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Old 16th April 2004, 12:49 AM   #1
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Default dB into voltage gain????

what is the formula for conversion???

What is 12dB of gain?

Also.. what is a good gain to go for in a pre-amp?

Thanks!!!
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Old 16th April 2004, 01:19 AM   #2
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A little db tutorial:
Voltage Gain (dB) = 20 log (Volt. Gain)
ex: Vin = 0.1V, Vout = 1.0V, Vgain = 1/0.1=10,
Vgain (db) = 20 log 10 = 20

dB are logaritmic relational unity. There is a lot of different dB relations, here just a few:

dBW=dBWatt, dB difference referenced to 1W, 1W=0dBW, 10W=10dBW, 100W=20dBW equation: dBW = 10 log (x/1W)
Used often to express speaker sensitivity. Just a note:
In power measurements, 3dB is double. So a speaker of 90dBW is two time more sensitive than one of 87 dBW!
In voltage measurements, 6dB is double.

dBm = dBmilliwatt, dB difference referenced to 1mW. Also the impedance of the system where the power is measured is often mentionned. For example 0dBm in 600 or 50 ohms. Often used for small signal, example phone line level, receiver maximum input, RF measurements, etc...

dBuV = dBmicro-Volt, often used to express radio receiver sensitivity.

dBA dBB or dBC (also called dB A-Weighted, etc...): you see them in acoustic measurements. Refer to Sound Pressure Level measured using different response filters A, B & C.
A-weighting takes into account the typical human ear response, reducing the frequency response of the SPL meter to the 500 - 10,000 Hz range, where our ears are mostly sensitive, while the B-weighting measures the whole range from 32 to 10,000 Hz. The C is like the B but with peak response.
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Old 16th April 2004, 01:30 AM   #3
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Quote:
What is 12dB of gain?
About 4X

Wayne
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Old 16th April 2004, 04:14 AM   #4
billr is offline billr  New Zealand
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you may have noticed that the voltage and power db expressions are different for the same ratio, ie. doubling power gives a 3db gain, whereas doubling voltage gives rise to a 6db gain

some find this confusing, allow me to explain.

power db ratio is given as 10log(p1/p2)

p = V*V/R [this is a clumsy way of saying V squared divided by R]

therefore the new ratio =

10log (V1*V1/R)/(V2*V2/R)

the R's cancel out giving

10log(V1*V1)/(V2*V2)

or put another way 10log(V1/V2)*(V1/V2) or in english

ten log (v1 divided by v2) squared

using the rules of logarithms, this can be rewritted as

20log(V1/V2)

hence the voltage doubling factor being shown as 6db.

if you already knew this sorry to teach you to suck eggs.

kind regards

bill
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Old 17th April 2004, 04:20 AM   #5
hermanv is offline hermanv  United States
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Quote:
What is formula to convert dB into Voltage ratio?
You find the antilog (base 10) of the ratio divided by 20 (Voltage) or by 10 (power).

Voltage ratio = 10^(dB/20)

Power ratio = 10^(dB/10)

Why 10 for power and 20 for Voltage? When you double the Voltage in a circuit you get four times the power. Or put another way to double the power, multiple the Voltage by 1.414.
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Old 17th April 2004, 04:50 AM   #6
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Quote:
Originally posted by hermanv
Why 10 for power and 20 for Voltage? When you double the Voltage in a circuit you get four times the power. Or put another way to double the power, multiple the Voltage by 1.414.
Think in terms of the difference.

sqrt(2)*voltage = double the power

Double the power vs. Double the voltage.

Double the voltage = 4x the power

4x the power = 2x double the power

So doubling the voltage is the same as doubling the power twice hence 20 in the formula instead of 10 for power ratio.
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Old 17th April 2004, 01:25 PM   #7
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If a data sheet say a certain opamp has an open loop dc gain of 120dB then that means it has a power gain of 10^12 but a voltage gain of 10^6.
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