Resistor value = ? db reduction

8thnerve · « **on:** 30 Sep 2004, 01:50 pm »

Is there a chart online somewhere that would express db reduction for a line-level signal based on resistance?

i.e. 10K resistor = ? reduction in db

Thanks!

Occam · « **Reply #1 on:** 30 Sep 2004, 02:46 pm »

Nathan,

An arbitrary resistor put in series with a voltage yields no attenuation, I believe what you are asking about is attenuation associated with a voltage divider, typically the attenuation assosicated with a series resistor followed by another resistor (or component with a frequency depedant reactance) which is usually a shunt to ground. You are concerned with the voltage at the junction of these 2 resistors.

For example -
Input via a 10k series resistor connected to a 50k shunt resistor to ground. The output voltage Vout = Vin( Rshunt/(Rshunt + Rseries))
and for this example Vout = Vin*(5/6)

You can convert that voltage ratio to dB with the equation
dB = 20*log(votage ratio) = 20*log(Rshunt/(Rshunt+Rseries)) =
20*log(5/6) = -1.583 dBv

You can get far better explanations by using google and entering 'voltage divider' which will offer discussion and calculators -
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html

and doing the same for 'decibel calculator'
http://www.mcsquared.com/dbframe.htm

I'm sure there are calculators out there that combine both funtions, or its quite easy to do your own in Excel with the formulas above.

WerTicus · « **Reply #2 on:** 6 Oct 2004, 03:06 pm »

i find that even the mathematically correct level isnt quiet right

resistors are cheap anyways just buy a few 'near' the value and switch them around till you think you have the right one...

i wouldnt solder them for a couple weeks till your sure either.

i also fine that two x resistors in parrellel of double the value sound better than one due to increased bandwidth.

mgalusha · « **Reply #3 on:** 6 Oct 2004, 09:49 pm »

You might want to check out: http://www.electronics2000.co.uk - He offers a free program called "Electonics Assistant" which contains various electronics calculators, one of which is for calculating voltage divider values. It's labeled "potential divider calculator" in the program. Works great if you don't like doing the math or plopping the values into a spreadsheet.

A handy little program and the price is certainly right.

8thnerve · « **Reply #4 on:** 12 Oct 2004, 09:30 pm »

Ok, before I do this, perhaps some of you experts can tell me if this will work.

I want to drop the volume by 12db. Using a potential divider to reduce the V from 2 to .5, requires R1 at 30 Ohms, and R2 at 10 Ohms, yes? My next question is, will this give me a 40 Ohm output impedance at the divider?? The output impedance of the Lynx TWO card is 50 Ohms, and the input impedance of the Flying Mole amps is 10 Ohms, how does this relate to this divider?

Thanks.

8thnerve · « **Reply #5 on:** 14 Oct 2004, 01:51 pm »

Can someone help me on this?

Dan Banquer · « **Reply #6 on:** 14 Oct 2004, 02:31 pm »

"and the input impedance of the Flying Mole amps is 10 Ohms, "
Are you absolutely sure about this? For typical audio line level unbalanced inputs the input impedance is typically 10k ohms minimum. Or am I missing something?
d.b.

Occam · « **Reply #7 on:** 14 Oct 2004, 05:10 pm »

Quote from: 8thnerve

Can someone help me on this?

It would be easier if your info was a bit more accurate....
Per the Flying Mole website the M100 and M300 have 10,000 (10k) ohms input impedance.

WTF are R1 & R2?

If you use a 10Kohm resistor as your shunt element in a voltage divider and a 30Kohm resistor as the series element, you will get that -12.04dbV attenuation.
(Indeed, if you maintain that same ratio, your 10ohm & 30ohm elements, you will get that same -12dbV attenuation, its just that your Lynx sure as heck wont be able to drive it, let alone the 10K input impedance of the Mole that is in parallel.)

There are websites that are far more lucid as to calculation of input and output impedances than I can. Simply do a google search on 'Thevenin' and actually use a caculator and work through the examples.You might do the same with 'Kirchhoff's Law', or simply break down and buy a copy of Horowitz and Hill 'The Art of Electronics'.

PEB · « **Reply #8 on:** 20 Oct 2004, 01:23 pm »

Hi,
If this was really important to you, then you've probably figured it out by now. If not, I'll take a stab at it.

-12dB = 1:4 ratio voltage drop

Voltage divider => 4:1 ratio of Rs:Rp, where Rs=series R, Rp=parallel R

Assume your amp input impedance is 10k, and you want to use it for Rp,
Then Rs = 4 * 10k = 40k.

However, if you want a -12dB drop for any other amp, then you can drop the resistance values, such that most all amps input impedance appear large compared to your V-divider values.

e.g., choose Rp = 1k
Then Rp,eq = 1k || 10k ~ 890ohm (only 10% lower)
Rs = 4x 890 = 3.6k

Now say you connect a 47k input-Z amp to your 890/3600 V-divider.
Rp,eq = 1k || 47k ~ 980ohm.
3.6k / 980 = 3.67, or about -11.3dB drop.

Note that the impedance load presented to the card is 40+10 = 50k for the first case, and 0.9+3.6 = 4.5k for the second case.

Given the card's 50ohm output Z, neither case will be a difficult load for it to drive.

You can use a DVM and 1kHz test tone to verify the voltage presented to the amp is 1/4 that without the V-divider in circuit.