pre-attenuator

gh0st · 7719

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Offline gh0st

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on: March 22, 2012, 01:44:32 PM
Hi Guys,

I've been playing with using resistors to pre-attenuate the input on my Crack (volume control got too loud too quick) and I've hit a weird snag someone might have an idea on...

Following @Laudanum's suggestion I went over to http://www.goldpt.com/mods.html and decided to implement a -10db attentuation and try to maintain the 100k input impedance, which required a 68k and 47k resistor as a voltage divider before the volume pot.

For sheer convenience I installed them at the input RCAs - it had the desired effect on the volume control and measured 100k between the RCA centre pin and ground. Trouble was, there was a slight intermittent buzz in one of the channels, and the noise floor generally seemed to be higher than it had been. I guessed maybe putting the resistors next to the RCAs was increasing the onward input wiring's susceptibility to picking up noise.

So I moved them to the volume pot end... and the effect has been to lower the resistance between centre pin and ground to 36k. It seems to have fixed the noise issue, but has decreased the increased control I got with the volume dial, and I'm wondering a) why this has happened and b) is 36k an acceptable figure for input impedance?

In physical terms, I've soldered the ground end of the 48k resistors (Rp2 in the goldpoint diagram) to the wire that bridges between the grounds of the left and right pots, could this be the reason??

Dom V


Offline Laudanum

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Reply #1 on: March 22, 2012, 02:57:29 PM
Yes, but input to center post should still be right around 100K with volume control all the way down.   I think thats how/where the pre attenuation cicuit maintains the same input impedance (100K) as the pot.  I think that measured resistance will be different with voluime pot all the way up as well compared to no pre-atten.  I think you give up some maximum "volume" as a trade off for wider/better control.  But someone will correct me if Im wrong on any or all of this.  I messed around with resistors and a pot before I ever got the kit, just experimenting in anticpation of needing it.  I read what I could easily find and, admittedly, didnt understand it all.  So what Im giving you is what I think I remember of the very basics.  Better/more accurate explanation surely will be forthcoming.

Desmond G.


Offline Grainger49

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Reply #2 on: March 23, 2012, 01:06:35 AM
Dom,

Are you using the stock pots or a Goldpoint?

Moving the "pre-attenuator" to the volume pot should not make a difference except in noise picked up.  That is it should not lower the resistance between the center pin and ground.  It is the same circuit, only the location changed.  So if you did indeed get something different the circuit changed.

The input sees a 67k in series with the parallel combination of the 100k pot and the 47k.  And I'm not close to a calculator now so I think it is pretty close to the 100k that you started with.

What I am not clear about is the buzzing.  Does it happen intermittently when you are listening and not adjusting the volume?  Does it only happen when you adjust the volume or touch the Crack?  Is there a computer or phone in the vicinity? 



Offline gh0st

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Reply #3 on: March 23, 2012, 02:23:16 AM
I'm using a TKD ladder stepper, 100k.

The only difference between where they are now (next to the pot) and where they were before (next to the RCA)  I can possibly think is relevant is the location of the connection to ground of the 47k, but I still don't see why that would create such a dramatic change in the resistance.

Funny you should mention phones and computers - the buzzing was intermittant (not on or off, it varied in level as well) and it did vary depending on proximity to my phone, or me full stop. I could even vary it depends on how I sat in the chair next to it! It was definitely better when I was further away though. I want to fix the 100k input load first, then see if that comes back.

I think I'll temporarily rewire the connection of the 47k back to the RCA grounds, see if that makes a difference.


Dom V


Offline Grainger49

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Reply #4 on: March 23, 2012, 03:51:08 AM
Leave the phone in another room for a trial. 

It is odd that you have a higher noise floor with the resistors where I would prefer to put them.  This location allows the stronger signal to travel further into your Crack.  The higher the signal the lower background noise should be. 

Does this make sense?



Offline Laudanum

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Reply #5 on: March 23, 2012, 03:57:25 AM
Ghost ... In my best, non-techie understanding ... a 100K pot or stepped attenuator presents a constant 100K input impedance to a source.   Turning the volume knob doesnt change the actual input impedance seen by the source, it varies the series resistance applied to the signal and therefore the amount of attenuation.  If you wire up that pre-attenuation circuit with the resistors listed, the constant 100K input impedance is maintained.    Measuring from pin 1 (ground) to pin 3 (input), the 100K resistance shouldnt change when you rotate the shaft, it should remain constant.  That represents the input impedance.  Measuring from input (pin3) to the wiper (middle pin), varies when you rotate the shaft (varies resistance in series with the signal).  If you measure this with a 100K pot, from input to wiper, you should measure 100Kohms with the shaft all the way counter-clockwise and 0 ohms with the shaft all the way clockwise.  With the pre-attenuation circuit in place, you should still measure 100K with the shaft all the way counter-clockwise but you will now measure some resistance with the shaft all the way clockwise.  Im bad at this math and dont understand it completely so Im not sure what this resistance should be).  But anyway, that is the tradeoff ... with the pre-attenuation circuit you trade maintaining the constant, same 100K input impedance plus more tracking range for some maximum volume.  
Smarter folks will surely correct me if Im wrong.

Desmond G.


Offline gh0st

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Reply #6 on: March 23, 2012, 05:18:42 AM
I agree with most of what you've said, but I don't believe that the installation of the pre-attenuator circuit would mean that the input impedance alters when the volume dial is turned. Using the values I gave in my first post, the load as seen by the source should still be ~100k regardless of the position of the volume control.   

Dom V


Offline Laudanum

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Reply #7 on: March 23, 2012, 06:01:55 AM
I agree with most of what you've said, but I don't believe that the installation of the pre-attenuator circuit would mean that the input impedance alters when the volume dial is turned. Using the values I gave in my first post, the load as seen by the source should still be ~100k regardless of the position of the volume control.   

No, the input impedance doesnt/shouldnt change when you turn the volume knob.  But that measurement would be between pin 3 and pin 1 (input to ground), that should remain constant.   You stated that you measured from the center pin (wiper) to ground.  I dont think that changes in that measurment when turning the knob is indicate that the actual input impedance is changing. 

Desmond G.


Offline gh0st

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Reply #8 on: March 23, 2012, 06:04:52 AM
ahhhh I see the misunderstanding.... I meant the centre pin of the RCA, not the pot. Will try to be more clear in future.

Dom V


Offline Laudanum

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Reply #9 on: March 23, 2012, 06:45:13 AM
Ahhh, I got it now.  :-)

Desmond G.


Offline Grainger49

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Reply #10 on: March 23, 2012, 08:16:55 AM
I just calculated the 67k in series with the parallel combination of 100k and 47k, it is 98.9k.  Pretty close to the stock 100k number.

If the stepped attenuator is a dual ladder type you have combinations of resistance values at each step that add to 100k.  So the input sees 100k and the wiper feeds different voltages to the tube grid.
« Last Edit: March 23, 2012, 08:18:38 AM by Grainger49 »



Offline gh0st

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Reply #11 on: March 23, 2012, 09:14:31 AM
Thanks Grainger, for educational purposes, could I trouble you to show your working on that one?

Dom V


Offline Grainger49

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Reply #12 on: March 23, 2012, 09:22:48 AM
Rtotal=Rs + Rp

1/Rp=1/R1+1/R2=31.972k ohms
Above R1=100k and R2=47k

If your calculator doesn't do reciprocals do it this way:

Rp= R1*R2/R1+R2

So add the 68k to 31.972k you get 99.972k ohms.
« Last Edit: March 23, 2012, 02:49:52 PM by Grainger49 »



Offline gh0st

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Reply #13 on: March 23, 2012, 09:46:30 AM
thanks, wish I'd kept up more maths since leaving school...

I'm still super confused though, aren't I effectively putting a 68k resistor in series with the 100k of the attenuator's input (thereby summing them to 168k), and making that parallel with 47k? If that's so, then wouldn't it be:

Rtotal = 1 / (1/(100+68)) + (1/47)

which comes out to 36.7, which is pretty much what I'm now getting?

Dom V


Offline Grainger49

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Reply #14 on: March 23, 2012, 10:04:28 AM
Nope, the 100k of the attenuator is in parallel with the 47k.  That combination is in series with the 68k.

The combination of the 100k and 47k is a little below 47k.  That added back to 67k gives my result, unless I punched the keys wrong.

Let me see if I have this right.  The resistor in series with the signal, the first one is the 68k, and the one that goes to ground from the input of the attenuator is the 47k.   

Is that right?