Below is the “Math for electronics” section of the 2014 edition of the No-Nonsense Technician Class License Study Guide. As always, comments welcome…Dan
When dealing with electrical parameters, such as voltage, resistance, current, and power, we use a set of prefixes to denote various orders of magnitude:
- milli- is the prefix we use to denote 1 one-thousandth of a quantity. A milliampere, for example, is 1 one-thousandth of an ampere, or .001 A. Often, the letter m is used instead of the prefix milli-. 1 milliampere is, therefore, 1 mA.
- micro- is the prefix we use to denote 1 millionth of a quantity. A microvolt, for example, is 1 millionth of a volt, or .000001 V. Often you will see the Greek letter mu, or μ, to denote the prefix micro-. 1 microvolt is, therefore, 1 μV.
- pico- is the prefix we use to denote 1 trillionth of a quantity. A picovolt is 1 trillionth of a volt, or .000001 μV.
- kilo- is the prefix we use to denote 1 thousand of a quantity. A kilovolt, for example, is 1000 volts. Often, the letter k is used instead of the prefix kilo-. 1 kilovolt is, therefore, 1 kV.
- mega- is the prefix we use to denote 1 million of a quantity. A megahertz, for example, is 1 million Hertz. The unit of frequency is the Hertz. (T5C05) It is equal to one cycle per second. Often, the letter M is used instead of the prefix mega-. 1 megahertz is, therefore, 1 MHz.
Here are some examples:
- 1,500 milliamperes is 1.5 amperes. (T5B01)
- Another way to specify a radio signal frequency of 1,500,000 hertz is 1500 kHz.
(T5B02) - One thousand volts are equal to one kilovolt. (T5B03)
- One one-millionth of a volt is equal to one microvolt. (T5B04)
- If an ammeter calibrated in amperes is used to measure a 3000-milliampere current,
the reading it would show would be 3 amperes. (T5B06) - If a frequency readout calibrated in megahertz shows a reading of 3.525 MHz, it would
show 3525 kHz if it were calibrated in kilohertz. (T5B07) - 1 microfarad is 1,000,000 picofarads. (T5B08) (Farad is the unit for capacitance.)
- 28.400 MHz is equal to 28,400 kHz. (T5B12)
- If a frequency readout shows a reading of 2425 MHz, the frequency in GHz is 2.425 GHz. (T5B13)
When dealing with ratios—especially power ratios—we often use decibels (dB). The reason for this is that the decibel scale is a logarithmic scale, meaning that we can talk about large ratios with relatively small numbers. At this point, you don’t need to know the formula used to calculate the ratio in dB, but keep in mind the following values:
- 3 dB is the approximate amount of change, measured in decibels (dB), of a power increase from 5 watts to 10 watts. (T5B09) This is a ratio of 2 to 1.
- –6 dB is the approximate amount of change, measured in decibels (dB), of a power decrease from 12 watts to 3 watts. (T5B10) This is a ratio of 4 to 1.
- 10 dB is the approximate amount of change (actually it is the EXACT amount of change), measured in decibels (dB), of a power increase from 20 watts to 200 watts. (T5B11) This is a ratio of 10 to 1.
David, VE7EZM/AF7BZ says
Dan, you state “At this point, you don’t need to know the formula used to calculate the ratio in dB” and that’s fine, but I hope later in the new edition you do give the definition. And here, a power increase by a factor of 10 is EXACTLY 10 dB, not just approximately, so maybe you should give that example first without using the word “approximate” and the other two examples afterwards, using the word. Put in one with a factor of 100, or 20 dB, too before you give the approximate ones for 3 dB etc. That way some alert readers who don’t already know what the definition of dB is will enjoy having been correct when, later on, they find out that their guess was in fact correct. Having taught math for many years I know that students like it when you set things up so that before you tell them, they guess, and their guesses turn out to be right. That seems to make them willing to work harder when it gets more complicated. Carrots, not sticks, etc.
Dan KB6NU says
Thanks for your comments, David. Sorry about the troubles with entering this comment. I’d like to remove that captcha altogether, but for some reason my site gets bombarded with so much spam that I have to leave it in.
You’re right about 10 dB being the exact amount of change, but that’s not the way that the question is worded, and I like to stay as close to the wording of the question as possible. What I will do is mention this to the question pool committee and get them to change the question. Perhaps you could do that as well. You can do this by going to the NCVEC feedback page. If they don’t change the question, I can add in parentheses that it’s EXACTLY 10 dB.
While normally I might agree with you about adding other examples, one of the principles behind this study guide is to only include material that is covered on the test. That’s the reason I also disagree about including the formula for calculating dB. I think that adding that formula will just confuse most people.
Thanks again for your comments.
Neil Fountaine says
I am studying for the Ham Technical exam. 1 question in the online practice is –6 dB is the approximate amount of change, measured in decibels (dB), of a power decrease from 12 watts to 3 watts. (T5B10) This is a ratio of 4 to 1. I am having a problem in how to calculate the -6 db without a calculator. Can you help? Thank you
Dan KB6NU says
The formula for calculating the power ratio in dB is: dB = 10log(P2/P1). No need to calculate it, though. Just remember that a power ratio of 4 is 6 dB. If P2 > P1, meaning that the signal has been amplified, then it’s +6dB. If P1 > P2, meaning that the signal has been attenuated, then it’s -6dB.