I have a couple of questions related to the alignment procedure for the amplifier of Klipsch SW-12, a powered subwoofer.

Its service manual (link below) says the following:

(page 3)


The compression circuit consists of U5, U6, and Q28 and is adjusted via R30. If adjusted correctly, this circuit will limit the amplitude of the signal so the minimum amount of distortion in the form of clipping will occur at the output of the amplifier. R34 is used to set the maximum output level for the amplifier so it doesn’t underamplify or clip. Next, the signal goes through a buffer amplifier U2, which provides a gain of 4. Then the signal goes on to the driver circuits.

The alignment procedure (page 5) is the following:


Equipment required:
A signal source capable of supplying a 30Hz sine wave at 300mVrms.
A true RMS Voltmeter such as the Fluke 8060B.
A 16 ohm load rated for at least 200 Watts.
An oscilloscope (optional).

To totally align SW12 series 1 amplifiers, follow this procedure:
1. Disconnect power from the UUT (unit under test).
2. Connect the UUT (unit under test) to a 16 ohm load.
3. Connect a signal generator to the RCA input of the amp.
4. Set all controls on the UUT to their full clockwise position.
5. Set the signal generator for 30Hz and 60mVrms output.
(be sure and measure the output of the generator for 60mVrms.)
6. Connect the voltmeter leads to the output of the UUT.
7. Apply power to the UUT.
8. Adjust R34 for 33Vrms. Range is from 32.1 Vrms to 34 Vrms.
9. Change signal level to 1.5mVrms @ 30Hz .
10. Measure the output voltage. Should be between .94 Vrms and 1.06Vrms.
Adjust R30 if necessary.
* NOTE: Some interaction between adjustments is common. Recheck steps 8 and 10 for proper voltages.
11. Alignment of the UUT is now complete. Disconnect power and other connections from the UUT.


  1. I don't have a 16-ohm load. I want to use an 800-watt space heater that is 17.5 ohm. What should the target voltages for R34 and R30 be? Do I need to change them? Or should they remain the same?

  2. As to adjusting R30, to make the compression lower (to make less compression), should the output voltage be raised or lowered?

================================= The service manual for SW-12 is here:


Update (9/28/2018)

Accepted Tony EE rocketscientist's answer and built a 186.23-ohm resistor, consisting of 4 of the 16 750-ohm power resistors I had had since I started working on the subwoofer last year. Their not being exactly 750 ohms helped.

Their values and the combined (calculated) resistance:






a resistor with 4 750-ohm resistor

resistance of the above resistor

The space heater's resistance with/without the above resistor

resistance of the space heater

resistance of the space heater with the above resistor


1 Answer 1


Rev A

it could be a 220Ohm 10 Watt resistor if it is stable. An audio power Amp is just a voltage source near 0 Ohms ( 10mOhm ish)

Old answer

  1. You need 187 Ohm 20W to make 16 Ohms

Figure out what's easiest for you.

e.g. 100x 1/4W axial resistors on a "Digi-reel" shunted in parallel 18.7k Ohm

or one 20W resistor or whatever...

  1. For 30Hz 1.5mVrms input, output "Should be between .94 Vrms and 1.06Vrms." Thus CW rotation increases signal input and more compression occurs in the log amp.

    • So turn CCW slightly. to the minimum level of this range.
    • Going outside the range, of course, increases risk to cone-wire overtravel stresses or excess compression.
  • \$\begingroup\$ If the 17.5 W space heater was all I got, what would be the best values of the output voltage to seek? Unless I get a 16-ohm load, is the above alignment procedure meaningless? \$\endgroup\$
    – zeron
    Commented Sep 6, 2018 at 18:29
  • \$\begingroup\$ ideally the Amp has an output impedance <1% of load and thus may be current limited with 4 Ohms. so 16 or 8 Ohms wont matter unless you happen to have a 4 Ohm woofer. it could be a 220Ohm 10 Watt resistor I think. It still is the same voltage \$\endgroup\$ Commented Sep 6, 2018 at 19:25
  • \$\begingroup\$ Zout is 10 milli Amish lol or less \$\endgroup\$ Commented Sep 6, 2018 at 21:41
  • 1
    \$\begingroup\$ Thanks, Tony EE rocketscientist, for your answers and comments. \$\endgroup\$
    – zeron
    Commented Sep 29, 2018 at 7:21

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