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Lecture 2. Decibels and EMC Units, Government EMC Requirements and some Typical EMC Test Facilities. EMC Units. Voltage in volts (V) Current in amperes (A) Power in watts (W) Electric field in volts per meter (V/m) Magnetic field in amperes per meter (A/m)
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Lecture 2. Decibels and EMC Units, Government EMC Requirements and some Typical EMC Test Facilities
EMC Units • Voltage in volts (V) • Current in amperes (A) • Power in watts (W) • Electric field in volts per meter (V/m) • Magnetic field in amperes per meter (A/m) • Power Density in watts per square meter (W/m2) • NOTE: These units tend to vary over a very wide range, so decibel units are commonly used to compress EMC data.
Decibel Unit (dB) • The decibel (dB) unit of measure was first used to compare audio power intensities in telephony. • The dB was originally defined as a ratio of two power levels. For example, consider an audio amplifier that yields an output sound power of 1 kW when an input sound power of 10 W is applied to its input, then its power gain is Pout/Pin = 1000/10 = 100. Similarly, its power gain in decibels is:Power GaindB = 10 log10(Pout/Pin) = 10 log10(1000/10) = 20 dB • It may be said that the output power of this amplifier is “20 dB above its input power level”.
Why capitalize the “B” in dB? • The “dB” unit was first proposed by Bell Telephone Labs, and it was named in honor of the inventor of the telephone, Alexander Graham Bell. That is why the “B” in “dB” is capitalized; it stands for a man’s last name! • 1 dB represents the amount of audio noise power increase that is just “barely discernible” as a change in volume by the typical human ear.
Because 10 log10(2) = 3 dB, a 3 dB rise in power level corresponds to a doubling of the power. Because 10 log10(10) = 10 dB, a 10 dB rise in power level corresponds to a 10-fold increase in power. Since the response of the human auditory system is approximately logarithmic, a 3 dB rise, or a doubling, in signal power represents the same increase in perceived loudness, no matter what the starting power level. Thus, starting with a sound level of 0.1 W and doubling it to 0.2 W will sound to the human ear like the same incremental increase in volume as starting with a sound level of 1 W and doubling it to 2 W, or starting with 4 W and doubling it to 8 W. In each case, the increase in sound level is a constant 3 dB!
Nonlinear compression provided by the decimal unit of measure
Sound files to show the size of a decibel What happens when you halve the sound power? The log of 2 is 0.3, so the log of 1/2 is -0.3. So, if you halve the power, you reduce the power and the sound level by 3 dB. Halve it again (down to 1/4 of the original power) and you reduce the level by another 3 dB. That is exactly what we have done in the first graphic and sound file below. (From http://www.phys.unsw.edu.au/~jw/dBNoFlash.html) Click here to listen to the 3-dB stepped white noise (The noise power is successively halved between each sample)
Broadband noise decreasing by 1 dB steps. Click here to hear the barely perceptible sound level difference between 1 dB stepped white noise samples, yet by the end of the demo, you can tell the noise level has been greatly reduced! (Like a missionary placed by Fiji cannibals in a gradually heated pot!)
Broadband noise decreasing by 0.3 dB steps You may notice that the last is quieter than the first, but it is difficult to notice the difference between successive pairs. Solving 10*log10(Pratio) = 0.3 yields Pratio = 1.072, so to increase the sound level by 0.3 dB, the power must be increased by 7.2%. Likewise, solving 20*log10(Vratio) = 0.3 yields Vratio = 1.0351, hence or the voltage amplitude must be increased by 3.51%. Click here to play the 0.3 dB stepped white noise sounds… note that the loudness of each step is imperceptible from the one before it.
Note how dB units of measure act to compress a wide range of power ratios: Power Ratio dB = 10 log10(Power Ratio) 106 60 dB 103 30 dB 10 10 dB 2 3 dB 1 0 dB 10-1 -10 dB 10-3 -30 dB 10-6 -60 dB
Voltage Gain, Current Gain, and Power Gain of a “Terminated” Amplifier Voltage Gain = vout / vin Current Gain = iout / in Power Gain = Pout / Pin= (vout*iout) / (vin*iin) = (Voltage Gain)*(Current Gain) Assuming all current and voltage quantities (vin, iin, iout, vout) are expressed in root mean square (rms) units, Input power = Pin = vin2/Rin = iin2Rin and Output power = Pout = vout2/RL = iout2RL Power Gain = Pout/Pin = (vout2*Rin) / (vin2*RL) = (iout2RL) / (iin2Rin) Using dB units: Power GaindB = 10 log10(Pout/Pin) = 10 log10((vout2Rin) / (vin2RL)) = 10 log10((iout2RL) / (iin2Rin))
In many practical applications, the source and load terminations are Rin = RL = 50 Ω. If this is the case, the power gain in dB may be expressed as Power GaindB = 10 log10(Pout/Pin) = 10 log10(vout2 / vin2) = 20 log10(vout/vin) = 10 log10(iout2 / iin2) = 20 log10(iout/iin) Thus the power gain in dB may be expressed as 10 times the base-10 log of a power ratio (power gain), or equivalently as 20 times the base-10 log of a voltage ratio (voltage gain) or 20 times the base-10 log of a current ratio (current gain). The above observation motivates us to extend the definition of the decibel to include the ratio of two voltages and the ratio of two currents, in addition to the ratio of two power levels. We shall now define the decibel in 3 ways: dB = 10 log10(power ratio) = 20 log10(voltage ratio) = 20 log10(current ratio)
Using dB to indicate absolute quantities • We have seen that a decibel is the ratio of two quantities. • However, absolute power, voltage, or current levels are also often expressed in decibels by giving their value above (or referenced to) some base quantity. The reference level used is tacked on to the dB unit label. For example, a value expressed in dBµV would be the number of dB above a 1 µV “reference value”. Examples: • Voltage V1 = 10 V in dBµV = 20 log10(V1/1 µV) = 20 log10(10 V/1 µV) = 140 dBµV • Voltage V1 = 10 V in dBmV = 20 log10(V1/1 mV) = 20 log10(10 V/1 mV) = 80 dBmV • Current I1 = 0.01 A in dBmA = 20 log10(I1/1 mA) = 20 log10(0.01 A/1mA) = 20 dBmA • Current I1 = 0.01 A in dBA = 20 log10(I1/1 A) = 20 log10(0.01 A/1A) = -40 dBA • Power P1 = 10 mW in dBmW = 10 log10(P1/1mW) = 10 log10(10 mW/1 mW) = 10 dBmW • Power P1 = 10 mW in dBµW = 10 log10(P1/1 µW) = 10 log10(10 mW/1 µW) = 40 dBµW • Power P2 = 1pW in dBµW = 10 log10(P2./1 µW) = 10 log10(1pW/1 µW) = -60 dBµW • NOTE dbmW is used so often that it is commonly shortened to “dbm”
Converting a power level in dBm to an equivalent voltage level in dBµV or an equivalent current level in dBµA • When converting between units of volts, amperes, and power, assume a 50-ohm load if the problem does not state otherwise. • ExampleGiven: A power level P1 = -12 dBmW = -12 dBm across 50 ohm load Find: (a) the corresponding voltage V1 across this load in absolute volts, (b) the voltage in dBµV, and (c) the corresponding current I1 through this load in dBmA. • -12 dBm = 10 log10(P1/1 mW) => P1 = 63.0957 µW • P1 = 63.0957 µW = V12 / 50 Ω => V1 = 0.056167 V • V1 in dBµV = 20 log10(0.056167 / 1µV) = 94.99 dBµV • I1 in dBmA = 20 log10((0.056167 / 50Ω) / 1mA) = 1.01 dBmA
Analyzing Cascaded Transmission Path using dB RL=50Ω Load + Vin = 20dBµV= 10 µV - • 6 dB • Volt. Gain “x2” • 20 dB • Volt. Gain “x10” • 40 dB • Volt. Gain “x100” + Vo - Vin The overall voltage gain of cascaded amplifiers shown above is the product of their individual voltage gains, therefore Vo = 10 µV *2*10*100 = 20 mV Alternately, the overall voltage gain in dB is the sum of their individual voltage gains in dB. To prove this, let us calculate the output voltage “Vo” in dBµV units 20log(Vo / 1 µV) = 20log([10 µV *2*10*100] / 1 µV) = 20log(10) + 20log(2) + 20log(10) + 20log(100) = 20+6+20+40 = 86 dBµV => Vo = 20 mV (same answer as before!) Thus we have shown that VodBµV = VindBµV+6dB+20dB+40dB = 86 dBµVSimply add the individual dB gains of each part of a cascaded path to find the overall gain!
Example Problem (From our textbook.) Solution dBμV dBm
Equipment sold in USA: Radiation requirements from 9 kHz – 3 GHz by Federal Communications Commission (FCC) Title 47 Code of Federal Regulations Part 15(b) for “unintentional radiating” unlicensed electronic equipment - (Class A equipment – industrial environment, Class B equipment – residential environment.) Requirements established in 1979. For industrial digital devices (10 m range) For analog radio receivers For residential digital devices (3 m range)
For products sold outside the USA – tending toward an international standard: CISPR – “International Special Committee on Radio Interference” (French acronym) Publication 22 (1985). Commonly called “CISPR 22” Note FCC requirements measured at 3m and 10m, while CISPR requirements measured at 10m and 30 m. Assume field intensities fall off as “1/distance” in order to scale EMC measurements to compare FCC and CISPR requirements.
Line Impedance Stabilization Network (LISN) • Used for measuring conducted emissions. Couples RF noise (450 kHz – 30 MHz) from BOTH phase and neutral ac power lines of the product under test to the spectrum analyzer. • Prevents noise external to the test (on the common ac power network) from contaminating the measurement) L1 and C1. • Presents a constant impedance (50 ohms) over the frequency range and from site to site to the product between “phase (hot) and ground” and between “neutral (cold) and ground”.
Example 2: PCB Radiation - Device • Two Data Terminal Equipment (DTE) ports (EIA RS422A, time multiplexed) • One Ethernet 50- coaxial cable (IEEE 802.3)
