dB Equations

[Earl K]

Hi

- Here's a .gif copy of the "dB" wheel .

- One may notice that the function of, 10log Z1/Z2 ( or R2/R1 ) suggests that a person should be able to directly calculate ( say ) pad attenuation from a set of given impedances.

(i) I can't make this basic formula work for me. I've been struggling with this for a while but to no avail. I'm forced to measure real-time voltage drops, which while accurate , is a real PITA with it's own challenges .

(ii) Can anyone help ?

If so , please demonstrate by example, showing the dB drop for a 13.5 ohm resistor strapped across an 8 ohm load / along with the presence of a 2 ohm buildout resistor .

Thanks <. Earl K

[Zilch]

dB expresses a ratio.

Consider the resistances as comprising a voltage divider:

Your 13.5 and 8 in parallel is 5.0233 ohms.

The total resistance is 5.0233 + 2 = 7.0233

The output/input voltage ratio is 5.0233/7.0233 = 0.7152

Log 0.7152 = -0.1456

20 times that = -2.9111 dB attenuation.

*******

Try it with a known 6 dB attenuation:

8 Ohms in parallel with 8 is 4 ohms.

The total with 4 ohms in series is 8 ohms

The ratio is 4/8 = 0.5

Log 0.5 = -0.3010

20 times is -6.0206 dB attenuation.

[B&KMan]

Hello Earl,

I try to explain my informations found in book, maybe help maybe not...

is explain formula for a constant impedance.

see pict...

:cheers:

[Zilch]

Jean shows how to compute a constant-impedance L-Pad (a special but common case) knowing the speaker impedance and the parallel resistor value.

A = the voltage ratio, assuming total circuit impedance = R = speaker impedance.

dB attenuation = 20 Log A.

Alternatively, he also shows how BOTH resistor values may be derived from a desired attenuation and a known speaker impedance.... :thmbsup:

Final exam now: How do we derive the requisite voltage ratio [A] from a desired dB attenuation?

[Robh3606]

Nice posts

Rob:)

[Earl K]

Thanks guys for answering my call for some instruction . :)

- Hopefully others' have learned something . I know I did : )

- Now, the fact that there are two working formulas seems to be the source of confusion for what has bedevilled me for sometime .

- Zilch, the ESP formula appears to be expressed as a reciprocal of your expression. I'm still trying to understand Rods' favoured approach. Both formulas will give the same results / though the necessary numerators & denominators are different. In both cases they are derived from the same working impedances . It's just which two of three values are chosen to be used with either formula .

- You'll see that Rod E. has added a 1 to this formula. That action results in arriving at consistantly positve dB level changes. I don't understand why this has been done as opposed to just working with straight voltage ratios ( be they plus or minus ).

- One working formula as given at Rod Elliotts' ESP site is ; ( this is paraphrased by me )

Fomula #1

dB = 20log [ (Rs/Rp) + 1 ]

Rs representing the series resistance
Rp representing the "parallel" load resistance , comprising only the original resistance and the additional parallel resistance ( ie; no series resistance ) .

So for my example of adding a couple of padding resistors to an 8 ohm load ;

Rs = 2 ohms
Rp = 5.0233 ohms ( From paralleling of 13.5 ohms with 8 ohms, equaling 5.0233 ohms ).


Running the Formula ;

dB = 20log [ (2/5.0233) + 1 ] or 20*log 1.3981 or 20*.14553826 or 2.9110 dB


Formula #2

dB = 20log ( Rp/Rt )

Rp = 5.0233 ohms ( again, from the paralleling of 13.5 ohms with 8 ohms )
Rt = 7.0233 ohms ( the total resistance, as "seen" by the source, 2 & 5 ohms )

Running the Formula ;

dB = 20log ( 5.0233 / 7.0233 ) or 20*log .7152 or 20 * -.1456 or - 2.9110 dB


- I understand the switch to a " 20log function " since 10log functions are reserved exclusively for power ratios. I still don't understand the "dB" wheels inclusion of a 10log function that seems to be workable with known resistances. I guess that's for another day .



Final exam now: How do we derive the requisite voltage ratio [A] from a desired dB attenuation?

Directly from Rods' site ;

Vr = 1 / [antilog (dB / 20)]





<> Thanks ( Now, I have some other problems to sort out, such as ; why are my bench measurements out of whack )

ps;
You might also want to build yourself an Excel spreadsheet and put all these equations in it.

That's a good idea, since I'm likely to get parts of these 2 formulas intermixed . :p

[B&KMan]

Final exam now: How do we derive the requisite voltage ratio [A] from a desired dB attenuation?

as you wish ZilchMaster

:D

[Zilch]

Heh.

Both Earl and Jean pass the final exam!

:cheers:

[Zilch]

You'll see that Rod E. has added a 1 to this formula. That action results in arriving at consistantly positve dB level changes. I don't understand why this has been done as opposed to just working with straight voltage ratios ( be they plus or minus ).Yes, as you have observed, one approach works with the ratio of the two resistances, whereas the other works with the ratio of just one of them and the total resistance. Both give the same correct result, as you've shown.... :thmbsup:

[Mr. Widget]

That's all interesting and all, but I am too lazy when I can just use these:

http://ccs.exl.info/calc_cr.html#second

I haven't tried their Zobel calculator, but the padding formulae and the LR and Butterworth networks have all worked perfectly.


Widget

[jak]

Ok, forget making any goddamn spreadsheets. Posts deleted. You guys probably couldn't handle the math behind the applets anyway.

Hi Giskard,

Sorry for the off topic question I can't seem to find a way to PM you with this. In your avatar could you tell me the JBL model # of the tall 1/2 triangular speakers if you know. I had a friend that had these a long time ago and they were my favorite JBL product.

Thanks much!

Newbie....Jeff

[Mr. Widget]

Hi Newbie... that is one of JBL's best vintage designs... it is called the 250Ti, here is some info about the LE version: http://www.lansingheritage.org/html/jbl/specs/home-speakers/1990-250ti.htm


Widget

[jak]

Thanks for the info!

Jeff

[Baron030]

The Db Wheel and its related discussion posts have been a great resource. :applaud:
But, like Mr. Widget, I am just too lazy and would rather go to a web site calculator.
The following site not only calculates the Db loss, it also calculated the wattage rating for the resistors.

http://www.lautsprechershop.de/tools/t_w_teiler_en.htm

Also, check out this site's impedance correction calculator. It really demystifies those complex RLC circuits.

Baron030

[4313B]

Yeah, there's quite a few of these little applets these days. My whole point was to stimulate those who have a desire to understand the math behind the applets. I use the easy methods too these days but I also know from whence it all came. ;)

[Zilch]

Yeah ... I use the easy methods too these days but I also know from wence it all came. ;)Would you put the ESP link back up so readers can see that part too, then, please?

[Seems a buncha folks are benefiting from this little thread.... :thmbsup: ]

[4313B]

http://sound.westhost.com/lr-passive.htm

[frank23]

great site giskard,

has anyone ever made a corresponding network of a compression driver and horn combination? the text says it is very difficult for a tweeter alone, so it must be almost impossible for a driver / horn combination.

I made it easy on myself by using a 1.2ohm resistor in parallel with a 2420 and a series resistor of about 6 ohm to get the desired attenuation and higher impedance which deviates almost nothing from the resistive 7.2 [=6+1.2] ohm. The total crossover works a lot like the N300t crossover, only I have an extra kink in the compensation for my 2344.

frank

ps, the below picture is from the website mentioned by giskard, it is not my crossover, it represents the possible electric model of a tweeter