Reducing output of pre-amp

qguy

Member
2005-03-14 7:25 am
Burnaby
I have 2 power amps being fed by a single pre-amp. The power amp does not have gain controls, I need to reduce the output of one of the power amps ? how do I do this ? Can I use a zobel network for this like what is done for tweeters ? if Yes, what values do I need to reduce the output by say 3,4,5 and 6 Dbs ? I was thinking of using a pot, but someone said it would change the frequency.

Other suggestions welcome as long as it cheap :)
 
So add an input level control to the loud amp and don't make alterations in the output section. If you don't have room in the amp or prefer not to make changes in the amp you can add a level controlled second output of the preamp. This could be a simple as a POT(entiometer) from your local supplier (Radio Shack) or if you know the number of dB down you want, a pair of resistors in a pad for each channel. Concept is identical to the pot but not easily adjustable.

The pot will have virtually no effect on frequency response if the impedance is kept low. That _could_ be a problem with tube preamps but wouldn't be a problem with solid state.

 

tinitus

Ex-Moderator R.I.P.
2005-11-24 1:47 am

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Mooly

Administrator
Paid Member
2007-09-15 8:14 am
Thanks

What do you mean by "The pot will have virtually no effect on frequency response if the impedance is kept low."

Using a Solid state pre and power amps

It means use a pot of around 10K resistance.

Why not just add a fixed attenuator to the inputs of the power amp. Two resistors per channel wired as a divider. You could do it directly on the back of the sockets.
 

qguy

Member
2005-03-14 7:25 am
Burnaby
It means use a pot of around 10K resistance.

Why not just add a fixed attenuator to the inputs of the power amp. Two resistors per channel wired as a divider. You could do it directly on the back of the sockets.


What is the disadvantage of using a pot vs a fixed attenuator ?

Can you please provide the values of the resistors for say 3 - 6 db and which one is in series with the positive side and which one is across the negative and positive. I can not seem to find an idiots guide to calculating the resistor values on the web.

Thanks
 
Does this just choose a single resistor in series with the positive side of the signal ?

The single resistor solution WILL be a problem for response, possible hum pickup and not very accurate level.
See
Attenuator (electronics) - Wikipedia, the free encyclopedia)

Don't let the math scare you it's not nearly as nasty for simple audio attenuators. See the configuration of the "L" pad. If you made Rp and Rq both 4700 Ohms (common value) signal at the output would be 50% of the input (slightly over 6 dB down) Load to the preamp would be around 9500 (varies with the amplifier input impedance) and the output impedance to the amp would be around 2300 Ohms (varies with the source drive impedance of the preamp). Changing the RATIO of the resistors while keeping the SUM of the values constant will change the attenuation.

A "POT" is one big resistor that continuously varies the 'tap off' point. If the math really scares you you could do the super cheesy method of adjusting the pot to the level you like and the measure the resistance from slider to top' (signal input) and slider to 'bottom' (ground/earth) and buy those values of fixed resistors and substitute those for the adjustable pot. You would NOT be the first person to do this.

 

Mooly

Administrator
Paid Member
2007-09-15 8:14 am
What is the disadvantage of using a pot vs a fixed attenuator ?

Can you please provide the values of the resistors for say 3 - 6 db and which one is in series with the positive side and which one is across the negative and positive. I can not seem to find an idiots guide to calculating the resistor values on the web.

Thanks

The fixed resistors are simple, cheap, reliable and absolutely the best way of providing an attenuator.

I can tell you from experience that you won't know what level of attenuation you need untill you actually try it.

The resistors are wired in series. One end goes to ground, The other end goes to the input socket. The input connection to the amp now goes to the junction of the two resistors.

Try something like 6K8 and 2K2 for a start (use what you have as long as they are below say 20K) The 2K2 would be at the ground end.

Just for you,
calculators for electronic circuit design