Decibels and how they relate to Voltage

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large_ghostman

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i was wondering as my (ok dads) function generators outputs in several formats, you can use mV and V and Decibels (and some others), i always think of Decibels as relating to sound, but things like Oscilloscopes and such have attenuation (not sure what that is) normally marked in Decibels, so how does it relate to voltage and electronics in general?
 

I don't think I've ever seen a scope marked in dB's?.

Decibels are a ratio - so are meaningless without knowing the reference point it's related to.

Voltage dB's are different to power dB's as well - have a read here:

https://en.wikipedia.org/wiki/Decibel
 
ok so when i press the 30db attenuation button on the FG what two values is it the ratio of?
 
ok so when i press the 30db attenuation button on the FG what two values is it the ratio of?

I'm more familiar with db relating to power, but 10db = one order of madnitude, so 30db attenuation would divide the signal by 1000.

having checked the wiki reference I'll amend that to say the power is divided by 1000, so the voltage is divided by 31ish (power =V^2/R).
 
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The output now to the Max output
I think it is comparing to the telephone reference level.
I agree it is a comparison of two levels. Unless other wise stated, and no reference is given, I think 0db is what the phone company calls 0.
 
From my college days I think 0db in audio terms is 775mV into a 600ohm load giving a power of 1mW.
Commercial power amps tend to be rated for full power with a 775mV i/p, so you'd be able to calculate the o/p power from the attenuator, if you trust the method that is.
 
The equation for voltage is:

dB = 20 * Log (Vout / Vin)

Power is DIFFERENT. A calculator is here:

https://www.crownaudio.com/db-volts.htm

-30 db is a reduction of 31.6, or divide the original output by 31.6

There are all sorts of reference values, like dbm, dbu etc. They fix a reference value. You do need to know if you are dealing with power or voltage.

The cool thing about db is that they add or subtract rather than multiply.

An application where it's useful is say designing a MATV distribution system. Each connection has a loss associated with it, each tap/splitter has a loss, the cable has a loss in db/100 ft. Thus you don't have to do messy math.

Log is log base 10. so LOG(10) is 1, log (100) is 2, log (1000) is 3; so it's just the exponent e, of n^e. e.g. 10^1, 10^2, 10^3.
Yep, exponents of 1.2 exists and so do exponents of 1/2. 10^(1/2) is the sqrt(10).

LOGs can be of any base and Ln is the natural logarithm, which has an irrational base.

LOG you may see written as LOG() and LOG10()

So, this is the quick lesson.
 
In summary, a dB is the logarithmic ratio of two power levels calculated as 10LOG (P2/P1). The power can be in any form -- sound, electrical, light etc. Note that the ratio is about power, not voltage. But you can use voltage to express the two power levels if the two values are measured at the same impedance. Because the power is proportional to V squared then the calculation is 20LOG(V2/V1) (since multiplying a LOG value by 2 is the equivalent of squaring the number). Note that, even though it is not technically correct, it is common to calculate the voltage dB even when the two impedances are not the same (such as an audio amplifier voltage gain from input to output).
 
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Decibels are used frequently to show the voltage gain of an amplifier or an opamp and the input and output impedances are COMPLETELY different.
A voltage gain of 10 times is +20db. 100 times is +40dB. 1000 times is +60dB etc.
Some opamps have a DC and low frequency voltage gain of 120dB which is 1 million times.
 
Just about everything has been said already, but I want to give you this link to a great online book https://www.dspguide.com/, and what is says about desibels:

"A bel (in honor of Alexander Graham Bell) means that the power is changed by afactor of ten. For example, an electronic circuit that has 3 bels of amplification produces an output signal with 10 × 10 × 10 = 1000 times the power of the input. A decibel (dB) is one-tenth of a bel. Therefore, the decibel values of: -20dB, -10dB, 0dB, 10dB & 20dB, mean the power ratios: 0.01, 0.1, 1, 10, & 100, respectively. In other words, every ten decibels mean that the power has changed by a factor of ten.

Here's the catch: you usually want to work with a signal's amplitude, not its power. For example, imagine an amplifier with 20dB of gain. By definition, this means that the power in the signal has increased by a factor of 100. Since amplitude is proportional to the square-root of power, the amplitude of the output is 10 times the amplitude of the input. While 20dB means a factor of 100 in power, it only means a factor of 10 in amplitude. Every twenty decibels mean that the amplitude has changed by a factor of ten.

Since decibels are a way of expressing the ratio between two signals, they are ideal for describing the gain of a system, i.e., the ratio between the output and the input signal. However, engineers also use decibels to specify the amplitude (or power) of a single signal, by referencing it to some standard. For example, the term: dBV means that the signal is being referenced to a 1 volt rms signal. Likewise, dBm indicates a reference signal producing 1 mW into a 600 ohms load (about 0.78 volts rms)."

Direct link to that chapter: https://www.dspguide.com/ch14/1.htm
 
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Most of what has been said already is correct, but let me explain one odd little trap which could catch the unwary.

dBm is decibels with respect to 1mW, so 0dBm is 1mW, +10dBm is 10mW and -6dBm is 0.25mW.
So far so good.
In "audio" circuits the characteristic impedance is usually 600 Ohms and so 1mW corresponds to a voltage of 0.775 volts.

However, RF circuits usually have a characteristic impedance of 50 Ohms. In this case 1mW corresponds to a voltage of 0.223 volts.

JimB
 
thanks for the links and explanation's i am slowly getting my head around it! was a bit hard at first as i have always thought of decibels relating to sound and nothing else! but i guess it's the kind of thing it's better for me to learn about now , funny thing is i went to school today and ask my science teacher to explain it talk about being fobbed off! he could only realy relate it to sound then had the nerve to try and tell me that although it applied to other things, sound was really the only thing it was relevant to! thanks to you all i have a better understanding and maybe once i have it securely nailed down in my brain i might educate my poor science teacher in the hope that if anyone else asks him to explain it he will give a more accurate answer i am only joking i would never dream of correcting a teacher again they kind of give you a funny look and when you mention internet experts in electronics they give you the other look adults have so when i have questions i now ask on here otherwise i am going to end up thanks again ALL

LG (he who knows too little and asks too much)
 
I think that all signals "live forever" they just attenuate as they "go" When you play music, or shout out loud, the sound goes on forever. But, at some point it has attenuated enough so that it becomes unrecognizable. It becomes part of the noise that is everywhere. That happens to all signals.. light, radio waves etc. The signal never "goes flat" it just attenuates forever.
 
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on the one hand that sounds daft as logic would say sound or any other signal would eventually just diminish to nothing. but from what i have read things change but never actually go, so there is as many atoms in the universe now as there ever was! was really is hard to get the head around
 
Hi,

One of the reasons for the decibel is that in making certain measurements not much changes over small ratios like 1.5, 2, 3, etc., and large ratios are hard to write down like 1000, 10000, 100000, etc. With the decibel it makes it possible to understand the relative importance of the change (ratio in this case) by looking at the numbers and comparing them like we would ordinary numbers, even though they are calculated differently then we usually do.

For example, a voltage gain of 10 in decibels is 20, while a gain of 100 is 40, and a gain of 1000 is 60, and a gain of 10000 is 80, and 100000 is 100. So now lets look at the original numbers first as a group. We have:
10, 100, 1000, 10000, 100000.

What does that tell us? That tells us that the gain changed quite a bit from 10 to 100000, in fact it looks like it changed a REAL lot because 100000 is much much greater than 10.

Now lets look at the decibels we calculated above as a group:
20, 40, 60, 80, 100

What does this tell us? This tells us that the gain did increase from one to the next, but it did not change by an exceeding large amount. Yes it did change, but comparing the two ends we see 20 and 100, which looks like a much smaller change than before. And that's for a good reason, because the human ear detects changes differently than the absolute numbers we first looked at, the ear detects changes more like changes expressed in decibels. So that's one reason why we started using decibels.
The other reason is because if you look at the two lists, the absolute numbers and the decibels, you'll see that writing out the ratios in decibels is a little easier than writing out those numbers with all the zeros following the "1". So we adopt the decibel again, and in this case it is similar to just showing the power of a common base rather than the whole number.

Of course another reason for using decibels today is because the decibel appears in much of the literature on audio and other applications. To understand what the author is talking about we have to know decibels.

BTW, for calculators/programs that do not have the log base 10 function, you can calculate it from the log base 'e' function also written as "ln(x)" with:
log10(x)=ln(x)/ln(10)

and that ln(10) is a constant so you only have to calculate it once and then store it.
 
And here is cool use of that:

Suppose we want to do a binary search and we have 2047 elements. We would like to know the next power of 2 which is 2048. We take the ln(2047)/ln(2) and get 10.998xxx.

If we up that to the nearest Int(10.98)+0.5) we get 11

2^11 is 2048

Now we can split at 1024 and do the binary search.
 
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