As we progress through the fold of technological developments, be it in commercial products, military advancements, automotive electronics, or any of the other industries where electromagnetic devices are developed, EMI will be present and EMC testing will be required.

For those technical personnel who are walking into a laboratory for the first time, learning EMC because their boss said so, or are taking up an interest in the industry, a few simple yet fundamental math lessons need to be known to understand how things are calculated in an EMC laboratory environment.

This article is intended for the beginner or anybody who needs a refresher on the beautiful math of the EMC world.

**Ohm’s Law and Impedance**

Any electrical engineer knows Ohm’s Law. If not, a refund should be issued from the institution attended, because it’s literally the first thing learned in an electrical engineering class. Ohm’s law is utilized to derive power (Watts), voltage (Volts), current (Amperes), and resistance/impedance (Ohms). Two simple equations are given to find any of the above values:

*Power (Watts) = Voltage (Volts) x Current (Amperes)*

*Voltage (Volts) = Current (Amperes) x Impedance (Ohms)**

**RF, including EMC Testing, typically uses 50 Ohms as the defining impedance*

From these two equations, we can find any of the above values utilizing simple algebra. Once we know how to get from Power to Volts to Amps to Ohms or any combination in between, we can start talking about more important things we might see.

**Bring on the Decibel**

EMC testing, including limits, field strengths, transfer functions, shielding effectiveness, and a myriad of other measurements and notations are almost always noted utilizing the decibel (dB). The decibel is a way of expressing units of measurement in a logarithmic notation. This is a very useful way of expressing numbers that are increasing quickly (exponentially) as they relate to what is happening in an EMC system or during an EMC test.

In many EMC test and measurements, the results are often very large or very small, and the dB gives a very convenient way of presenting the data. If two very dissimilar numbers need to be shown on a graph, the dB is a good way of doing it.

For example, looking at power:

- 1 Watt = 30 dBm

- 10 Watts = 40 dBm

- 1 Kilowatt = 60 dBm

Seeing those values should beg the question of how they are derived. Converting between the dB values of power, voltage, and current is pretty simple. Remember that our impedance is 50 Ohms, so it’s a constant in these equations.

And you really only need to know 3:

- Power (dBm) = 10*log(power in milliwatts(mW))
- Voltage (dBuV) = 20*log(Voltage in microvolts (uV))
- Current (dBuA) = 20*log(current in microamperes(uA))

Since this is a 50 Ohm system, we also need to remember what the impedance of 50 Ohms is in logarithmic terms:

Impedance (dBOhms) = 20log(impedance) = 20log(50) = 33.979 = 34

Something to remember is that we can’t just add and subtract these logarithmic terms like we would linear terms. In fact, adding and subtracting in logarithmic terms is the same as multiplying and dividing in linear terms.

For example:

13 dBuV + 44 dBuV = 57 dBuv

If we convert these terms back to linear and perform the inverse of the Voltage equation above:

uV = 10^{(dBuv/20)}^{ }

uV = 10^{(13/20)} = 4.47 uV and uV = 10^{(44/20)} = 158.49 uV

4.47 uV * 158.49 uV = 708.45 uV

20*log(708.45) = 57.01 dBuV

Conversely, subtracting in logarithms is the same as dividing in linear:

dB1 – dB2 = Linear1/Linear2

**How to Convert Between dBuA, dBm, and dBuV**

Often enough, an engineer will need to convert between current, voltage, and power for several different reasons, whether it be converting limits, measurements, or understanding data. Simple addition and subtraction will get us wherever we want to go.

Remembering that our impedance is 50 Ohms:

Impedance (dBOhms) = 20 log (50) = 33.979 = 34

dBm = dBuV – 107 dBuV = dBm + 107

dBuV = dBuA + 34 dBuA = dBuV – 34

dBm = dBuA + 73 dBm = dBuA – 73

As a quick exercise, let’s convert 0 dBm to dBuV:

0 dBm = x dBuV

0 dBm = 0.001 Watts or 1 mW

0 dBm = 0.224V

0.001W = (Volts^{2})/50 Ohms

0.001W x 50 Ohms = Volts^{2}

Sqrt(0.001Wx50Ohms) = Volts = 0.224V = 0.224×10^{6} uV Volts

20 log (0.224×10^{6 }uV) = 106.987 dBuV

**Decades and Octaves**

Decades and octaves are typically used in frequency ranges, which are found throughout the EMC sphere. A good understanding of these will enable you to read limits and graphs when looking at data.

An octave is defined as a doubling or halving of a value. So if we are at 1 GHz, an increase of 1 octave is 2 GHz. an octave above 2 GHz is 4 GHz, and so on.

A decade is defined as ten times (or a tenth) of any value. If we are at 1 GHz, an increase of 1 decade is 10 GHz. A decade above that is 100 GHz and so on.

Limits will sometimes be expressed in terms of decades. For example, an emissions limit might be expressed as “starting at 10 kHz and increasing 10 dB per decade to 1 GHz.”

**A dB is a dB**

When you convert linear terms to logarithmic terms (dB), you’ve created a situation where all of the units are relatable. This is important to note because as power levels increase and voltage levels increase, an engineer might be interested to know how much power it takes to get to a certain field level.

To increase an octave in Voltage (or field level), you need to add 6 dB. To increase an octave in power, you need to add 3 dB. And this is where this knowledge becomes important.

If we have a field level in a test of 100 V/m, and we need to increase that to 200 V/m, we need to add 6 dB. Well, that goes for our power also. We need to increase the power by 6 dB coming from our test system, mainly our amplifier.

So, for example, let’s say we need 150 Watts (51.76 dBm) to reach 100 V/m (163.52 dBuB/m) in our system. Since we need to add 6 dB to the field, we also need to add 6 dB to the amplifier, so we need 51.76 + 6 = 57.76 dBm. From what we’ve learned so far, that’s:

10^{(57.76/10)} = 597,035 mW = 597 Watts, which is a big difference from 150 Watts.

**Story Time**

Given everything above, and with a little practice, the simple math of EMC can be learned and applied in different ways. During the author’s time in the laboratory, a customer thought they knew ways to get their device to “pass” by applying some of the math above, albeit incorrectly.

While testing CE101, the measurement was 3 dB above the limit for current harmonics. All measurement system integrities were verified, and the author was able to see the measurement on an analyzer. The customer representative asked them to “subtract out our ambient” since the laboratory could have been noisy. While the author and his colleagues agreed, they reminded the customer that actually subtracting the dBuA levels was inappropriate and the numbers MUST be converted to linear terms.

The limit level was 100 dBuA, and the measurement was approximately 103.50 dBuA. The noise floor of the analyzer when everything was connected was about 73 dBuA.

So, they showed the customer the following math.

Measurement → 10^{(103.5/20)} = 149,623.56 uA

Ambient → 10^{(73/20)} = 4,466.84 uA

Subtracting → 149,623.56 uA – 4,466.84 uA = 145,156.72 uA

Go back to dB → 20log(145,156.72) = 103.24

STILL 3 dB ABOVE THE LIMIT

So, just remember how to apply things correctly, and you’ll have success in any EMC laboratory you may walk into.