to contain the fields and prevent them from interfering with the test equipment and other nearby electronic devices. Some radiated emissions tests also specify an absorber-lined shielded room, although useful emissions measurements could still be made outside the room. A shielded room is not required for conducted emissions and transient immunity measurements.
12. Conducted Emissions, Bulk Current Injection, Lightning Immunity and Electrical Fast Transient Immunity tests can only be done on products with an attached cable.
Chapter 2
1. @ DC: R ≈ 170 mΩ/m. @ 1 MHz: R ≈ 330 mΩ/m. @ 1 GHz: R ≈ 10 Ω/m.
2. C ≈ 52 pF/m
3. L ≈ 52 nH/m
4. At DC, we'll assume that the plane resistance is negligible relative to the trace resistance. The trace resistance is 4.0 Ω/m or 40 mΩ/cm.
At 10 MHz, the plane current is focused under the trace, so we'll assume that the plane resistance is approximately equal to the resistance of the trace. Skin depth is 21 μm, which is slightly thicker than the half-ounce copper. Resistance of the trace is still about 40 mΩ/cm. Resistance of the trace and plane is about 80 mΩ/cm or a little less.
5. C/ε = G/σ therefore G = C(σ/ε) = 10-8 F/m x 5.4 x 10-4 S/m/(4.3 x 8.854 x 10-12 F/m) = 0.14 S/m (or 1.4 x 10-3 S/cm).
6. Inductance of each loop is 0.146 μH. Leakage inductance is approximately the inductance per unit length of the wire pair (edge-to-edge spacing is 0.5 mm) times the circumference of the loop. This is 0.52 μH/m x 0.16 m = 0.083 μH. So mutual inductance is approximately 0.146 - 0.08 = 0.066 μH. Note that his is a coupling coefficient of only k = 66/146 = 0.45 (about 45%). Even though the loops share a lot of area, most of the flux is near the wire and the wire-to-wire spacing is too large to get efficient coupling. Using the same method, reducing the wire insulation thickness to 0.025 mm would increase the coupling coefficient to 80%.
