[Jmmmmm]

I've got a resistor that's supposed to lower the volume by 25db. Anyone have a guess as to how many ohms the resistor is? Does anyone know a spl for how much the volume is lowered with any resistor?

 

[wang228]

i'm not an expert in electronics...and on a hight dose of alcohol right now... but i think the value of the resistor would depend on 1) the impedance and 2) the efficiency of the headphones. So there's no specific ohm that would decrease the volume by 25 db for all headphones..... at least if you mean adding a single resistor in series with the headphone.

 

[PTheD]

Here is what I am thinking. the dB level that a resistance adaptor/volume attenuator is going to take a signal down is directly related to the impedence of the headphones that are at one end of it &to the voltage level coming out of the source of signal on the other side of it. If you have an RMS level of 1.000 volts out from the source & a set of headphones that are 16 ohms then this is what would happen. A 16 ohm adapter would cut the signal level in half, the SPL would be halved (Lets say with the source to the headphones at 1.000 volts it was 100 dB, with a 16 ohm adapter in the middle of the two that would bring the systems resistance up to 32 (which is a doubling in resistance which halves the signal level by 3 dB which is a halving in SPL (as you may or may not know the decible scale is logrithmic)), at 32 ohms the SPL coming out of the headphones would drop to 97 dB from the 100 that it was at before when you only had the 16 ohm headphones & the 1.000 volt RMS source. For 25 dB you just keep halving & doubling untill you get down 25 dB...hence for a 16 ohm headphone a 25 dB drop would be, I think, a 4080 ohm resistor would have to be tacked on into the signal path.

 

[PTheD]

Quote:

Originally Posted by wang228 i'm not an expert in electronics...and on a hight dose of alcohol right now... but i think the value of the resistor would depend on 1) the impedance and 2) the efficiency of the headphones. So there's no specific ohm that would decrease the volume by 25 db for all headphones..... at least if you mean adding a single resistor in series with the headphone.

If I am not mistaken the efficiency of the headphones does not need to be takin into account because it will remain the same no mater how much resistance or lack there of that is put in the signal path before the signal gets to the headphones. If the headphones are at an efficiency of 100 dB/mW before you add a resistance adaptor they are still going to be 100 dB/mW after, its just that they will be getting less power due to the resistance of the adaptor, the headphones efficiency doesn't change at all. That is how I understand it.

 

[wang228]

oops.. i think PTheD is right ^_^ Suppose the headphone effiiciency doesn't vary with the absolute voltage or mW applied, simple voltage dividing should be good. i'm not sure why i mentioned efficiency as a factor in the first post. please pardon my carelessness.

 

[PTheD]

Quote:

Originally Posted by wang228 oops.. i think PTheD is right ^_^ Suppose the headphone effiiciency doesn't vary with the absolute voltage or mW applied, simple voltage dividing should be good. i'm not sure why i mentioned efficiency as a factor in the first post. please pardon my carelessness.

Oh don't do that to yourself! You weren't careless, you just said something that we now both think was incorrect. A simple mistake, that for all of two people being ignorant of the complexitys of electrical mathamatics is only that little bit away from being correct. Of little importance such trivialitys are. (And Yoda I like to talk like

)

 

[wang228]

haha.. thanks PtheD.

 

[PTheD]

ahhhh, you bet. I am one of those easy going sorts. Welcome to the forum btw. I would have sent you a PM but my inbox is getting so full with this Shure E4/E4C canalphone buisness I thought I would drop you a "ho do you do" on this thread instead.

 

[Jmmmmm]

Quote:

Originally Posted by PTheD For 25 dB you just keep halving & doubling untill you get down 25 dB...hence for a 16 ohm headphone a 25 dB drop would be, I think, a 4080 ohm resistor would have to be tacked on into the signal path.

Hmmm....ok, the reason I ask is because the ue 5pro's come with a 'volume attenuator' that is supposed to lower the sound by 25db. While what you said makes sense, I doubt they include a 4080 ohm resistor. I was thinking it would be somewhere in the low hundreds, somewhere between 75 and 200, but I have no idea. Any other idea?

 

[mephisto]

25dB sounds like a lot of attenuation. Are you sure this thing is a resistor and not a mechanical filter to reduce ambient sound levels?

 

[PeterR]

Quote:

Originally Posted by PTheD A 16 ohm adapter would cut the signal level in half,

Yes.
Quote:

the SPL would be halved

No, it would be 1/4. Half the voltage means a quarter of the power.
Quote:

(Lets say with the source to the headphones at 1.000 volts it was 100 dB, with a 16 ohm adapter in the middle of the two that would bring the systems resistance up to 32 (which is a doubling in resistance which halves the signal level by 3 dB which is a halving in SPL (as you may or may not know the decible scale is logrithmic)), at 32 ohms the SPL coming out of the headphones would drop to 97 dB from the 100 that it was at before when you only had the 16 ohm headphones & the 1.000 volt RMS source. For 25 dB you just keep halving & doubling untill you get down 25 dB...hence for a 16 ohm headphone a 25 dB drop would be, I think, a 4080 ohm resistor would have to be tacked on into the signal path.

-25dB means 10^-2.5=0.056 times the voltage, so R would have to be ((1/0.056)-1)*16 Ohms, that's about 270 Ohms.

 

[hawkfire]

hello,

I think jmmm is confusing the variable pot performance on the attenuator with actual resistors.