Tech Square Technical and speculative discussion of amps, cables and other topics.

---

[ Follow Ups ] [ Post Followup ] Thread: [ Display  All  Email ] [ Tech Square ]

---

[ Alert Moderator ]

---

Re: Attenuation with PLLXO?

Posted by beermanpete@socal.rr.com (A) on March 27, 2007 at 22:53:55

In Reply to: Attenuation with PLLXO? posted by ggrdos@hotmail.com on March 23, 2007 at 16:29:06:

You need to make a resistive divider and place it at the input to the amplifier. To calculate the resistors needed you need to know the input impedance of the amplifier and the attenuation needed.

The formulas are:

db/20
R2 = Rt(10 )

R1 = Rt - R2

R2A = 1/((1/R2)-(1/Rt))

where
Rt = input impedance of amp
R1 = series arm of divider
R2 = shunt arm of divider
R2A = resistor in parallel with original Rt to achieve R2
db = attenuation

Example

Rt = 100k ohms
db = -3

-3/20
R2 = 100,000(10 ) = 70,795

R1 = 100,000 - 70,795 = 29205

R2A = 1/((1/70,795)-(1/100,000)) = 242,402

This will allow you to add two resistors to the amplifier, leaving the original input resistor. Reversing the change is easy if you don't like it or want to change it later.

Use resistors that you like the sound of that have a tolerance of 1 percent or better.

This post is made possible by the generous support of people like you and our sponsors:

VH Audio

---

[ Top of Page ] [ Contact Us ] [ Support/Wish List ] [ Copyright © 2025, Audio Asylum, all rights reserved ]

---

[ General ] [ Speakers ] [ Tubes ] [ Vinyl ] [ Digital ] [ Hi-Rez ] [ Video Asylum ] [ Cables ] [ Tweaks/DIY ] [ Music ] [ Films ]

---