dB (decibel)

[meowth08]

for voltage:

20*log (v1/v0)

for power:

10 *log(p1/p0)

how was 10 for power and 20 for voltage derived?

thanks in advance

m8

 

[JimB]

The basic unit is the "Bel", which is rather large for practical use.
So, divide the Bel by 10 and get the deci (10th) Bel.
So there are 10 deciBels in a Bel, and that is where the 10 comes from.

The Bel is a logarithmic ratio of two powers.
If we measure the powers directly, we can just put the numbers into the 10xLog(P/p) equation.

It is often more convenient to measure voltage rather than power.
Power is related to the square of the voltage (W = V^2 /R).
When working with logarithms to find the square of a number, just multiply the log of the number by two.

So we are squaring the voltage so the decibel equation becomes 2 x 10 x Log(V/v) = 20 Log(V/v).

Easy when you know!

A caveat.
When measuring voltage and calculating dB, the impedance at the input and output of the circuit should be the same, otherwise the power relationship is not correct.
Sometimes a gain of an amplifier will be expressed in dB when the input and output impedances are not the same.
This is not a correct thing to do, but it is done.

JimB

 

[meowth08]

I GOT IT!!! Thank you.

	y=log(x)^n
	y=nlog (x)

where:
	'a' is the base
	'n' is the exponent

when: n=2

then: y=2log (x)

from here:

	W=V^2/R

for power:

	10log(P/p)

	P/p=W/w=(V^2/R)/(v^2/R)

then it becomes:

	(V^2)/(v^2) or (V/v)^2


10log(V/v)^2 or 2*10log(V/v)

 

[meowth08]

A caveat.
When measuring voltage and calculating dB, the impedance at the input and output of the circuit should be the same, otherwise the power relationship is not correct.
Sometimes a gain of an amplifier will be expressed in dB when the input and output impedances are not the same.
This is not a correct thing to do, but it is done.

JimB

Thanks for the information.
Suppose the input and output impedances are not equal, what should be done to get the correct power relationship?

after a while:

some amplifiers are cascaded, is this applicable to single amplifiers only?

m8

 

[JimB]

Suppose the input and output impedances are not equal, what should be done to get the correct power relationship?

Find the input and output impedances, measure the voltages, calculate the input and output powers (W= V^2 /R), and the calculate the dBs using 10 x Log(P/p).

some amplifiers are cascaded, is this applicable to single amplifiers only?

I am not completely sure what you are asking here.
If you have two amplifiers, one with a gain of 14dB and one with a gain of 10dB, the overall gain of the cascaded pair is the sum of the individual gains in dB, in this case 24dB.

JimB

 

[meowth08]

I am not completely sure what you are asking here.
If you have two amplifiers, one with a gain of 14dB and one with a gain of 10dB, the overall gain of the cascaded pair is the sum of the individual gains in dB, in this case 24dB.

JimB

In the schematic shown on the website below, can I just use 10log(P/p) after getting powers on the input and output?

https://www.hobbyprojects.com/junction_transistors/amplifiers_in_cascade.html

m8

 

[JimB]

Yes.

JimB