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Old 23rd January 2012, 02:05 PM   #1
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Default Frequency Response of WW Pots?

I have looked and looked and looked and I can't find much of anything about the FR of wire wound potentiometers. I assume this depends on the resistive value and other factors, but still, it is important information for me. I bought a couple of precision wire wound pots with the intention to use them as volume controls but cannot find this information. It can't be that bad, since variable L-pads used in passive loudspeaker crossovers have wire wound elements in them. Also, the resistors used in typical passive crossovers are wire wound types, but no one seems to specify the inductance of these resistors and L-pads. I read some of Bourns information about this and they say that the equivalent circuit for a wire wound pot for AC circuits is complex. They mentioned that they aren't good above a few kilohertz without specifying much else. I'm confused about this issue. I sent an email to ETI Systems, since they made the pots I bought and am waiting for a reply. Does anyone know this information?

The advantage that I can see of wire wound pots is that they have the lowest possible equivalent noise, always less than 100ohms regardless of the actual value for the pot.

Last edited by dirkwright; 23rd January 2012 at 02:10 PM.
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Old 23rd January 2012, 02:17 PM   #2
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I don't have any numbers, but they're vastly better than you might think. If you pull apart a Bournes ten turn pot, you'll find the diameter of the wire coil is quite small. That reduces the inductance. It's also wound around a piece of heavy copper wire. That kills the inductance ever further- look up the "tuning wands" used by old radio hands. One end was ferrous and would increase the inductance of a coil. The other end was brass or similar and would reduce the inductance. They're certainly good well beyond "a few kilohertz". I wouldn't use a wire wound pot in the feedback system of a wide bandwidth amp, due to stability concerns at the MHz level, but anywhere else they should be fine.
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Old 23rd January 2012, 02:25 PM   #3
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Thanks Conrad. I lack the proper test gear to figure this out myself, hence my questions. I'm encouraged by your remarks. The ones I have are 10Kohms, ETI Systems MW22B-3, general purpose precision wire wound potentiometers.
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Old 23rd January 2012, 06:01 PM   #4
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Quote:
Originally Posted by dirkwright
The advantage that I can see of wire wound pots is that they have the lowest possible equivalent noise, always less than 100ohms regardless of the actual value for the pot.
No, they have thermal noise from their resistance, just like any other resistance. They may have very low excess noise, like other metal resistors.
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Old 23rd January 2012, 06:17 PM   #5
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Quote:
Originally Posted by DF96 View Post
No, they have thermal noise from their resistance, just like any other resistance. They may have very low excess noise, like other metal resistors.
Bourns consistently lists all of their wire wound potentiometers as having the equivalent noise resistance (ENR) of 100ohms. You can look it up yourself. Maybe I'm mistaken, I don't know. They only have thermal noise.

Vishay say the same thing. Maybe I don't understand ENR, but here's a snap from a data sheet.
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Old 23rd January 2012, 06:28 PM   #6
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I can only assume they mean excess noise, perhaps from the sliding contact. Thermal noise is an unavoidable property of resistance.

Rethink: assume they mean ENR just like a valve, then 100R means the noise at the output is a bit higher than you would expect from a normal resistive attenuator at room temperature. The extra noise is equivalent to thermal noise from a 100R resistor at the input. Note that a resistive attenuator both reduces signal and adds noise.

Last edited by DF96; 23rd January 2012 at 06:32 PM. Reason: rethink
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Old 23rd January 2012, 06:41 PM   #7
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ENR definition: "A quantitative representation in resistance units of the spectral density of a noise-voltage generator, given by Rn = (Wn) / (kT0,) where Wn is the spectral density, k is Boltzmann's constant, T0 is the standard noise temperature (290 K,) and kT0 = 4. 00 × 10-21 watt-seconds. Note: The equivalent noise resistance in terms of the mean-square noise-generator voltage, e2, within a frequency increment, f, is given by Rn = e2/ (4kT0f. )"

I still don't understand what they mean.
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Old 23rd January 2012, 06:43 PM   #8
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Quote:
Originally Posted by DF96 View Post
I can only assume they mean excess noise, perhaps from the sliding contact. Thermal noise is an unavoidable property of resistance.

Rethink: assume they mean ENR just like a valve, then 100R means the noise at the output is a bit higher than you would expect from a normal resistive attenuator at room temperature. The extra noise is equivalent to thermal noise from a 100R resistor at the input. Note that a resistive attenuator both reduces signal and adds noise.
They don't provide those kinds of details.
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Old 23rd January 2012, 06:47 PM   #9
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The ENR is added to the output terminal 2, which is the wiper. Still, I believe it's far lower for wire wound than other types of pots.
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Old 23rd January 2012, 07:19 PM   #10
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Yes, because excess noise will be low. However, most volume pots are used with no DC so their excess noise will be low whatever construction is used.

My guess is that the frequency response problems will more than outweigh any minor noise advantage.
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