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/m

Lecture 2. Decibels and EMC

Units, Government EMC

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/m

)



NOTE: These units tend to vary over a very wide

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 Gain

= 10 log

(Pout/Pin) = 10 log

(1000/10) = 20 dB



It may be said that the output power of this amplifier is “

20 dB

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

Because 10 log

(2) = 3 dB, a 3 dB rise in power level

corresponds to a

doubling

of the power.

Because 10 log

(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

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.

One decibel is close to the Just

Noticeable Difference (JND) for

sound level. As you listen to these

files, you will notice that the last is

quieter than the first, but it is rather

less clear to the ear that the second

of any pair is quieter than its

predecessor. 10*log

(1.26) = 1, so

to increase the sound level by 1 dB,

the power must be increased by

26%, or the voltage by 12%.

Broadband noise decreasing by 1 dB steps.

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

Note how dB units of measure act

to

compress

a wide range of power

ratios:

Power Ratio dB = 10 log

(Power Ratio)

10

60 dB

10

30 dB

10

10 dB

2

3 dB

1

0 dB

10

-10 dB

10

-30 dB

Voltage Gain, Current Gain, and Power Gain of a

“Terminated” Amplifier

Assuming all current and voltage quantities (v_in, i_in, i_out, v_out) are expressed in root mean square (rms) units,

Input power = Pin = v_in2_/R

in = iin2Rin and Output power = Pout = vout2/RL = iout2RL

Power Gain = Pout/Pin = (v_out2_{*Rin) / (v}

in2*RL) = (iout2RL) / (iin2Rin)

Using dB units:

Power Gain_dB = 10 log₁₀(Pout/Pin) = 10 log₁₀((v_out2_{Rin) / (v}

in2RL))

= 10 log₁₀((i_out2_R

L) / (iin2Rin))

Voltage Gain = vout / vin

Current Gain = iout / in

Power Gain = Pout / Pin = (vout*iout) / (vin*iin)

In many practical applications, the source and load terminations are Rin = R_L = 50 Ω. If this is the case, the power gain in dB may be expressed as

Power Gain

= 10 log

(Pout/Pin) = 10 log

(v

/ v

in2

) = 20 log

(v

/v

)

= 10 log

(i

/ i

in2

) = 20 log

(i

/i

)

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:

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 dB

µ

V would be the

number of dB above a 1

µ

V “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

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.



Example

Given: 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 dB

µV, and (c) the corresponding current I1

through this load in dBmA.



-12 dBm = 10 log

10

(P1/1 mW) => P1 = 63.0957 µW



P1 = 63.0957 µW = V1

/ 50 Ω => V1 = 0.056167 V

V1 in dB

µV

= 20 log

10

(0.056167 / 1µV) = 94.99 dBµV

Analyzing Cascaded Transmission

Path using dB

6 dB

Volt. Gain “x2”

20 dB

Volt. Gain “x10”

40 dB

Volt. Gain “x100” + Vo -+ Vin = 20dBµV= 10 µV

-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

Vo

= Vin

+6dB+20dB+40dB = 86 dBµV

Simply add the individual dB gains of each part of a cascaded path to find the overall gain!

Vin

Example

Problem (From our textbook.)

Solution

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 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

dBV EantdBV 25.968

EantdBVVantdBV 3.522

Therefore the measured field strength at the antenna is:

dB 20 log Vant

Eant

  

 3.522

Gain of Antenna in dB = =>

Vant Eant 1.5

We are given the antenna "gain", or the ratio of the antenna's output voltage "Vant" to field strength at antenna "Eant"

dBV VantdBV 29.49

VantdBV VrxdBV 8 200 100   

VrxdBV 13.49 dBV VrxdBV 20 log Vrx

1 10 6

        V Vrx4.726 10 6 VrxdBm 10 log

Vrx2 50.0

1 10 3

         dBm VrxdBm93.5

freq = 220 MHz Class B Equipment (Residential)

CISPR22 Class B at 220 MHz = 30 dB_V/m maximum at 10 m

But our measurement was taken at 20 m, so we must use the "inverse distance rule" to convert 30 dB_V/m at 10 m to 30 - 20log(20/10) = 30 - 6 = 24 dB_V/m at 20 m.

=> FAILS CISPR22 by 2 dB

FCC Class B at 220 MHz = 46 dB_V/m at 3 m. Converting to a 20 m specification as above, 46 - 20*log(20/3) = 29.52 d_V/m at 20 m.

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.

(1) Prevents noise external to the test (on the common ac power network) from contaminating the measurement) L1 and C1.

TEM Cell Test Results, 150 V/m, CW -0.450 -0.375 -0.300 -0.225 -0.150 -0.075 0.000 0.075

0 50 100 150 200 250 300 350 400 450 MHz  V O U T Control 1 Control 2 Control 3 nc1 nc2

Original Control Sample Most Recent Control Sample

(Unlabeled)

Example 2: PCB Radiation

- Device

 Two Data Terminal Equipment (DTE) ports

(EIA RS422A, time multiplexed)

E Field through the Middle

Hole at 20 GHz