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If I have an amplifier that provides +3dB, or an attenuator that provides -20dB, then how do I know if that applies to voltage or power?

In other words, if I have a 0dBV signal and I run it though a +3dB amplifier, do I get 3dBV?

Similarly, if I have a 0dBm signal and it runs through +3dB amplifier, do I get a 3dBm signal?

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  • \$\begingroup\$ decibel always assumes a reference thus making it not ambiguous. if no reference is used, the context is that the reference = 1. \$\endgroup\$ Mar 13, 2015 at 17:56
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    \$\begingroup\$ @steveverrill I think the original title was fine - I don't buy the reasons for changing it. In fact the title now contains a misunderstanding - do decibels scale by voltage or power - no option for both v and p together and this misses the point. Bad work in my book and I wouldn't be surprised if the OP changes it back to the original title. \$\endgroup\$
    – Andy aka
    Mar 13, 2015 at 19:33
  • \$\begingroup\$ I have no idea what "scaling by voltage or power" means. In fact, I feel it introduces an... ambiguity \$\endgroup\$
    – Cuadue
    Mar 13, 2015 at 20:13
  • \$\begingroup\$ @Cuadue - I think you can change it back by editing. At least you should expect some clarification of what they mean. \$\endgroup\$
    – Andy aka
    Mar 13, 2015 at 20:15
  • \$\begingroup\$ dB = 10 x log_10(Power ratio). | Power = k x V^2. | dB = 10 x log_10(V^2 ratio). | log(A^n) = nlog(A). | log(V^2) = 2 log(V) | 10 log(V^2) = 2 x 10 Log(V) = 20 Log(V) \$\endgroup\$
    – Russell McMahon
    Mar 18, 2015 at 4:24

9 Answers 9

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Deci-Bels always express a power ratio. Specifically, dB is defined as 10Log10(pwr2/pwr1). Therefore "20 dB" is exactly the same thing as "100 times as much power". This power ratio is never ambiguous, but sometimes what it applies to can be. However, this is no different than for a statement like "100 times as much power". It may be ambiguous what has 100 times more power than what else, but the ratio itself is clear.

dB are sometimes used to specify gains of amplifiers that work on voltages. Since dB always specifies a power ratio, and power is proportional to the square of the voltage, dB in this context can be thought of as 10Log10((V2/V1)²), which is the same thing as 20Log10(V2/V1).

Power depends not only on the voltage but also the impedance that voltage is driving. Sometimes those impedances aren't know, or the system works inherently with voltages and the actual power isn't relevant, so the simplification is made that the power ratio is the square of the voltage ratio. This is often the case in audio circuits. In other applications, like RF, the impedances are known and important, so then they are taken into account and dB represents the actual power ratio.

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  • \$\begingroup\$ When examining amplifiers, filters, etc., impedances are often ignored not because they're unknown but because the amount of power actually dissipated in part of a circuit is small and not of interest. If an amplifier's input impedance drops by a factor of two with each doubling of input frequency, but the input impedance is 10M at 1KHz, a 10KHz 1V signal would require more power than a 1KHz 2V signal, but in most cases the fact that the latter signal has twice the voltage would often be significant; the fact that the former requires feeding twice as much power would not. \$\endgroup\$
    – supercat
    Mar 13, 2015 at 19:56
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The original title to the question was : -

Are Decibels Inherently Ambiguous?


It unambiguously applies to both.

If voltage increases by 3.0103dB it is \$\sqrt2\$ times bigger. If the power increases by 3.0103dB it is 2 times bigger.

Absolutely no ambiguity there in any shape or form.

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  • \$\begingroup\$ So is it true that 0dBm+3dB==3dBm and 0dBV+3dB==3dBv? \$\endgroup\$
    – Cuadue
    Mar 13, 2015 at 18:01
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    \$\begingroup\$ Only if the amplifier is designed for the same impedance on input and output. \$\endgroup\$
    – The Photon
    Mar 13, 2015 at 18:31
  • \$\begingroup\$ @Cuadue yes it is true \$\endgroup\$
    – Andy aka
    Mar 13, 2015 at 18:36
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    \$\begingroup\$ What's 3dB between friends, eh? \$\endgroup\$ Mar 13, 2015 at 18:49
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The decibel is a measure of a power ratio. I always assume it is about power unless something specifically indicates otherwise. The calculations you get into when working with voltages are really just about converting their effect back into the power ratio.

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The decibel simply specifies the ratio of the magnitudes of like quantities, and is never ambiguous because, for example, when referring to voltage, - or current, or weight, or pressure, or any other scalars:

$$ dB = 20 \log_{10} \frac{V1}{V2} $$

However, when time gets involved and movement of a force through time becomes power,

$$ dB = 10\log_{10} \frac{P1}{P2} $$

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Standing alone, yes, decibels are ambiguous. But decibels are as ambiguous as saying "times two" without referencing what you're doubling. There should always be a reference.

The gain listed in dB is referring to the input type. So...

What's your input?

I'm not sure I've seen something as ambiguous as a device called "amplifier" with its gain in dB and no other information. You'll usually have a voltage amplifier or a current amplifier; an amplifier with the input being voltage or current respectively. That is, if you have a voltage amplifier, then the input is voltage and the output is the voltage with gain. A voltage amplifier with 3dB gain will turn a 1V input into 1.4125V (0dBV to 3dBV).

If your input is dBm, then the input is power, and the output is power with gain. For your example, a device with a 0 dBm input and a +3dB gain, the output will be 3dBm. Similarly, if the input is 0dBW the output will be 3dBW.

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  • \$\begingroup\$ Assuming the input is referred to in the same units as output, why does it matter? Presuming I use some bizarre third-order Voltage unit V^3, then does the dB relationship still hold? \$\endgroup\$
    – Cuadue
    Mar 13, 2015 at 18:39
  • \$\begingroup\$ Obviously it holds. It doesn't matter what you call the units or what operations you want to do with them. If you think the values aren't coming out right, then check again, because you're breaking the distributive, associative, or commutative rules of math. \$\endgroup\$
    – Samuel
    Mar 13, 2015 at 18:58
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    \$\begingroup\$ Why the down vote? \$\endgroup\$
    – Samuel
    Mar 13, 2015 at 20:00
  • \$\begingroup\$ I agree that it is obvious, however I'm looking for a relatively rigorous proof. FWIW, I gave you +1 \$\endgroup\$
    – Cuadue
    Mar 13, 2015 at 20:00
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The voltage gain of an amplifier, expressed in dB, is 20log(Vo/Vi). That's clear and unambiguous. You cannot relate that to power gain of the same amplifier unless you know the input impedance (say, resistance Ri) and the load impedance (say, resistance RL).This is because output power is Vo^2/RL and input power is Vi^2/Ri and the power gain is thus:

10log[(Vo^2/Vi^2).(Ri/RL)] = 10log(Vo^2/Vi^2)+10log(Ri/RL) = 20log(Vo/Vi)+10log(Ri/RL)

Therefore, if input and load impedances are not equal then power gain in dB is not equal to voltage gain in dB; they differ by 10log(Ri/RL).

If, however, input and load impedances are equal then log(Ri/RL) = log(1) = 0 and the power gain and the voltage gain, in dB, are equal. This is the case in, eg, transmission line repeater amplifiers where the inputs and outputs are connected to 50 Ohm transmission lines and input and output impedances must both be matched to the line to prevent relections and maximise signal transfer. In audio power amplifiers power gain is somewhat meaningless, as the input impedance is normally very high and input signal is very small, and the output impedance is only a few of ohms and output signal is large. So power output is the normal parameter quoted by manufacturers.

It's worth noting that dBm and dBW, essentially, represent specific powers, since they refer strictly to ratios with respect to 1mW and 1W, respectively. Thus for example, 3dBm = 2mW and 20dBW = 100W; the UK amateur radio licence states that 'maximum output power on HF bands is 26dBW', and that equates to 400W (10log(400) = 26dBW)

So, in summary, if you quote gain in dB, you must state whether you refer to voltage gain, power gain, or current gain.

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As others have said, decibels are the unambiguous expression of 10x the log of the power ratio. The problem is, usage over the years has been inconsistent and very often just plain wrong. Some examples:

  • Expressing the gain of an amplifier in db using 20xlog(Vo/Vi) and ignoring the fact that the input and output impedances are hugely different. I don't think that "expressing the voltage gain in dB" is correct usage of the unit.
  • A widespread belief that 0dBm is referenced to "one mW in 600 ohms". This came from old VU meters that were calibrated in dBm. They were effectively voltage measuring devices, and only read correctly when across a 600 ohm line. 0dBm is of course referenced to 1mW - the impedance does not have to be 600 ohms. Unfortunately some meters had the words "0dbm = 1mw in 600 ohms" written on their face...

And please don't get me started on PMPO and other abuses of electrical units...

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  1. dB is used to quantify ratio between two intensity or power values while dBm is used to express an absolute value of power.

  2. dB is a dimensionless unit while dBm is an absolute unit.

  3. dB is relative often relative to the power of the input signal while dBm is always relative to 1 mW signal.

ref: Read more: Difference Between dB and dBm | Difference Between http://www.differencebetween.net/science/difference-between-db-and-dbm/#ixzz5T098jHjz

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Representing a voltage ratio in dB form is a little ambiguous. The dB form of any ratio is X,dB = 10*log(X), but when you are working with voltage, we assume both voltages appear across the same impedance value and force the returned value to be a power ratio by multiplying the log by 20 instead of 10 (P = V^2/R).

So, we derive the voltage ratio in dB form VR,dB = 20*log(V2/V1) using a power equivalance, then break the power equivalence and treat this as a logarithmic ratio of voltages multiplied by 20.

When deriving dB forms of equations, if you have a constant multiplier, then you have to plug this into the K,dB = 10*log(K) function. But if you are working with voltage ratios, you use a 20 in front.

You can take into account different input and output impedance levels by adding an extra term to the dB voltage ratio equation that will make "voltage dB's" and "power dB's" equivalent.

VR,dB = 20log(V2/V1) + 10log(R1/R2)

If we didn't have the 20 in front of the voltage ratio in dB and used X,dB = 10*log(X) for everything, we would always have to know if we were talking about a power ratio or voltage ratio. A lowpass filter would have a -3dB power cutoff frequency that would be the same frequency as the -1.5dB voltage cutoff frequency.

When you have to deal with an amplifier with voltage gain specified in dB and you need to worry about input and output loading, you have voltage dividers on the input and output. You have to remember you are really working with voltage ratios and need the 20 out in front of the logs even for the ratios of resistances.

Total Voltage Gain,dB = 20log(Rin/(Rsource + Rin)) + Av,dB + 20log(Rload/(Rout+Rload)).

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