8 ohm L-pad total resistance - diyAudio
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Old 2nd October 2008, 08:03 PM   #1
sreten is offline sreten  United Kingdom
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Default 8 ohm L-pad total resistance

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Hi,

you know the sort of thing above. As an 8ohm L-pad it is only ~
correct and there is an ideal total resistance for each attenuation.

I'm curious at to what total value is used in these pads, the ideal
value varies with the attenuation you want, e.g. if 6dB then the
total would be 12R, 4Rpad + 8R_Driver||8Rpad = 8R.

But this gives 5R at no attenuation, 8R_Driver||12Rpad.

I'm guessing around 20/25R is used, any measurements ?.

/sreten.
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Old 2nd October 2008, 11:59 PM   #2
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Wouldn't measuring the resistance between the outer legs tell you the series resistance? Turning the dial while doing this would tell you if I have the right schematic in my head if the value does not change, and with min/max attenuation figures you could work out the parallel resistor too.

These contain two wire-wound rheostats right? Anyone measured the series inductance of these toroids? Maybe it's low, since there's almost nil core area anyway.
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Old 3rd October 2008, 02:06 AM   #3
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~40 ohms on 2 different ones I measured
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Old 3rd October 2008, 07:12 AM   #4
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To get the actual values of the 'legs' of the L-Pad, you measure from the center to one side with the 'dial' turned to both extremes. Then do the same from the center to the other side.

Usually, what you have is an open high resistance in (more or less) parallel with an open 8 ohm resistance. When the dial is turned in one direction, the 8 ohm resistor is in parallel with the input, and the tweeter is shorted to ground by the high resistance leg.

In the other, high level position, the 8 ohms is out of the circuit, and the high resistance is in parallel with the tweeter.

The low resistance in this case, is 8 ohms.

The high resistance is 10x 8 ohms or higher, so as to have no effect on the circuit or speaker impedance when it is fully in the circuit.

Sometime one leg is something like 120 ohms or 150 ohm, sometimes 600ohms , and the other leg is the rated impedance (8 ohms, 16 ohms, whatever).

Attached is a diagram that shows the internal workings of an L-Pad.

Note that each leg is 'open' in that only one end and the potentiometer wiper is in the circuit. The other end of the resistance leg is not connected. That's why you have 3 connection points instead of 5.

Steve/bluewizard
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File Type: gif l-pad_hi_lo.gif (58.4 KB, 346 views)
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Old 3rd October 2008, 12:31 PM   #5
sreten is offline sreten  United Kingdom
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Hi,


Just assumed the basic ones are a single track, many c/o
diagrams with L-pads imply a single track potentiometer.

Two tracks linked together, I have seen that in some
crossover diagrams, so my question is too simple if
most L-pads are made with two tracks.

(comes with using 20R 5W wirewound " speaker volume
controls"for messing about with tweeter levels and never
buying "proper" L-pads ......)

In that case measuring the total resisitance it would vary
as you move the wiper rather than be simply constant.

For the simple case total resistance does not vary.

for the two track case :
At full volume measured total = the output side track.
At zero volume = the input track = 8ohms presumably ?

Hmmm .....

Its even more complicated depending on whether you
assume the the two tracks are linear tracks or not.
600R is way too high for a linear track, but high power
wirewound potentiometers are never logarithmic AFAIK.

/sreten.
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Old 3rd October 2008, 03:35 PM   #6
Pano is online now Pano  United States
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Good question.

I've measured the total DCR on several cheap L-Pads with a driver attached. It did not stay at one value. It did vary by a few ohms at different postions. Always wondered about that.....

I'll try again and note what I find.
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Old 3rd October 2008, 06:28 PM   #7
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The Variable L-Pad should work just like a normal Fixed L-Pad. You have a series component which is some percentage of 8 ohms, and you have the parallel (parallel to the speaker) component. When the parallel component is combined with the parallel speaker impedance, and the series resistance is added in, the result should always (more or less) be 8 ohms.

Consequently, these would reasonable represent potential resistor values -

-0.25db= 0.25 series, 274 parallel = 3.1% reduction in voltage signal
-0.5db = 0.45 series, 135 parallel = 6.3%
-1.0db = 0.86 series, 65.6 parallel = 10%
-2.0db = 1.65 series, 30.9 parallel = 20%
-2.5db= 2.00 series, 24.0 parallel = 25%
-3.0db = 2.34 series, 19.4 parallel = 30%
-4.0db = 3.00 series, 13.7 parallel = 37.5%
-6.0db = 4.00 series, 8.04 parallel = 50%
-8.0db = 4.82 series, 5.29 parallel = 60%
-10db = 5.47 series, 3.70 parallel = 68%
-12db = 5.99 series, 2.68 parallel = 75%
-14db = 6.41 series, 1.99 parallel = 80%

I think you get the picture.

First, it doesn't seem to a linear taper on the potentiometers, or at least not in the calculated resistor values.

Next, I'm not sure how small a signal reduction you can hear. If -0.25db is unhearable, then the highest value of the large value potentiometer is probably, like I said, about 125 to 150 ohms.

If -0.25db is detectable, then perhaps the highest value is closer to 300ohms.

At any rate, I think the large-value section of the L-Pad has to be at least 10 times the value of the speaker impedance for it to have minimal affect on the circuit when it is fully in parallel with the speaker.

Perhaps, I've misunderstood the original question, but none the less, hopefully, there is some value to what I have added.

And for the record, I didn't actually make the calculations, I used in on-line L-Pad calculator.

http://www.sengpielaudio.com/calculator-Lpad.htm

Steve/bluewizard
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Old 6th October 2008, 01:36 PM   #8
sreten is offline sreten  United Kingdom
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Quote:
Originally posted by BlueWizard

Perhaps, I've misunderstood the original question ...
Steve/bluewizard
Hi,

The point is using one (simple pot) or two linear wirewound
tracks you cannot maintain the theoretical 8 ohms total.

I know how to build a proper 8 ohm switched attenuator.

But you cannot build an equivalent with linear tracks.

So what are the measured values of an L-pad ?
how is it fudged ? is the question .......

/sreten.
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Old 6th October 2008, 02:29 PM   #9
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Perhaps I'm still confused, it's been known to happen.

But, I concede you are right; you don't get a perfect 8 ohms across the range of motion of the L-Pad. From what I remember, it varies a fraction of an ohm from a perfect 8 ohm. Perhaps even a large fraction of an ohm. So, it is not perfect, but it is certainly within necessary tolerances to be functional.

Just a few thoughts.

STeve/bluewizard
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Old 6th October 2008, 04:36 PM   #10
rabbitz is offline rabbitz  Australia
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I have a Visaton L-pad with pins as per a normal pot. Looking from the font, left pin 1 (ground), centre pin 2 (wiper), right pin 3 (input).

It has 2 linear resistors with 8R attached to pin 3 (input), 35R attached to pin 1 (ground), pin 2 (wiper) contacting the resistors at the same angle.

Some measured values between pins,
LEVEL: 3-2 (input to wiper), 3-1 (input to ground), 2-1 (wiper to ground), (Z out)
MIN : 7R8, 8R0, 0R1 (wiper shunts to ground)
MIN+ : 6R8, 7R7, 1R0, (0R87)
: 4R7, 15R2, 10R7, (3R26)
MID : 3R6, 22R, 19R5, (3R04)
: 1R9, 27R9, 26R1, (1R77)
MAX-: 1R0, 31R6, 30R7, (0R97)
MAX : 0R1, 35R, 35R (input open to wiper)

So the output impedance varies (if I've calculated correctly) and could never be 8R on this L-pad.

BTW, I was always told a normal 2 resistor L-pad does not to alter the impedance of the filter..... true or false.

Hope this info helps.
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