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understanding decibel

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sathiyanarayananL

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hi,

can anyone explain or share web link about decibel? I cant able to understand it clearly. also what to know about 3db line mentioned in all graphs.

thank you
 

bel is a factor of 10 (it is a log scale). decidel is 10th of a bel. 0 decibel = 0 bel = 1 (no amplification)
 

Many opamps have a voltage gain at low frequencies without negative feedback of 2,000,000 times which is 126dB. It is much easier writing 126 instead of 2,000,000.
 

To pass a magnitude to decibels for example the voltage gain as Audioguru said, is used this formula: Av (dB)= 20·log|Av|. (for intensity gain too). But power gain is defined like this: Ap (dB)=10·log|Ap| (those "20" from Av come from changing the power gain in the "10" formula and get the voltage gain).

As they said is just another form to write easier things.
 

To pass a magnitude to decibels for example the voltage gain as Audioguru said, is used this formula: Av (dB)= 20·log|Av|. (for intensity gain too). But power gain is defined like this: Ap (dB)=10·log|Ap| (those "20" from Av come from changing the power gain in the "10" formula and get the voltage gain).

For voltage gain, the formula to be used is db(gain)=10 log (voltage gain); this much is simple. Sometimes we use power gain and the formula to be used is db(power)=20 log (power gain). Confusing?

Power is quadratic in voltage or current; P is proportional to v^2 or i^2 or is simply v.i; after taking the log, the power 2 becomes multiplicative.

Clear enough? Not quite. logarithm can be taken only of a number and log (100V) is meaningless. Gain is a pure dimensionless number and is suitable for logarithm. Not quite do for power.

For power, we need a reference power source: db(power)= 20 log ((power indicated)/(reference power)). For power gain we must use an unit: db (power) = 35 (mW reference). This is often indicated in various fashions.

For an opamp with a voltage gain of 2 000 000, we have a db (gain) of 63. This is equivalent to decibel 63 = bel 6.3; antilog 6.3 = 2 *10^6 =2 000 000
 

For voltage gain, the formula to be used is db(gain)=10 log (voltage gain); this much is simple. Sometimes we use power gain and the formula to be used is db(power)=20 log (power gain). Confusing?

For an opamp with a voltage gain of 2 000 000, we have a db (gain) of 63. This is equivalent to decibel 63 = bel 6.3; antilog 6.3 = 2 *10^6 =2 000 000

I am not following you...

Where did you see that Voltage Gain (dB)=10·log| Voltage gain | ?
 

Where did you see that Voltage Gain (dB)=10·log| Voltage gain | ?

chuckey has given a decent reference: https://en.wikipedia.org/wiki/Decibel

You need to be careful whether the intended reference is a power (decibel number will be double) or a potential (you need to use a multiplier of 10).

The decibel is also used in acoustics and other areas. I was wrong above: It should have been db(power)=10 log (power gain)= 20 log (voltage gain)

Simple gains are ratios and are a pure number. But the rules are not followed consistently.
 

Many opamps have a voltage gain at low frequencies without negative feedback of 2,000,000 times which is 126dB. It is much easier writing 126 instead of 2,000,000.

:lol: It is easier writing a gain of 2M than a gain of 126dB :lol:
 

Hi,

8 replies, but no comment from the OP...Are you satisfued with the answers or do you need more information?

3dB line:
Imagine a low pass RC filter.
From left to right you see a line.
At the beginning it is straight and horizontal, then around cutoff frequency it is bent, after that it is straight with a constant slope.

Now extend the horizontal line to the right, and the line with the constant slope to the left.
They cross each other at cutoff frequency. And you can see the gain of the bent line at this frequency is -3dB.

Klaus

- - - Updated - - -

Added.

While a simple RC has an exact rate of 20.000dB/decade....
The commonly known 3dB are axactely 3,010299956dB.

Klaus
 

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