Someone explain to me how
3 db increases power by two (doubled)
6 db increases power by four (quadrupled)
9 db increases power by eight (octupled)
YET
10 db increases from 1W to 10W (factor of 10)

I'm studying for my HAM Technician license. There's not a big section on decibels and I'm getting a little conflicted that 9 db takes 1W to 8W but 10 db takes it to 10W.
10 db's take 1W and makes 10W.
20 db's take 10W to 100W
30 db's take 100W to 1Kw
and on and on.

 

I'm studying for my HAM Technician license.

Use this link!!! HamStudy.org: Technician class Ham Radio practice tests, ham radio flash cards, and ham radio question pool.
It is a "Flash Card" study program using the actual question pool and practice tests. After less than a week of using it you should Ace the exam!

And, with your electronics knowledge, you shouldn't have a problem of going right on up to the Extra Class exam using it. There are several others but I think this one is the best and used it myself to get my Extra Class ticket! I only missed 2 questions for a 96 score.

 

I'm getting a little conflicted that 9 db takes 1W to 8W but 10 db takes it to 10W.

It's just the result of the math that defines a Decibel (dB = 10log(P1/P1).
What is that confusing?

 

What you need to appreciate is that the bel is the logarithm of a power ratio.

log(2) = 0.3
log(3) = 0.48
log(4) = log(2^2) = 2log(2) = 0.6
log(5) = 0.7
log(6) = 0.78
log(7) = 0.84
log(8) = 3log(2) = 0.9
log(9) = 2log(3) = 0.95
log(10) = 1
log(100) = 2
log(1000) = 3

1 bel = 10 decibel = 10 dB

Hence a 10-times power ratio is 10 dB
100-times power ratio is 20 dB.

20 dB = 10 x 10 power ratio
30 dB = 10 x 10 x 10 power ratio

 

1 bel = 10 decibel = 10 dB

Hence a 10-times power ratio is 10 dB
100-times power ratio is 20 dB.

20 dB = 10 x 10 power ratio
30 dB = 10 x 10 x 10 power ratio

That clears up a bunch for me. It's a bit like the "Power" function (powers & roots). 10^2 = 10 x 10 = 100. It's something I've never before had the need or reason to pursue understanding decibels. I have always thought of db's as the noise I made playing rock & roll drums for 8 years and now have tinnitus because I didn't protect my hearing. Darn db's!

Use this link!!! HamStudy.org: Technician class Ham Radio practice tests, ham radio flash cards, and ham radio question pool.

Been on https://www.hamradioschool.com/technician-learning for days. Going through chapter 8.2 is where the db is briefly discussed. I will have another look later at the link you suggest. For now I'm zipping through the electrical basics questions. What got me was the decibel.

 

That clears up a bunch for me. It's a bit like the "Power" function (powers & roots). 10^2 = 10 x 10 = 100. It's something I've never before had the need or reason to pursue understanding decibels. I have always thought of db's as the noise I made playing rock & roll drums for 8 years and now have tinnitus because I didn't protect my hearing. Darn db's!

Been on https://www.hamradioschool.com/technician-learning for days. Going through chapter 8.2 is where the db is briefly discussed. I will have another look later at the link you suggest. For now I'm zipping through the electrical basics questions. What got me was the decibel.

Let's look just a bit further to perhaps help explain some other related things you might wonder about.

Your original question is how decibels relate to increases in power.

So let's define the original power as Po and the final power as k·Po. Clearly, the final power increased by a factor of k. We can certainly talk about k being 2 or 10 or 0.5 and that makes pretty obvious sense to us humans. But it turns out we often need to talk about very large and very small power ratios, particularly in RF work. Humans are as good thinking in terms of k being 0.005 or 20,000, especially if we have to combine those ratios with other ratios, which is often the case when signals are processed through a chain of elements.

So how can we express k in decibels instead of as a raw numerical scaling factor?

As noted previously, the definition of the bel is

x = log(P2/P1) B = 10·log(P2/P1) dB

If

P2 = k·P1, we have

x = 10·log(k·P1/P1) dB = 10·log(k) dB

k  x
0.000001 -60 dB (one one-millionth of the power)
0.1 -10 dB
0.5 -3.010 dB
1 0 dB
2 +3.010 dB
3 + 4.771 dB
5 +6.990 dB
8 +9.031 dB
10 +10 dB
1,000,000 +60 dB (a million times the power)

As you can see, when we say that 3 dB represents a doubling of power, that is only an approximation, but it's a pretty good one.

Now for two related questions?

Why do we like expressing power ratios in decibels?

Aside from the compression it allows, meaning that we don't have to work with huge and small numbers, something that humans aren't very good at, it lets us use addition (something we are pretty good at) instead of multiplication (something we struggle with) when combining processing elements.

Using completely made up figures off the cuff, say I've got an input signal from a microphone that is amplified to give a hundred times the power. It is then fed into an antenna system that loses half the power (only half the input power makes it into the transmitted signal). The signal is spread out in space so that, by the time it reaches the receiving antenna, only one ten-thousandth of the transmitted power is seen and only half of that survives to get to the front end amplifier, which then boost the signal to a thousand times the remaining power.

What is the output power of the receiver compared to the input power at the transmitter?

We could figure out the gain for each block and then multiply them all together, which would involve multiplying huge numbers, modest numbers, and tiny numbers together. Or, we could express each gain in decibels, giving us a bunch of modestly sized numbers, and simply add them all up.

Why is this so?

Let's say that I have several blocks, each with its own power gain. In the above description I have five blocks, so let's call them K1 through K5. The final output, P_out, relative to the initial input, P_in, is

P_out = k·P_in
P_out = K5·(K4·(K3·(K2·(K1·P_in))))
P_out = (K5·K4·K3·K2·K1)·P_in
k = K5·K4·K3·K2·K1

But if I express k in decibels, this becomes

k = 10·log(k) dB = 10·log(K5·K4·K3·K2·K1) dB
k = 10·[log(K5) + log(K4) + log(K3) + log(K2) + log(K1)] dB
k = 10·log(K5) dB + 10·log(K4) dB + 10·log(K3) dB + 10·log(K2) dB + 10·log(K1) dB

Each term above is the gain of one of the blocks, expressed in dB

Second question: Where does the factor of 20 come from when using voltages in the decibel equation?

You've almost certainly seen a similar equation

Av = 20·log(V_out/V_in)

to express the gain of a voltage amplifier in dB.

Why is it a factor of 20 and not 10, which is what is used in the definition of the decibel? After all, deci is the prefix of 1/10th.

It's because the bel is defined as a ratio of power. A voltage that is twice as large as another voltage does not represent a doubling of power, because power is related to the square of the voltage.

Assume out two voltages are applied across a resistor, R.

P_in = V_in² / R
P_out = V_out² / R

P_out/P_in = (V_out² / R) / (V_in² / R) = (V_out/V_in)²

Now, when we convert this to decibels, we have

k = 10·log(P_out/P_in)
k = 10·log((V_out/V_in)²)
k = 10·[2·log(V_out/V_in)]
k = 20·log(V_out/V_in)

 

The Tech exam is made up from a question pool of 423 multiple choice questions. The test is only 35 of those questions with a passing score of 70. I think that allows 10? questions @ 3pts each to be missed? The ARRL printed book study guides are far more in depth than the exam questions used. The online study link puts each of those questions on a "flash card" so it is easy to go through the questions. It keeps a lot of metrics such as %correct and provides random 35 question practice tests as well. There are a few questions where you will have to calculate the answer. It is an excellent test prep course! When you have gone through the test pool and can ace the practice test you are ready to take the real one. The course will also help you find a local test location and sign up for it online. It is also possible to take the test online but there are a lot of hoops to jump through to do so.

 

The other thing to note is that when you have multiple amplifiers in a system, you need to multiply the gain factors, whereas you add the dB numbers.

For example, if you have a gain of 10 followed by a gain of 2, the total gain is 10 x 2 = 20.

In dB, gain of 10 = 10 dB, gain of 2 = 3 dB, total gain is 10 + 3 = 13 dB.

 

10 db's take 1W and makes 10W.
20 db's take 10W to 100W
30 db's take 100W to 1Kw

No. I suggest you revisit this and correct it.

 

Hello,

Perhaps the attached pdfs will help.

Bertus

 

Excellent help all. Thanks.