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Old 16th October 2002, 11:30 AM   #1
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Angry Ladder attenuator: Where to solder resistors

I have been "inspired" to build a ladder attenuator for my active pre-amp. I have a Seiden 23 position 4 deck 4 pole shorting switch with no instructions on where to solder resistors. Before I order resistors I would like to know what it is I am to do with them. There are four bands of 23 circular solder tags (one around each deck) to receive one end of each resistor. The question is: "What do I connect the other end of each resistor to?"

I can't see anything else which the resistors might connect to save for a single tag on each deck which connects the wiper to the resistor. Having gone through this experience, have you any clue? A picture of a 2 deck version of my 43NEG switch appears at: http://www.tachyon.co.jp/seiden/seiden.html

I don't have any problem with working out the resistors I need as I am using the Goldpoint chart for 24 positions and leaving out one position to make up a 64db 100k attenuator. Would I be correct in thinking I need 92 resistors?


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Old 16th October 2002, 01:01 PM   #2
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These switches are very open. You must be able to see the wiper tag. Do you really have 23 solder tags, not 24? One of the contact spring may be longer than the rest.
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Old 16th October 2002, 02:07 PM   #3
Kermit is offline Kermit  Norway
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Judging from this: http://www.goldpt.com/schm_ml.html I'd guess you connect all the other ends of the resistors together and connect it to the input / last position on one of the deck and to ground / last position of the other one. Then connect the wipers together and take the output from there.
Wouldn’t 23 positions leaving the last one open require 22*2 resistors? 88 total for 2 channels?
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Old 17th October 2002, 06:58 AM   #4
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Thank you Peranders & Kermit. I think it is getting clearer.

The wiper moves over 23 positions. Each has a single solder tab. In between either end of the wiper's sweep, between positions 1 and 23, is a position over which the wiper cannot rest. I will call it "position 24". It has a large gold solder tab.

Kermit as I understand it, for each channel, I need a loop of wire connecting each resistor around each deck together. On the first deck, one end of the loop is connected to input and the other to output. On the second deck the loop similarly connects resistors to each other, and one end of the loop is connected to ground(but what is the other end connected to?). Then I join the two decks together using position 24. Does this sound correct?

Number of resistors: 23X4=92

Thanks again guys.
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Old 17th October 2002, 04:36 PM   #5
Kermit is offline Kermit  Norway
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Quote:
Kermit as I understand it, for each channel, I need a loop of wire connecting each resistor around each deck together.
Yup, but you don’t need a wire, just connect the ends of the resistors together.

Quote:
On the first deck, one end of the loop is connected to input and the other to output.
Here I believe you are wrong. Look closely at the link I posted. You will se that the resistors are connected to input on the first deck and the last position of the switch. You take the output at the wipers (“position 24”).

Quote:
Then I join the two decks together using position 24
Yup, and output comes here too.

Hope this clarifies things a bit. Good luck.
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Old 20th October 2002, 08:55 PM   #6
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Default L TYPE ATTENUATOR

Setriode,

I hope you haven't started your soldering yet since there may be a few things you need to consider.

-What are you driving with this volumecontrol?
A tube stage? An interconnect to your amplifier?And what input impedance does it have?

-You need to know the input impendance of the next stage and it's capacitance.

-There is no need for that many resistors,you can use a voltage divider with 1 series R per step keeping the grid to ground R constant.

-Indeed : pos. 24 is a stop and your signal input.

-Also it is a lot easier to work on if you had two rotaries of the single pole single deck make before break type.In which case you can than group all R together at the outputleads and connect your output wire there.

You state 4 deck 4 poles? I assume 4 decks with one pole per deck all poles in the same position?

You see if you vary the shunt R you also alter the polarisation of the next stage and the output impedance.
As a consequence of this the freq.responce will vary depending on the volume setting.
This is ofcourse not desirable.

Best regards,
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Old 20th October 2002, 10:28 PM   #7
dice45 is offline dice45  Germany
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Frank,
welcome back.

setriode,
before you start designing, consider that if you manage to get your attenuator having an impedance of 600Ohm (instead of, say, 50kOhm) working, sounds muuuch nicer. Just, linestage has to handle full swing all the time.
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Old 20th October 2002, 10:45 PM   #8
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Hi,

Thks,Bernhard.Sure feels good to be back.

You have me wondering with that 600 R attenuator now.
I've been tinkering with that too together with LC RIAA correction.
But short of buffering both in- and output with transformers or using a variable transformer and rotary switch I can't find an easy solution.
Mindboggling....

Cheers,
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Old 21st October 2002, 12:58 AM   #9
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setriode,

The answers to your questions depend on which type of switch, you have, exactly, and whether you want to build a single ended or balanced attenuator, and have that implemented in either series, ladder or shunt mode.

A breif synopsis:

<b>Series</b>
+ only one switch contact directly in the signal path.
+ fewest number of resisitors
+ constant Zin
+ only needs two decks
- IME worst sounding, esp as you go to greater attenuation because of all the resistors and solder joints in the path.

<b>Ladder</b>
+ only two resistors in the signal path
+ constant Zin
- two switch contacts in the signal path at all times
- 4 dechs and twice the number of resisitors required than a series.
+ IME much better sounding than a series.

<b>Shunt</b>
+ One resistor in the series signal path, so can be very high quality like Vishay S102
+ two decks and same number of resistors as series.
+ all other resistors are in the shunt path so are less important sonically.
? Zin will vary from a max at lowest attenuation to a min value of the series resistor at max att. If your pre can drive a modest value series resistor it should not be a problem. eg Pass pre's, some tube gear won't like it at all, esp the "traditional" crap with no drive capability at all.
+ usually best sounding of the resistive att's if your pre can drive it.

All the schematics for these are on the Goldpoint page, <a href="http://www.goldpt.com/diy.html">here</a> as are the resistor lists.
46 - series
96 - ladder
46 - shunt

Having heard all of them, the S&B transformer volume controls are <i>vastly</i> superior to all the resistive types I've tried.

HTH
Cheers




PS: Frank. Good to see you back. Your sin-binning was a travesty.
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Old 21st October 2002, 01:56 AM   #10
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People,

"Shunt
? Zin will vary from a max at lowest attenuation to a min value of the series resistor at max att. If your pre can drive a modest value series resistor it should not be a problem. eg Pass pre's, some tube gear won't like it at all, esp the "traditional" crap with no drive capability at all."

And that is where the problem is:drive capability.
This is especially true when incorporating in an existing design as an "upgrade".
The culpritt is that nine out of ten will not have interstage isolation (buffering).
Hence my suggestion of keeping RgIn constant and only vary the series R.
Any frequency response deviation then is only caused by capacitance (which should be kept at a minimum).
Another important point is to keep in mind is that Vg- for the next stage is often derived by RgIn + Rk,the shunt resistor in // with RgIn will alter this operating point for all but one position namely where series R and RgIn are identical in value (6dB att.)
By using dual mono rotaries and a single deck one can arrive at excellent results provided a good cathode follower is driving the interconnect to the amp with low Z.
Moreover two rotaries in a dual mono config. will allow for balance adjustment provided the steps are not too coarse.
Or, if it must be a double deck single rotary you can simply tie the resistor output leads from each deck and take it from there to the cf.(IMHO only the White is good enough.)

Using it this way will also eliminate the expense of a second deck and avoids yet another switch contact in series with the signal.
At the same time it will then also isolate the attenuator from the load the amp see which is kept constant.

Another way to sonic bliss is to use the good old cermet linear pot with a law faking network.
Beats all log.pots I've heard so far.
The trick is to put a system together taking all possible factors into account where you design for minimal drive requirements at normal listening levels.
Not an easy task!


Quote:
Having heard all of them, the S&B transformer volume controls are vastly superior to all the resistive types I've tried.
Yep,this boy is saving up for a custom made one with my own custom made silver wire.
Any bankers among you?

Cheers folk,
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