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Message started by aaron_do on Mar 23^rd, 2010, 12:00am

Title: dBΩ
Post by aaron_do on Mar 23^rd, 2010, 12:00am

Hi all,

just wondering if we express a TIA resistance in decibels, do we take 20logR or 10logR to get dBΩ? My thinking is power = I²R or 20logI +10logR so dBΩ should be 10logR. However, I can see that if V=IR 20logV = 20logI + 20logR...so which is correct?

thanks,
Aaron

Title: Re: dBΩ
Post by buddypoor on Mar 23^rd, 2010, 4:21am

Hi Aaron, I understand your question, however I think it is not appropriate to use equations like P=I*I*R or V=I*R to find an answer how dBohm is defined.

The reason is as follows: It is simply a definition and it is "dangerous" to use dB within formulas, because DB is NOT a unit (like volts, seconds, meter,..). It is only an indication that "something" has been done with this quantity - in this case: Divide by 1 ohm, transfer the number into log, and multiply by 20.
Example: 100ohms are identical to 40 dBohm and if you multiply this resistance by 10 you arrive at: 40 dbohm+20dB=40dBohm.
(You add dbohm and dB and the result is dBohm!)

Title: Re: dBΩ
Post by aaron_do on Mar 23^rd, 2010, 7:45am

Hi Buddypoor,

thanks for the response. Of course I am aware that dBΩ + dB gives dBΩ...i'm really just trying to find out what is the most common definition of dBΩ, and why is it defined like that. Apparently it is 20logR. I was mainly concerned with a TIA, and the explanation I got was that Pout/Pin = (Vin²/Rout)/(Iin²Rin) or in dB, Pout/Pin = 20log(transimpedance) - 10log(RinRout)...

cheers,
Aaron

Title: Re: dBΩ
Post by buddypoor on Mar 23^rd, 2010, 8:08am

aaron_do wrote on Mar 23^rd, 2010, 12:00am:

.............
do we take 20logR or 10logR to get dBΩ? My thinking is power = I²R or 20logI +10logR so dBΩ should be 10logR. However, I can see that if V=IR 20logV = 20logI + 20logR...so which is correct?

Hi Aaron, perhaps it's good to argue from the other side:
Of course, your formulas are correct:
The first result (including the expression 10logR) gives the power and the second one (including the expression 20logR) gives a voltage.
But the main question remains: Shall I call one of the logarithmic expressions "dBohms"? That`s a free decision and it has been made long time ago. Nevertheless, good point!

Title: Re: dBΩ
Post by loose-electron on Mar 23^rd, 2010, 10:56am

Yeah, this ones messy.

dB, dbV, dBm (better stated as dBmW) are pretty well defined, but if you are going to go outside the mainstream, I would ask for explicit formula definition and not let them fling a poorly defined concept at you.

Guessing is that its with reference to 1 ohm, but since its not widely recognized, if I ued it in a specification, I would spell it out someplace.

Title: Re: dBΩ
Post by buddypoor on Mar 23^rd, 2010, 11:27am

Yes, loose-electron, I agree with you.
However, there are some cases/applications in which it makes really sense to use this dB convention also for resistances, e.g. by noise voltage calculations and for link budgets (when all contributing parameters are expressed via dB).

Title: Re: dBΩ
Post by RFICDUDE on Mar 29^th, 2010, 3:54am

Perhaps we can simply apply the typical use of the LOG for measuring differences in power or intensity in terms of resistance.

Log10 is used for comparing order of magnitude differences (on a uniform spaced plot) between two power or intensity quantities.

ΔP(dB) = 10×LOG₁₀(P₂/P₁)

dBV is somewhat loosly defined because it does not stipulate anything about the load resistances of the voltages being compared, so it doesn't give a true comparison of power difference (unless the resistances just happen to be the same. Nonetheless, dBv is still a very useful measurement of order of magnitude differences between voltage levels in a circuit (especially for analog circuits).

So, how can we use this for resistance?
ΔP(dB) = 10×LOG₁₀(V²₂R₁/V²₁R₂)

Using the law of logarithms

ΔP(dB) = 20×LOG₁₀(V₂) - 20×LOG₁₀(V₁) + 10×LOG₁₀(R₁) - 10×LOG₁₀(R₂)

So, it would seem that 10×LOG₁₀ of the resistance is maybe more appropriate?

It is interesting to note that if you are calculating power difference based on current then the sign of of the log resistance difference changes.

Comments are welcome ...

Title: Re: dBΩ
Post by aaron_do on Mar 29^th, 2010, 6:34am

I have seen it used as 20logR without any definition at all, so I thought it must be standard...However, I also thought 10logR sounds more appropriate. Unless someone can say for sure that 20logR is the standard, then I think its important to define the calculation before writing it...

cheers,
Aaron