Line Level attenuation

[habana]

Does anyone know the resister values needed to attenuate the CD line-in on my amplifier. I want to attenuate it by about 12dB.

Thanks.

 

[EC8010]

Do you know the input resistance of your amplifier?

 

[habana]

attenuation

Hi EC8010

I'm sorry I don't. Is that important? I was thinking along the lines of the Kimber attenuators, they are general purpose units so I understand.

 

[peranders]

Make a voltage divider with the ratio 1:4

=>

10 k + 3.3k these values will work if the input impedance is >10 kohm (most likely) and if not lower the values, 5.6k + 1.8 k maybe?

 

[habana]

Line attenuation

Thanks for the reply.

I've actually built a passive pre amp using stepped attenuators and so the CD output goes straight into the power amps. Does this make any difference?

Excuse my ignorance but in the example you gave (10K, 3.3K) which value goes on the line and which on the return to line?

Thanks.

 

[dhaen]

The 10k goes in series with the line. The 3k3 goes from across the input of the preamp.

 

[EC8010]

It helps if you know the input resistance. The resistor values suggested will work, but the input resistance loading your CD player will be roughly 10k. Most CD players use an op-amp output stage, and most op-amps produce higher distortion into lower resistance loads. Admittedly, 10k should be just outside that range, but if you can have a higher input resistance, why not? If you had known that the input resistance of your pre-amplifier was 10k (for instance), you could have simply added a 30k series resistor, and your new input resistance would have been 40k. You might prefer to keep the input resistance constant when adding attenuators, just in case the input resistance affects the sound.

 

[jan.didden]

atten

Don't forget that adding these largish resistors in series does worsen your signal-to-noise ratio (which will be worsened anyway by the attenuation).
I don't know any CD output stages (except possibly weakish tubed ones) that are hurt by a couple of kilo-ohms load. I would go even lower, 3k series and 1 k to gnd.

Jan Didden

 

[dhaen]

A final point...

The attenuator should be at the destination end of the cable.

Cheers,

 

[fdegrove]

Brrr...

Hi,

A final point...

All data given is way too vague but I understand D'haen's point when he suggested the attenuators at the end of the cable.

All in all a middle of the the road decision is called for here, vere to either end and you'll face a frequency abberation.

I agree with D'haen, best put the attenuator at the amp input.

Poor Q poor A I'm afraid,

P.S. Yet another case for transmissionline transfer theory.

 

[Aud_Mot]

A handy calculator for both balanced and unbalanced resistor pads

http://www.mcsquared.com/dbframe.htm

Clock on T and H-Pads link in the upper right hand corner of the page.

They have other good information on this site.

Enjoy,

Aud_Mot

 

[peranders]

A handy calculator for both balanced and unbalanced resistor pads
http://www.mcsquared.com/dbframe.htm
Clock on T and H-Pads link in the upper right hand corner of the page.

This was a good link but I think habana should stick to a simple voltage divider and also forget about the increased noise, we are talking about CD quality not 24-bit sound!

 

[habana]

Thanks for your contributions everyone.

The reason I want to attenuate the CD output is because I understand it will improve performance.

With this in mind, which do you think is the best option? Mindful of the fact that my electronic knowledge is somewhat hazy beyond Ohm's Law but I follow instructions well!

Thanks.

 

[fdegrove]

ATTENUATOR.

Hi,

The reason I want to attenuate the CD output is because I understand it will improve performance.

Not sure about that but here goes:

What you need is a high quality rotary switch such as Elma 23 positions, make before break.

In order not load down the CDP nor the amp I think it's probably best to use a series type attenuator here.

A 10K log would be fine, all resistors are normally soldered in series, you can do it differently as I usually do by adding the calculated values up per step and solder that value in individually...this avoids all the soldering contacts and resistors in series at higher attenuation setting but it requires precsion made resistors in a few cases.
In that case R2'= R1+R2 and so on.

Here's a sample:

Cheers,