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Replace audio pot with stepped attenuator - Need help!
Old 1 week ago
  #1
Gear Nut
 

Replace audio pot with stepped attenuator - Need help!

Hello there,

I try to replace the 1178's pots with stepped attenuators. The 1k input pots are wired with the screen to the pot's input and the two signal wires to the tapper and the out of the pot. So I understand that by this the pot works opposite direction: I turn up the pot (input) and the resistance lowers.

How would I replicate this behaviour with a stepped attenautor?

I bought single resistors in the row that it matches the pot inout and output, so rising values. But I'm afraid I should have gotten the values of the substracted values between tapper and output right?

Thanks for any help.
Attached Thumbnails
Replace audio pot with stepped attenuator - Need help!-img-9563.jpg  
Old 1 week ago
  #2
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brianroth's Avatar
 

FWIW, the red wire from the gray cable with the "1" sticker/label is the signal INPUT for the pot, the red wire from "2" is the wiper (hence OUTPUT from the pot).

The two bare screening wires are connected to the "low" (grounded) end of the pot which will be "off", with no signal to the wiper when the pot is fully counter-clockwise. I verified this via the photograph posted and the UREI 1178 service manual.

You don't say how many steps you desire for a stepped attenuator, nor how many dB per step. So, no way to make any specific suggestions.

Bri
Old 1 week ago
  #3
Gear Nut
 

Thanks Brian,

I got Elma 04 switches with 24 positions of which I use 22 to make the rotation less full circle. My question is more about the logic of where to put the wires.

I now have single resistors from zero to 1k. I found the values by getting a logarhythmic scale to 1k with 22 steps.

Quote:
Originally Posted by brianroth View Post
FWIW, the red wire from the gray cable with the "1" sticker/label is the signal INPUT for the pot, the red wire from "2" is the wiper (hence OUTPUT from the pot).

The two bare screening wires are connected to the "low" (grounded) end of the pot which will be "off", with no signal to the wiper when the pot is fully counter-clockwise. I verified this via the photograph posted and the UREI 1178 service manual.

You don't say how many steps you desire for a stepped attenuator, nor how many dB per step. So, no way to make any specific suggestions.

Bri

Last edited by shahstlz; 1 week ago at 11:06 AM.. Reason: changed 21 to 22 steps...
Old 1 week ago
  #4
Gear Nut
 

So what you say is that I should put "1" to the input of the switch, the "2" to the end of the resistors (so that it grabs the switched value), the screens are grounded.

But as I understand "1" and "2" don't grab the risng resistor value when turning the pot clockwise, but they grab the lowering value. Is that correct?

Quote:
Originally Posted by brianroth View Post
FWIW, the red wire from the gray cable with the "1" sticker/label is the signal INPUT for the pot, the red wire from "2" is the wiper (hence OUTPUT from the pot).

The two bare screening wires are connected to the "low" (grounded) end of the pot which will be "off", with no signal to the wiper when the pot is fully counter-clockwise. I verified this via the photograph posted and the UREI 1178 service manual.

You don't say how many steps you desire for a stepped attenuator, nor how many dB per step. So, no way to make any specific suggestions.

Bri
Old 1 week ago
  #5
Gear Nut
 

Maybe that helps with communication. Here's a pic of the Elma switch with the yellow clamp on the input and the red on the output.
Attached Thumbnails
Replace audio pot with stepped attenuator - Need help!-img-9620.jpg  
Old 1 week ago
  #6
Lives for gear
 

Quote:
Originally Posted by shahstlz View Post
Thanks Brian,

I got Elma 04 switches with 24 positions of which I use 21 to make the rotation less full circle. My question is more about the logic of where to put the wires.

I now have single resistors from zero to 1k. I found the values by getting a logarhythmic scale to 1k with 21 steps.
Do you mean that the individual resistor values range from zero ohms to 1k? If this is meant to be a constant input impedance attenuator that matched the impedances of the original pot, the total resistance of the ladder should be 1k, but none of the individual resistors would be 1k.

If you meant that the total output impedance ranges from zero to 1k and were not referring to individual resistor values then it sounds like you've done it correctly.

Ideally, the input impedance of the control will be 1k at any setting. The output impedance will range from zero ohms to 1k. This is achieved by using a series of resistors whose values total 1k.

Last edited by David Kulka; 1 week ago at 06:08 PM..
Old 1 week ago
  #7
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Quote:
Originally Posted by shahstlz View Post
Maybe that helps with communication. Here's a pic of the Elma switch with the yellow clamp on the input and the red on the output.
The assembly in the photo is not wired as a ladder attenuator. Since all the resistors are bussed at one end, it looks more like an EQ resistance selector, a feedback resistance selector, or a shunt type attenuator. I just want be sure the OP understands that a constant impedance attenuator for an 1178 would not be built that way.

In the photo it looks like the yellow lead is connected to the wiper of the rotary switch. If the OP is wiring the control as a constant input impedance ladder attenuator that would be the output, not the input.

To replace the 1178 input pot, the attenuator doesn't absolutely need to be constant input impedance. Personally, I'd wire it like that because I'm fussy that way, and I'd be confident that I'm not changing the performance of the input circuit at all. But, as Brian said in the first reply, we don't know what the OP is aiming for, so it's hard to give a totally correct answer.

Last edited by David Kulka; 1 week ago at 12:19 AM..
Old 1 week ago
  #8
Gear Nut
 

Yeah, that's what I just thought. I need a ladder. So for the 1178's input pot in 21 steps that would be:

0
0,0477
0,4288
1,356
2,9326
5,2349
8,3246
12,2547
17,0717
22,8178
29,5312
37,247
45,999
55,817
66,73
78,766
90,852
107,41
121,868
138,646
156,665

That's for a log audio 1k pot as it is in the Urei. The aim is to replace high tolerance left right pots with stepped switches with low tolerance Rs to make compression on stereo signals easier to adjust.

But does the wiring work with a Elma 04 switch?

Quote:
Originally Posted by David Kulka View Post
The assembly in the photo is not wired as a ladder attenuator. Since all the resistors are bussed at one end, it looks more like an EQ resistance selector, a feedback resistance selector, or a shunt type attenuator. I just want be sure the OP understands that a constant impedance attenuator for an 1178 would not be built that way.

In the photo it looks like the yellow lead is connected to the wiper of the rotary switch. If the OP is wiring the control as a constant input impedance ladder attenuator that would be the output, not the input.

To replace the 1178 input pot, the attenuator doesn't absolutely need to be constant input impedance. I'd wire it like that because I'm fussy that way, and I'd be confident that I'm not changing the performance of the input circuit at all. But, as Brian said in the first reply, we don't know what the OP is aiming for, so it's hard to give a totally correct answer.
Thanks for your help!
Old 1 week ago
  #9
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brianroth's Avatar
 

Check out this page:

https://www.electronics-tutorials.ws...ttenuator.html

Scroll down a little bit to the section "Simple Passive Attenuator" which shows the schematic for a stepped attenuator used to replace a simple potentiometer. Note that the resistors are NOT "bused", but go "loop-de-loop" from one switch contact to the next.

Bri
Old 1 week ago
  #10
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It would look something like the section of this switch that is closest to the shaft:

Old 1 week ago
  #11
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brianroth's Avatar
 

Thank you David!! I did a bit of looking for a photo like you posted, but didn't find anything nearly that clear to describe the proper wiring of the resistors.

Bri

Last edited by brianroth; 1 week ago at 06:02 AM..
Old 1 week ago
  #12
Gear Nut
 

Ok, I think I got the basic logic. Still the numbers to select are not really clear to me. Let's start like this:

I got a 24 position (15 degree each) switch. I will only use 22 to have 45 degree of 'dead space' if you know what I mean. That would be more like a usual pot.

To get the numbers and the correct logarithmic response I measured the pot value in middle position, which for all pots was about 10% of the full value. So I got three values to start:
- pot at full left: 0
- pot at middle 100 Ohms
- pot full right: 1000 Ohms

Based on this a math teacher friend of mine came up with this graph and these numbers that I have attached here (he did values from 0 to 10 though).

Except for he did it for 21 positions that seems like a good start. If I translate this to ladder values it's like in my post #8.

Now the problem I see is that five values are below 5 Ohms, which is basically not useful nor available as resistor values.

Does anyone know how to get useful values to replace these standard audio 1k pots?
Attached Thumbnails
Replace audio pot with stepped attenuator - Need help!-2019-11-10-10.55.45.jpg  
Old 1 week ago
  #13
Lives for gear
 

Welcome to the world of engineering. You need to make decisions, some maybe by trial and error. You need to weigh the accuracy and specs you want against the parts that are available and the time you're willing to spend on this.

The numbers on your table look about right to me. Some resistor values below 5 ohms are indeed available. You can get custom values with combinations of resistors. You may decide that some steps or levels at the CCW end of the control aren't actually needed. You might decide to play with the linearity of the control positions to accommodate the parts that are available. And don't get hung up on the accuracy of the resistor values -- 22.6 ohms will be fine in place of 22.8178. As long as the two channels match, precise db steps aren't super important.

Last edited by David Kulka; 1 week ago at 06:26 PM..
Old 1 week ago
  #14
Gear Nut
 

I'll try it with the values that I have. Thanks for your helps. I'll get back to this thread with the results.

Quote:
Originally Posted by David Kulka View Post
Welcome to the world of engineering. You need to make decisions, some maybe by trial and error. You need to weigh the accuracy and specs you want against the parts that are available and the time you're willing to spend on this.

The numbers on your table look about right to me. Some resistor values below 5 ohms are indeed available. You can get custom values with combinations of resistors. You may decide that some steps or levels at the CCW end of the control aren't actually needed. You might decide to play with the linearity of the control positions to accommodate the parts that are available. And don't get hung up on the accuracy of the resistor values -- 22.6 ohms will be fine in place of 22.8178. As long as the two channels match, precise db steps aren't super important.
Old 1 week ago
  #15
Gear Addict
 
samwinston123's Avatar
 

Quote:
Originally Posted by shahstlz View Post
Ok, I think I got the basic logic. Still the numbers to select are not really clear to me. Let's start like this:

I got a 24 position (15 degree each) switch. I will only use 22 to have 45 degree of 'dead space' if you know what I mean. That would be more like a usual pot.

To get the numbers and the correct logarithmic response I measured the pot value in middle position, which for all pots was about 10% of the full value. So I got three values to start:
- pot at full left: 0
- pot at middle 100 Ohms
- pot full right: 1000 Ohms

Based on this a math teacher friend of mine came up with this graph and these numbers that I have attached here (he did values from 0 to 10 though).

Except for he did it for 21 positions that seems like a good start. If I translate this to ladder values it's like in my post #8.

Now the problem I see is that five values are below 5 Ohms, which is basically not useful nor available as resistor values.

Does anyone know how to get useful values to replace these standard audio 1k pots?
Here is an excel spreadsheet that will give you the correct E96 resistor values for a series attenuator. You need to fill out all of the yellow boxes and the green coumn are the actual resistor values you need. R TOTAL is the total resistance of the stepped control (1k in your case). R LOAD is the impedance of the load attached to the wiper. The ATT dB column are your target attenuation values from left to right.

https://www.dropbox.com/s/o1n579bz6s...ator.xlsx?dl=1

Last edited by samwinston123; 1 week ago at 01:48 AM..
Old 1 week ago
  #16
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Paul Gold's Avatar
I don't trust my electronics or math skills. When faced with this situation I usually replace the pot with a multi turn trimmer. I then adjust the trimmer for the steps I want in sequence and measure the trimmer for each position.
Old 1 week ago
  #17
Gear Nut
 

Great, thanks!
How do I know R LOAD?

Quote:
Originally Posted by samwinston123 View Post
Here is an excel spreadsheet that will give you the correct E96 resistor values for a series attenuator. You need to fill out all of the yellow boxes and the green row are the actual resistor values you need. R TOTAL is the total resistance of the stepped control (1k in your case). R LOAD is the impedance of the load attached to the wiper. The ATT dB row are your target attenuation values from left to right.

https://www.dropbox.com/s/o1n579bz6s...ator.xlsx?dl=1
Old 1 week ago
  #18
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Paul Gold's Avatar
Quote:
Originally Posted by shahstlz View Post
Great, thanks!
How do I know R LOAD?
You don't need to know R load if you use a multi turn trimmer of the same value as the pot. I suggest a multi turn trimmer because the resolution for measurement will be better.

To know R load you would need to look at the schematic. It will generally be the value of the next R shunt or a series R if it's at the input virtual ground of an opamp. Sometimes it's complex because of circuit complexity. I don't trust myself to figure that out unless it's straightforward.
Old 1 week ago
  #19
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samwinston123's Avatar
 

Quote:
Originally Posted by shahstlz View Post
Great, thanks!
How do I know R LOAD?
On the 1178 input the load is 4.7k in series with the gain control FET, so I would just use 4.7k. Since the input control is only 1k the load isn't going to have a huge effect on the values.
Old 1 week ago
  #20
Gear Nut
 

What I don't understand: Why do I need to know the load, if I just replace an existing pot in the circuit? The pot measures around 100 Ohms at middle setting. So the rotary switch should have 100 Ohms in that position. From my understanding that is basically it - just for the whole range from 0 to 1000 Ohms wired as a ladder.

What part of the logic am I missing here?
Old 1 week ago
  #21
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Besides its actual setting, the amount of attenuation at the each step depends on the source impedance at the input and the load impedance at the output. To use the old comparison to plumbing, picture a water faucet connected to a long thin hose. The volume of water coming out of the end of the hose at different faucet settings depends on the source pressure at the input of the faucet, the setting of the faucet itself, and also the diameter (and vertical angle) of the hose connected to the output -- the "load" on the output of the faucet.
Old 1 week ago
  #22
Gear Addict
 
samwinston123's Avatar
 

Quote:
Originally Posted by shahstlz View Post
What I don't understand: Why do I need to know the load, if I just replace an existing pot in the circuit? The pot measures around 100 Ohms at middle setting. So the rotary switch should have 100 Ohms in that position. From my understanding that is basically it - just for the whole range from 0 to 1000 Ohms wired as a ladder.

What part of the logic am I missing here?
If you're simply trying to recreate a logarithmic control then you don't need to know the load. In that case just set the load as 1,000,000 or something and have uniform gain steps.

If you are trying to get specific and accurate gain steps then you do need to know the load because it will alter the result. The real advantage of the stepped control is that you can tailor the steps to your needs and target the most usable range

I just realized that the formula to calculate the E96 values doesn't work for values under 1 ohm. I'll update it later today.
Old 1 week ago
  #23
Gear Nut
 

Yes. that's the thing: The input control of the 1178 is absolutely okay. I want it that way, just stepped.

Thanks for clarification. I gues I have to experiment a bit.

Quote:
Originally Posted by samwinston123 View Post
If you're simply trying to recreate a logarithmic control then you don't need to know the load. In that case just set the load as 1,000,000 or something and have uniform gain steps.

If you are trying to get specific and accurate gain steps then you do need to know the load because it will alter the result. The real advantage of the stepped control is that you can tailor the steps to your needs and target the most usable range

I just realized that the formula to calculate the E96 values doesn't work for values under 1 ohm. I'll update it later today.
Old 1 week ago
  #24
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Paul Gold's Avatar
Quote:
Originally Posted by shahstlz View Post
Yes. that's the thing: The input control of the 1178 is absolutely okay. I want it that way, just stepped.
.
You could get a stepped pot as opposed to a switch. That would save a whole bunch of work. They are not as accurate as a switch but may be close enough. If you wanted to reduce the range of the controls for finer control this wouldn't be an option.
Old 1 week ago
  #25
Gear Nut
 

The only reason to do all this, is to have left and right input changes quick and accurate between the two channels. So it's about having steps with low tolerance. Stepped pots don't give me the 1% tolerance of single resistors I guess.

Quote:
Originally Posted by Paul Gold View Post
You could get a stepped pot as opposed to a switch. That would save a whole bunch of work. They are not as accurate as a switch but may be close enough. If you wanted to reduce the range of the controls for finer control this wouldn't be an option.
Old 1 week ago
  #26
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Paul Gold's Avatar
Quote:
Originally Posted by shahstlz View Post
The only reason to do all this, is to have left and right input changes quick and accurate between the two channels. So it's about having steps with low tolerance. Stepped pots don't give me the 1% tolerance of single resistors I guess.
That's true but if you want to cover the whole range of the pot in 22 steps then the steps will be coarse. Isn't the input range of an 1178 something like 40dB? That's about 2dB per step which would be too coarse for most. If it's sitting on the stereo buss and you can use the master fader for fine adjustments then I could see it working.
Old 1 week ago
  #27
Gear Nut
 

Exactly, I can fine adjust the line input externally.

Quote:
Originally Posted by Paul Gold View Post
That's true but if you want to cover the whole range of the pot in 22 steps then the steps will be coarse. Isn't the input range of an 1178 something like 40dB? That's about 2dB per step which would be too coarse for most. If it's sitting on the stereo buss and you can use the master fader for fine adjustments then I could see it working.
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