Decibels

[lowell]

Ok, can someone please clear something up for me? I've read about Db and logarithms but haven't been able to fully grasp this whole Db thing. And I'm not sure which Db reference I should use: DBu DBv etc...

I get that:
log10(100)=2
or
10^2=100
Would Db then be calculated as:
10log10(100)=20Db?

Here's an example maybe someone can solve for me.

Say I input 3v p-p ac. (ignore bandwidth and frequency for now)
My output is 27v p-p ac.
Therefore my gain is 9=27v/3v=3^3
Is the DB then 10log3(27)=30Db? (got this equation from some website)

[Robert M. Martinelli]

Nope, the voltage gain is defined as follows:

Av = 20.log(Av), that is, twenty times the log (NOT ln) of the measured gain.

In your case, 20 log (9) yields a gain ( expressed in dB ) of 19.08 dB.

Remember that the logarithm in base "n" of a certain number is the exponent you have to apply to your base (the number n) to get that number.

e.g. log(100) = 2, as the base for log is 10, and to get 100 you have to take the base (10) and elevate it by the logarithm (2) ---> 10^2=100.

Some practical examples:

Voltage gain = 10 ----> voltage gain (dB) = 20.log(10) = 20 dB

Voltage gain = 20 ----> voltage gain (dB) = 20.log(20) = 26 dB

Voltage gain = 40 ----> voltage gain (dB) = 20.log(30) = 32 dB

( note that every +6dB your gain doubles )

Voltage gain = 100 ----> voltage gain (dB) = 20.log(100) = 40 db

Voltage gain = 1000 ---->voltage gain (dB) = 20.log(1000) = 60 dB

( note that every +20 dB your voltage gain Av increases by a factor of ten, 20 dB=10, 40 dB=100, 60 dB=1000 )

Hope this helps

Best regards

Bob

[guitarmike2107]

ERRMmm

bob would you mind epanding that into -db. I always thought -3db was 1/2 voltage gain.. but seems that I am wrong?

thanks

Mike

[Wilder Amplification]

ERRMmm

bob would you mind epanding that into -db. I always thought -3db was 1/2 voltage gain.. but seems that I am wrong?

thanks

Mike

No...1/2 POWER gain = -3dB. 1/2 voltage gain is -6dB. The constant is different.

Voltage gain is figured via -

(Output/Input) log x 20 = Voltage Gain

For instance, you have a 10 volt output signal for a 1 volt input signal -

10 / 1 = 10

Log 10 = 1

1 x 20 = 20dB

Power gain is is figured via -

(Output/Input) log x 10 = Power Gain

[R.G.]

would you mind epanding that into -db. I always thought -3db was 1/2 voltage gain.. but seems that I am wrong?

Think of it as a trick of the math. Logarithms do this.

The logarithm (which is the *exponent* of the logarithm base to get the real number) of 1 is zero. That is, ten to the zero-th power is 1. Ten to the first power is 10.

Logarithms of numbers less than 1 are negative. Ten to the -1 power is 1/10th. So the logarithm of 0.1 is -1.

So for a loss of 10 to 1, that is a "gain" of 1/10th, the logarithm of 1/10th is -1. The decibel corresponding to 1/10 is 20 times that, or -20. Therefore, a voltage drop of 10 to 1 is -20db.

A voltage drop of 100 to 1, down to 1/100th of the original signal is -40db because ten to the -2 power is 0.01. -2 times 20 is -40, and the answer is -40db.

In the case of a loss of 1/2 the voltage, the voltage ratio is 1/2. The logarithm of 0.5 is -0.30102999..., and twenty times that is -6 db and a string of less-significant digits.

The reason it's -6db and not -3db is that decibels were originated for POWER ratios, not voltage ratios. Calculating DB for power ratios, you take the log(10) of the power ratio and multiply by ten. For a 2:1 power drop, that is -3.01029999db. Power is proportional to voltage squared, so if you're only talking about voltages and the same impedance on both sides of whatever is increasing or decreasing the voltage, then doubling the voltage increases the power by four times, so you multiply the db number by 20 (2*10) to get the power ratio.

For same impedance, +3db power is the same as +6db voltage. The factor of 10 for power and 20 for voltage was always the hardest thing for me to grasp.

[lowell]

Ok so I'm building a headphone preamp for a buddy. He wants 20+db from it. Do I use 10log or 20log to calculate this? I would think he's wanting +20db of volume right? So to get +20db I need to increase POWER not VOLTAGE right?

Input will be line level, or 1v. I'm not sure whether I'm shooting for a 10v or 20v output, or if I should use 20log or 10log.

10log(10)=10db or
20log(10)=20db

[Wilder Amplification]

Ok so I'm building a headphone preamp for a buddy. He wants 20+db from it. Do I use 10log or 20log to calculate this? I would think he's wanting +20db of volume right? So to get +20db I need to increase POWER not VOLTAGE right?

Input will be line level, or 1v. I'm not sure whether I'm shooting for a 10v or 20v output, or if I should use 20log or 10log.

10log(10)=10db or
20log(10)=20db

I think you're trying to relate electrical power to acoustic power and they don't quite coincide with each other like you think they do.

Acoustic SPL is something competely different.

When you say +20dB relative to electrical power, +20dB means that the output power will be 20dB above a reference, which would be your input signal. In terms of power, your unit of measurement would be dBW. dBW is relative to 1 watt, with 1 watt = 0dBW. +20dBW would be 100 watts in this case since 100 watts is +20dB above 1 watt.

dB is not an exact number. It's a unit of measurement used in the comparison of amplitude difference relative to a specific reference, be they power, voltage or SPL (Sound Pressure Level).

[mooreamps]

Ok so I'm building a headphone preamp for a buddy. He wants 20+db from it. Do I use 10log or 20log to calculate this? I would think he's wanting +20db of volume right? So to get +20db I need to increase POWER not VOLTAGE right?

Input will be line level, or 1v. I'm not sure whether I'm shooting for a 10v or 20v output, or if I should use 20log or 10log.

10log(10)=10db or
20log(10)=20db

We consider "line level" or 1 volt rms 0 db. So, for +20, use one gain stage of a 12ax7 ; 100K plate, and 1.5K cathode.

Also, if this helps any ; decibels are of a logarithmic scale, and voltage is of a linear scale.


-g

[lowell]

Gary,
Ok so a 12ax7 is going to increase 1v to 70-100v. I don't believe that's what I'm after here for +20db. Also you haven't stated any math behind this or clarified if you're using DBv or DBu, 10log, 20log etc... Can you please be more specific in your reasoning. Oh, and I'll be using a J201 Jfet for this preamp.

[R.G.]

Ok so I'm building a headphone preamp for a buddy. He wants 20+db from it. Do I use 10log or 20log to calculate this? I would think he's wanting +20db of volume right? So to get +20db I need to increase POWER not VOLTAGE right?

Input will be line level, or 1v. I'm not sure whether I'm shooting for a 10v or 20v output, or if I should use 20log or 10log.

10log(10)=10db or
20log(10)=20db

It's clear and concise, as long as you know what you're calculating. Does he want +20db POWER or +20DB VOLTAGE? I bet he doesn't know. When pressed, he'd probably say "you know, gain of ten".

The problem is that the headphone preamp is dead in the middle of the power versus voltage argument.

It being a headphone amp, it's probably intended to drive 32 ohm (maybe! that's a quasi-standard for headphones) loads. The input impedance is probably something like 10K or bigger. 1V into 10K is a power of P =V-squared/R = 1*1/10000, or 0.0001W, 100uW. Possibly a lot smaller if the input impedance is, say, 100K or 1M. In any case, the input to this thing is probably a voltage.

An output that's +20db *power* is 100 times that, or 10mW. That's a voltage of V = SQRT(P*R) =SQRT(0.01*32) = 0.565V. So a POWER +20db is a VOLTAGE loss of almost half because of the differences in the impedances.

If what he meant was a voltage gain of +20db, he meant a voltage ratio of 20db, divided by 20 to correct back to the power of ten, or 1, and then the voltage ratio is 10x, and the output is 10V, not 0.565V. Voltage ratios ignore the power the voltages cause in the impedances they drive. And this is why voltage "db" is not strictly something that has meaning unless the input and output impedances are the same.

Acoustic power is also measured in DB, but that's relative to some reference level, generally the threshold of human hearing if it's dbSPL.

What he probably meant is "make it have a voltage gain of ten". That's what I'd do, unless I knew that he knew what he was talking about and really wanted a POWER gain of 20db. That context is only common in people who've had some training.

dB is not an exact number. It's a unit of measurement used in the comparison of amplitude difference relative to a specific reference, be they power, voltage or SPL (Sound Pressure Level).

I'd say that a different way. Decibels are exact numbers. But they are exact **ratios**, not absolute levels. And just like the power/voltage/SPL/radiated RF power/solar light flux/etc, they only have meaning when comparing ratios of the same thing. Comparing voltage level to SPL is just as bogus as comparing rainfall rates to seconds left to play in db. +23db more seconds left to play than rainfall per hour doesn't mean anything.

[lowell]

Ok so a voltage gain of 10. I will look into what headphones he's using and what the output Z of the interface is.

So in DBv then.
10Vout/1Vin=10=gain
20log(10)=20Db?

Again sorry if I'm annoying or not getting it. Why is 20log for power and 10log for voltage?

[mooreamps]

It's clear and concise, as long as you know what you're calculating. Does he want +20db POWER or +20DB VOLTAGE? I bet he doesn't know. When pressed, he'd probably say "you know, gain of ten".

.

I'd prefer he do some research on his own, and then post to me what he believes the answer is.


-g

[Wilder Amplification]

Ok so a voltage gain of 10. I will look into what headphones he's using and what the output Z of the interface is.

So in DBv then.
10Vout/1Vin=10=gain
20log(10)=20Db?

Again sorry if I'm annoying or not getting it. Why is 20log for power and 10log for voltage?

A 10V signal is equal to 20dBV since 1V = 0dBV.

If you have a voltage gain of 10, the amount of gain you have expressed as dB is +20dB. This means that no matter what the input signal is, the output will be 20dB higher than the input signal, or 10x higher in voltage than the input signal.

[Wilder Amplification]

Ok so a voltage gain of 10. I will look into what headphones he's using and what the output Z of the interface is.

So in DBv then.
10Vout/1Vin=10=gain
20log(10)=20Db?

Again sorry if I'm annoying or not getting it. Why is 20log for power and 10log for voltage?

A 10V signal is equal to 20dBV since 1V = 0dBV.

dBV is a unit of measurement relative to a fixed voltage reference and has nothing to do with gain itself. The fixed reference is 1 Volt, which is equal to 0dBV. When you have a 1 Volt signal, the level of that 1 Volt signal is said to be at 0dBV. A signal at 10 volts would be a +20dBV signal since it is 10x higher (+20dB) than the 1 volt reference. A 100mV (millivolt) signal would be at -20dBV since it is 10x lower than the 1 Volt reference.

When using dB in terms of gain itself, if you have a voltage gain of 10, the amount of gain you have expressed as dB is +20dB. This means that no matter what the input signal is, the output will be 20dB higher than the input signal, or 10x higher in voltage than the input signal, and the gain will always be 20dB regardless of the actual amplitude of the input and output. When speaking in terms of actual signal amplitude, then you use the dBV reference.

[Robert M. Martinelli]

ERRMmm

bob would you mind epanding that into -db. I always thought -3db was 1/2 voltage gain.. but seems that I am wrong?

thanks

Mike

A lot of most knowledgeable friends chimed in while I was "away", so it's possible that you have found the answer you were looking for already, anyway...

Expanding into -dB is very easy, you only have to remember the logarithm definition I gave within my previous post. The trick a logarithm does, being an exponent, is to "transform" a multiplication ( or a division ) into a sum ( or a subtraction ). This means that, if + 6dB means that your voltage signal gets doubled, -6dB means that your voltage signal gets halved. Likewise, +20dB "expresses" in dB a voltage gain of 10, so -20dB means your signal's voltage gets divided by ten.

Oh, and BTW -3dB means that your output voltage is 0.7071 times the input voltage.

Hope this helps

Best regards

Bob