Peak-to-Peak Voltage to dB conversion

[juz_ad]

I have been searching for an equation to convert Peak to Peak Voltage to dB and I have found 4 slightly different equations that all claim to do the same thing:

1) dB = 20*log_10(Vpp/0.707)
2) dB = 20*LOG10(Volts_peak_to_peak/SQRT(0.008*Z))
3) db = (20log10) (RMS ) & RMS= 0.707*P-P
4) dB = 20 log10 (Voltage 1 / Voltage 2)

Can anyone comment on a preferred method or advise on any of these equations please?

Thanks.

 

[klystron]

Is the signal sinusoidal ?

 

[juz_ad]

Hi klystron,

Yes, for the purposes of this research we can consider that the input will be a Sine wave with attenuation between 0 and 10V PP (+/-5V PP).

Thanks - hope that helps.

 

[albbg]

dB is not a measurement unit like ,for instance, volts are. It is a ratio between two quantities of the same type.
The last equation is correct and can be used to represent a gain or a loss, in dB. In fact if you have a voltage gain of, lat me say, 1000 means the ouptut voltage Vo will be 1000*Vi (input voltage). Using the last equation we will have GdB = 20*log10(Vo/Vi) = 20*log10(1000*Vi/Vi) = 20*log10(1000) = 60 dB. Vo and Vi must be of the same type: both peak-to-peak or RMS or peak.
It could be you want, instead, to convert your voltage to dBV that is a logarithmic measurement unit that has a reference of 1Vrms. In this case, first of all you need to convert from peak-to-peak to RMS. In case of sinusoidal signal you just have to divide by 2*sqrt(2), then apply the dB formula using Vi = 1V:

dBV = 20*log10[ Vpp/(2*sqrt(2) ]

Or could also that you want to calculate the power in a logarithmic scale. In this case you also need to know the resistance (R) across which the voltage is applied. Also in this case we need the RMS voltage, that is Vrms = Vpp/(2*sqrt(2))
The power is given by: P = (Vrms)²/R. A logarithmic popular measurement unit for power is dBm (that means referred to 1 mW). The formula is:

dBm = 10*log10(P/1mW) = 10*log10[(Vrms)²/(0.001*R)] = 10*log10[(Vpp)²/0.008*R] (is your eq. 2, using the logarithm properties)

 

[juz_ad]

The last equation is correct and can be used to represent a gain or a loss, in dB. In fact if you have a voltage gain of, lat me say, 1000 means the ouptut voltage Vo will be 1000*Vi (input voltage). Using the last equation we will have GdB = 20*log10(Vo/Vi) = 20*log10(1000*Vi/Vi) = 20*log10(1000) = 60 dB. Vo and Vi must be of the same type: both peak-to-peak or RMS or peak.

Thanks for the reply - can I ask you to clarify your working on these 3 equations please. For a Vin of 1V PP and a Vout of 1000V PP (1000x Gain), using the 3 equations you have listed, I get the following:

20*log10(1000/1) = 20,000
20*log10(1000*1/1) = 20,000
20*log10(1000) = 20,000

Sorry if I've misunderstood your reply - but I don't see how you're getting 60dB.

Perhaps if you could also show a working for 50% attenuation?

<edit 30/12/2011>

I have added an edit to this post because my workings above are incorrect - the equations resolve to 60dB as albbg stated in the original post. Thanks.

 

[albbg]

Simply log10(1000) = 3 because 10^3 = 1000

then 20*3 = 60 dB

 

[zorro]

1) dB = 20*log_10(Vpp/0.707)
2) dB = 20*LOG10(Volts_peak_to_peak/SQRT(0.008*Z))
3) db = (20log10) (RMS ) & RMS= 0.707*P-P
4) dB = 20 log10 (Voltage 1 / Voltage 2)

1) Maybe there is a mistype? If you substitute Vpp/0.707 by Vpk*0.707, it is the expression of a voltage in dBV (dB realtive to 1V) assuming it is sinusoidal.
2) where does it came from? It may be related to voltage if dBu for telephony levels (but 0.008 would need more digits)
3) is the similar to 1) if you take Vpk instead of Vpp
4) is the ratio V1/V2 in dB; voltages must be rms values, of also may be peak values if both have the same vaweform.

Regards

Z

 

[juz_ad]

From the advice I have received here and elsewhere it seems that, for my purposes (comparing the gain response of two VCA circuits) - then:

GdB = 20log10(Vout/Vin)

...seems to work. Vin for both circuits is a sine from the same, known source and is considered, for the sake of comparison to be '0dB' (10V PP) - any change in gain/attenuation shows as a +/-dB change. Does this sound reasonable?

Technically, for voltages, I *think* I should be using dBu - but I *think* for basic comparison the above equation will do.

I could do with some help on one way to manipulate this equation. If GdB = 20log10(y / Vin) = x - where I know the value of 'x' and need to calculate the value of 'y' - e.g. GdB = 20log10(y / Vin) = 40 ?

Thanks again for all the help - much appreciated.

 

[albbg]

OK, 0 dB gain means the Vout = Vin.

dBu is one of the logarithmic measurement unit used for voltages. The reference is 0.7746 Vrms

The formula is dBu = 20*log10(V/0.7746) where V must be exspressed in rms value

To convert back from x = 20*log10(y/Vin) ==> x/20 = log10(y/Vin) ==> 10^(x/20) = y/Vin ==> y = Vin*10^(x/20)

By the way 10^(x/20) is the linear voltage gain.

In you example Glin = 10^(40/20) = 100 then y = 100*Vin