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Based on my very basic current understanding of decibels in electronics, it is said that I should use the db = 10Log(p1/p2) formula if I want to calculate the power ratio in db between two values and the db = 20Log(v1/v2) formula if I want to calculate the voltage (or current) db ration between two values.

Although I understand how these formulas are derived, I don't understand why we need two formulas (one for power and one for voltage), are two formulas really necessary? I am asking this because as far as I understand, the calculated db value by itself is just a ratio between two values and if that is the case, isn't a ratio just a ratio and nothing more?

In other words, in my head, a doubling of power or doubling of voltage is the exact same thing! Both values are being doubled so I would expect the same db ratio result from both comparison (power and voltage). Below are two examples:

Example for power:

P1 = 200;
p2 = 100;

db = 10Log(p1/p2) = 3 db.

Example for voltage:

v1 = 200;
v2 = 100;

db = 10Log(v1/v2) = 3 db ... (notice not using 20Log(v1/v2) formula here)

The logic above makes sense to me because I see doubling as doubling, it does not matter to me if we are doubling power or doubling voltage, both measurements are being doubled so why should't they have the same db value? Specially since we already specify watts or volts on the db units result (dBm, dBv dBu, dBc, etc)?

What am I missing?

Thanks.

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    \$\begingroup\$ Because \$P = {{E^2} \over R}\$. \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 10 '15 at 3:54
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    \$\begingroup\$ Your two examples must not be talking about the same circuit - if you double the voltage, you quadruple the power. \$\endgroup\$ – Greg d'Eon Feb 10 '15 at 3:57
  • \$\begingroup\$ @The Photon : Not really a duplicate. My question has to do with what is the point of multiplying by 10 when using power and by 20 when using volts (or current)? Multiplying by 10 makes sense because we are converting bells to decibels but why muddle the waters with the 20 factor when representing volts or current? Other scales such as the Richter magnitude scale for example don't introduce some new factor because it represents something else, instead, it sticks with basic logarithmic scale to represent magnitude. Double is double, triple is triple etc, no matter the units right? \$\endgroup\$ – T555 Feb 10 '15 at 5:25
  • \$\begingroup\$ @Ignacio Vazquez-Abrams : As I mentioned in my original post. I already know the technical reason for why we multiply by 20. The question is why bother substituting P = E^2/R on the power formula? Why not treat Voltage the same as Power and focus solely on the ratio. In other words, why not just 10Log(p1/p2) and 10Log(v1/v2)? Why do we have to get all fancy and do the P = E^2/R substitution on the power formula? \$\endgroup\$ – T555 Feb 10 '15 at 6:47
  • \$\begingroup\$ The decibel is always 10 log P1/P2. That is the definition. If you don't know the power, but you know the voltage, then you can calculate 10 log P1/P2 using the equivalent formula 20 log V1/V2. There is no dichotomy or dual standard. The 20 log formula is just a computational shortcut to avoid calculating the power. \$\endgroup\$ – mkeith Feb 10 '15 at 7:27
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I think the reason is historical. THE telephone company (Bel as in Graham Bell) used the decibel scale to characterize its equipments using the power definition. So even if the measurements came in the voltage scale, they were needed to be in the power equivalent decibel format for comparison to the specifications.

A simple made up example: The central office drives a cable with 1W of power. It has to guarantee that the cable has 3dB or less attenuation to the customer, giving the customer at least 0.5W of power. Assume the load is 1 ohm. A technician goes to make measurements to verify, he measures in voltage and gets 0.707V (which is indeed 0.5W). log(1/0.707) = 0.15. So if he multiplies that by 10 and calls it decibel, it would be 1.5. Making it difficult to compare to the specification. Multiplies it by 20, he gets 3dB and can make direct comparison to the specification.

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  • \$\begingroup\$ Although everyone helping me out did an excellent job answering the question. For some reason, your explanation made me connect all the dots and cleared my question. Thanks. \$\endgroup\$ – T555 Feb 12 '15 at 5:20
  • \$\begingroup\$ Although I think you meant to say log(0.707/1) = -0.15 since the reference is 1 volt. Is that not the case? \$\endgroup\$ – T555 Feb 12 '15 at 5:34
  • \$\begingroup\$ I was just being sloppy. Yes, the measurement is -3dB compared to the reference. I wrote "3dB .. attenuation", to match that, I flipped the calculation. \$\endgroup\$ – rioraxe Feb 13 '15 at 1:48
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In other words, in my head, a doubling of power or doubling of voltage is the exact same thing!

Incorrect - doubling the voltage quadruples the power. Moving on....

Decibels are all about power ratios.

In your question you say you want to calculate the voltage (or current) db ratio. This is where you are getting confused.

For instance if I said the "signal" across a resistor increased by 6.02 dB, the power will have gone up four times and the voltage will have doubled.

On the other hand, if I said the voltage had doubled, in dB terms this is: -

20log(2) = 6.0205 dB and if I said the power had quadrupled, in dB terms this is: -

10log(4) = 6.0205 dB. Does this make sense?

The basic problem is that you believe that there is such a thing as "voltage (or current) db ratio" and that somehow this is independent or different to power.

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