10dB Trim Pot

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ruffrecords

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Nov 10, 2006
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Often there is a need to include a 10dB gain trim control in a signal chain. I needed one for a project recently. It is not hard to limit a linear pot to a 10dB range just by including a suitable value resistor in its ground lead but this leads to a cramped law for the pot with most of the gain changing in the second half of rotation (you get about 3dB in the first half and 7dB in the second half). What we really would like is 5dB at the centre point. You can achieve this with a slugging resistor connected from the wiper to the ground leg of the pot and adjusted to give 5dB loss in the centre. The math is not hard but painful and boring. It is made worse by the fact that pots have a tolerance. In my project I was using 10K pots but they all measured just over 11K (within the 20% tolerance allowed). This alters the value of both the resistors meaning you really need to calculate the value for each and every pot you use.

The attached schematic is annotated with hopefully easy formula to follow. R1 is the value of the pot. R2 is the resistor that sets the trim limit to 10dB. R3 is the slugging resistor that sets the mid point to 5dB. The formulae do give odd resistor values but I have found that using the nearest preferred value rarely makes more than 0.25dB difference.

For the really math challenged I have made a spreadsheet to work it all out for you. All you need do is feed in the measured pot value. I will post that separately.

Cheers

ian
 

Attachments

  • 10dB-Trim-Pot.png
    10dB-Trim-Pot.png
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Spreadsheet attached. Change extension to xlsx to import into a spreadsheet.

Cheers

ian
 

Attachments

  • 10dB trim Control.txt
    14.1 KB · Views: 35
ruffrecords said:
The attached schematic is annotated with hopefully easy formula to follow. R1 is the value of the pot. R2 is the resistor that sets the trim limit to 10dB. R3 is the slugging resistor that sets the mid point to 5dB.
The formulas for R2 and R3 are wrong, IMO.
Spreadsheet is ok.
 
moamps said:
The formulas for R2 and R3 are wrong, IMO.
Spreadsheet is ok.

You are quite right. The only one that was right was Rx!! For some reason I worked them out again instead of just referring to the spreadsheet.

Attached is a corrected schematic embedded in the original spreadsheet.

Cheers

Ian
 

Attachments

  • 10dB trim Control.txt
    24.9 KB · Views: 54
Hi Ian,

Any advice on how this can be converted to a different value? Say 20dB or 30dB of attenuation?

thanks
w.
 
warpie said:
Hi Ian,

Any advice on how this can be converted to a different value? Say 20dB or 30dB of attenuation?

thanks
w.

This scheme works well for 10dB but I have a vague memory of trying it for 20dB without success. For 20dB. first R2 in the original schematic needs to be changed to R1/9. R1 is the pot value so for a 10K pot, R2 needs to be about 1.1K .

You then need to arrange the slug resistor R3 to give 10dB at the midpoint. So:

(R2+Rx +0.5R1))/ (R2+Rx) = 3.16 and hence

R2 + Rx + 0.5R1 = 3.16R2 + 3.16Rx => 0.5R1 - 2.16R2 = 2.16Rx => Rx =(0.5R1 -2.16R2)/2.16

which makes Rx about 1.2K which in turn makes R3  equal to 1.6K

Now I rememeber the limitation of this method. With the pot at the top the total load is R2 (1.1K)in series with the parallel combination of R3 and the pot which is about 1.4K. So the total load with the pot turned right up is only 2.5K which is too much for my tube designs. That's why I took it no further. If your circuit is happy driving a 2.5K load then you can go ahead with the design as is. Alternatively you can scale up the resistor values to scale up the worst case load.

For 30dB, the loss is 30 times so R2 is 10K/29 = 345 ohms

Presumably we want 15dB loss at the centre which is about 5.6 times, so Rx will be:

Rx will be (0.5R1 - 4.6R2)/4.6 which is about 742 ohms which makes R3 about 871 ohms

As you can see, the worst case load in this case is even smaller at around 1146 ohms

Cheers

Ian
 
I am not a fan of slugging pots in production, while I have slugged outlier quad pot sections before to make them match better at 50%.

Bulk resistance for most pot elements are something like 20% so do the math, and measure your result.

JR
 
Thanks Ian, I'll give it a try and see how it behaves. Minimum load requirement is 600 Ohms so I suppose it should be OK having  20dB attenuation? 
 
JohnRoberts said:
I am not a fan of slugging pots in production, while I have slugged outlier quad pot sections before to make them match better at 50%.

Bulk resistance for most pot elements are something like 20% so do the math, and measure your result.

JR

Absolutely. As I mentioned in the original 10dB trimpot post you need to calculate the resistor values based on the actual measured value of the pot you are using.

Cheers

Ian
 
Ian, can this  be connected right before an output transformer or it'll mess up with the output impedance of the unit? 
 
warpie said:
Ian, can this  be connected right before an output transformer or it'll mess up with the output impedance of the unit?

It will mess it up. It is intended to be fed into a relatively high impedance load - has to be several times greater than 10K if you use a 10K pot.

However you can do a LO-Z version with a 2K pot as long as you stick to the 10dB version. Just divide all the resistor values by five.

Cheers

Ian
 
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