Decibel Equation

[trem]

OK.....I need help with this. I am sure it is right, but I do not understand the math.
Man, I wish I knew how top type.....

The equation in question is:
dB=20 x log10 V2/V1
the example given is:
If V2 = 10V and V1 = 1V then ..... logV2/V1 = 1 and the difference between the two voltages is 20dB

another example given is:
V2 = 100V and V1 = 1V, the difference in dB is 40 (20x2). So, the difference in dB between 1V and 10V is the same as the difference between 10V and 100V (40dB - 20dB = 20dB)

I am completely lost.
1. The equation says log10 V2/V1, but then I do not see that appear again in the example, it is just logV2/V1.

2. how does log10/1 equal 1.?

Do you guys understand my confusion.?
Thank You

[g1]

I'll try to make a couple points, and I hope if you already knew them you don't take offence.
Your first quote of the equation included the 20 (used for voltage):
"dB=20 x log10 V2/V1". The extra 10 in there should be omitted.
Proper formula is db=20 x log (V2/V1).
Your second mention of the equation omitted the 20:
"The equation says log10 V2/V1".

So does that clear anything up?
Secondly, do you get what a logarithmic proportion is? It's non linear, so 20db would be voltage increase tenfold, 40db would be one hundred fold, 60db would be one thousand times.

For power, the multiplier is 10 rather than 20.
So the formula for power is dB=10 x log (P1/P2).

[trem]

Please....I am more likely to offed you guys with my Questions/Ignorance.
I should have broken this up into sections of ignorance.
I have no idea what that "20" represents.
My limited math skills see it as 20 x log 10 x V2/V1.
V2/V1 = 10/1 which is 10.
Then multiply whatever the heck 20 x log10 is by 10....and the example comes up with 1.?


Now do you feel offended.? I have no idea what any of this is.
For starts, my basic Algebra must be all wrong.
Can you see how much of this I do not understand now.?

Probably putting the cart before the horse at this point, but no, I do not know what a logarithmic proportion is.
I understand what is meant by a non-linear equation though.......
Thank You

[g1]

I edited my post as I found a few more errors in your data .
Please try again.

[Shadrock2]

Decibels

Well.. I have some difficulty thinking about decibels myself.
But I can explain logarithm's.
Logarithm's are exponents as in 10 raised to the 2nd power (i.e. 10^2) is 100. The logarithm of 100 is expressed formally as log10 100 = 2 and, since humans have 10 fingers we developed a number system based on 10 (decimal system), the logarithm of base 10 is assumed if one just uses the term "log". If using other base numbers you would specify. As in:
log5 25 = 2 (i.e. 5^2) log5 125 = 3 (i.e. 5^3) log2 16 = 4 (i.e. 2^4 or 2*2*2*2).

Other factoids of interest. log(any number) 1 = 0 (stated another way... any number raised to zero power = 1)

log (1/10) = -1 log (1/100) = -2 log 3.16 = approx. 0.5 (i.e. sqr rt of 10 is approx. 3.16 = 10^0.5)

That is all the math I can teach at the moment. hope that helps

[trem]

OK...I see. I had no idea what any of that meant. I get it now.
I thought there was just ONE thing I did not understand.
Thank You

[eschertron]

The equation in question is:
dB=20 x log10 V2/V1

Many scientific calculators, and my excel spreadsheet (which is how I found out!) will default to natural logarithms if the 'base 10' is not explicitly chosen. The 10 in the equation shows which log table the calculator is using. That's all I know about logarithms and that's all I want to know

Because of the way the decibel scale was created, one Bel (10 decibels) represents an increase of sound intensity of a factor of ten. Bear with me for a second... I'm talking right now about intensity (power), expressed in Watts. That equation is dB = 10 * log10(p2/p1) So, for example, 2W is bigger than 1W by 3.01dB (thumbrule for engineers, call it 3dB). But because power is proportional to voltage squared, the equation we use has a 20* when using voltage relationships.

So if we are looking at voltage gain, 2Vrms is bigger than 1Vrms by 6.02dB, as the power carried by the increased voltage is 2 squared, or 4 times. JM2C

[Shadrock2]

"Many scientific calculators, and my excel spreadsheet (which is how I found out!) will default to natural logarithms if the 'base 10' is not explicitly chosen."

When a "natural" logarithm is used it is represented by "ln" as in ln e =1 My character set doesn't quite represent these things properly. A natural logarithm is one of base e (greek epsilon) and e is approximately equal to 2.7 something. This number has some elegant significance when using calculus. Otherwise the representation is the same. But if using log by itself it can be assumed to be base 10.

My calculator has both a "log" and a "ln" button.

[trem]

Yes.....Thank You as well. I did not understand the ratio that was going on, or the use of the word log.
But I do get it now, and that brings the following Equations/Formulas/Problems within my reach of understanding.
Thanks Again

[g1]

I had also assumed the base 10 to be understood.
Having it typed out in line makes it confusing. Seeing it with the 10 lower than the word log takes away the confusion, but we can't type it out like that .

[trem]

I had also assumed the base 10 to be understood.
Having it typed out in line makes it confusing. Seeing it with the 10 lower than the word log takes away the confusion, but we can't type it out like that .
[ATTACH=CONFIG]33579[/ATTACH]

Yeah, definitely.
But I was, probably, a bit less than forthcoming with my lack of math knowledge. It can be embarrassing.....but I barely graduated high school, and only have a grasp of the most basic of Algebra formulas.
What I had meant to say in my OP was.......I have absolutely No Idea what any of this means, but want to be able to do the calculations, can one of you guys please take me through it a step at a time, or direct me to a good math link on The Internet that will walk me through this.?
That is what I SHOULD have said.
Sorry.......

[R.G.]

A slippery concept is tied up in that "times ten for power ratios, times 20 for voltage ratios" stuff.

Decibels are fundamentally a *power* ratio statement. It is accurate to speak about one power output being 23db bigger than another power output. It is natural to guys who work with signal voltages more than signal power to want to state signal voltages in terms of db ratios, too, and the math does work out right that if you have db = 10* log(pwr1/pwr2), then the ratio of the voltages can be expressed as db = 20* log (voltage1/voltage2), and all of this is consistent as long as the two voltages are driving the same load impedance.

It's a mathematical misstatement to compare volts driving one impedance with volts driving another impedance by referring to them as dbv - it doesn't have any real physical significance. Saying one voltage is 43db bigger than another without also saying what impedances they drive or better yet that the impedances are equal are much the same as painting on a yellow sign "DANGER!! 50,000 OHMS!!".

This doesn't keep all of us from committing the sin repeatedly, but it's always better if you know you're sinning than if you do it unknowingly, right?

[überfuzz]

A slippery concept is tied up in that "times ten for power ratios, times 20 for voltage ratios" stuff.

Decibels are fundamentally a *power* ratio statement. It is accurate to speak about one power output being 23db bigger than another power output. It is natural to guys who work with signal voltages more than signal power to want to state signal voltages in terms of db ratios, too, and the math does work out right that if you have db = 10* log(pwr1/pwr2), then the ratio of the voltages can be expressed as db = 20* log (voltage1/voltage2), and all of this is consistent as long as the two voltages are driving the same load impedance.

It's a mathematical misstatement to compare volts driving one impedance with volts driving another impedance by referring to them as dbv - it doesn't have any real physical significance. Saying one voltage is 43db bigger than another without also saying what impedances they drive or better yet that the impedances are equal are much the same as painting on a yellow sign "DANGER!! 50,000 OHMS!!".

This doesn't keep all of us from committing the sin repeatedly, but it's always better if you know you're sinning than if you do it unknowingly, right?

You're right, the concept of decibel is easier to use properly if we consider an amplifier to be a Thévenin source. And like you suggested, math, or physics, 'allows' us to set up transfer-functions for parts of an amplifier thus making the term it a bit more tricky.

[Shadrock2]

Eschertron and R.G. get to the main aspect of decibels that makes my head hurt. Thank you both for helping clarify that.

There is one more thing about decibels that I find frustrating. This is the use of decibels when talking about negative feed back. Many times you see an author writing "I use 20 db of negative feedback...". That seems impossible to me as the nfb signal would would be 10 times the original signal and completely swamp the source. Further, with tubes, you tap off the transformer secondary (low impedance source) and feed a high impedance driver stage. Should you use the 10X power formula or the 20X voltage formula.

Some enlightenment on this would be helpful.

[R.G.]

Eschertron and R.G. get to the main aspect of decibels that makes my head hurt. Thank you both for helping clarify that.

There is one more thing about decibels that I find frustrating. This is the use of decibels when talking about negative feed back. Many times you see an author writing "I use 20 db of negative feedback...". That seems impossible to me as the nfb signal would would be 10 times the original signal and completely swamp the source. Further, with tubes, you tap off the transformer secondary (low impedance source) and feed a high impedance driver stage. Should you use the 10X power formula or the 20X voltage formula.

Some enlightenment on this would be helpful.

Re: use of "db" in stating negative feedback.
In the world of feedback amplifiers, the not-quite-correct use of "db" to mean "voltage ratio" is almost universal. So whenever you hear of an amplifier's voltage gain in db, or feedback in db, it's voltage. The proper engineering world actually uses "dbv" to mean "voltage ratio in db" and "dbw" to mean power ratio in watts (or ratio compared to 1W; it varies). You'll see "dbm" which means "metered voltage relative to 1mW". The proper scientific world generally deplores the slew of db-whatsits.

But we're stuck with them in talking about feedback amplifiers. In this case, "db" means "voltage ratio", with the 20x multiplier. So an amplifier with 40db gain has a voltage gain from input to output of 100 (that is, 40/20 = 2, and ten to the second power is 100).

The db of feedback is most simply thought of as the ratio of the gain of the amplifier without feedback divided by the gain of the amplifier with feedback, then with the ratio expressed in db (volts in most cases). If you take an amplifier with a gain of 100 (i.e. 40db), and use feedback to reduce the gain to 10 (i.e.20db), then the db of feedback is 100/10 expressed as a db, or 20db.

If you have an ampifier with 63db open loop gain, and reduce its gain to 20db, then the db of feedback is its open loop gain ( ten to the power of 63/20 ) divided by the closed loop gain (ten to the power of 20/20 ) expressed as db, or 1412.5/10 = 141.25; log 141.25 = 2.15, times 20 = 43. So this amplifier has 43db of negative feedback.

You'll notice that this looks identical to subtracting the gain after feedback from the open loop gain; this is a good approximation that's not always perfectly accurate, but is good where the open loop gain is "big".

So the phrase "used 20db of negative feedback" most closely translates to "reduced the open loop gain by a factor of ten to get the final closed loop gain".