Simple crossover - deciphering a cap's writing

[johnnm]

I was examining a crossover today and had some questions. It belongs to a 3-way speaker. It's extremely simple: the only components are a pot, a cap, and a resistor.

I measured the cap at ~6uF, but have no idea if this is within its tolerance. It's a white axial-lead cap with black writing:
4-50VNP
110249-1
235-7242KG
Can anyone decipher that?

A somewhat unrelated question: Pots can get dirty, but can they "change value" similar to resistors and capacitors? [I realize pots don't have a value per se, but they do have upper and lower limits on their resistive properties.]

 

[bowtie427ss]

4uF 50 volt non-polarized

if "235" is an EIA code, it looks like a Mallory.

A somewhat unrelated question: Pots can get dirty, but can they "change value" similar to resistors and capacitors?

Short answer, yes.

 

[johnnm]

4uF 50 volt non-polarized

if "235" is an EIA code, it looks like a Mallory.

Short answer, yes.

Enough to ever require replacing? (I am not going to replace this one.)

I measured the extreme values of the pot and got 1 Ohm and 150 Ohms.

Having forgotten a lot from my physics class, I went to wikipedia and tried to calculate the frequencies passed to each driver.
The "crossover" isn't really a crossover at all, just a series of high pass filters.

• The largest driver receives ALL the frequencies, as it's directly connected to the speaker terminals.
• The middle driver has it's signal passed through the cap and then the pot.
  With the measured values of the cap (6uF) and the pot's range (1ohm to 150 ohms) I found the cutoff frequency to be between 177Hz and 26,525Hz, depending on where the pot is set.
• Finally, the smallest driver has its signal pass through the cap, the pot, and an additional 16 Ohm resistor. Thus this driver should have a cutoff frequency somewhere between 160Hz and 1560Hz, again depending on how the pot is set.

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Looking at the cutoff frequencies for the last two drivers I believe I have mistaken the leads for the mid driver and the tweeter. (Too much work right now to open up other speaker.)

Do these seems like reasonable values for cutoff frequencies? I'm pretty shocked at the range the manufacturer allows the user to manipulate the cutoff frequencies for the smaller drivers.
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To determine cutoff frequencies I used the relationship
f = 1/[2*pi*R*C]
If this isn't the right relationship to use then the numbers above are wrong.

 

[bowtie427ss]

I'm no expert here by any means, but the "bad" pots that i've encountered in crossovers are generally "open" and require replacement. Are you sure this is a "pot" and not an L-pad?

AFAIK, the pot shouldn't affect frequency at all, it should only change the voltage passing thru that circuit.

Can you post a pic of this network, or possibly a circuit drawing?

I'm sure one of our resident guru's like Ray W, Zilch, or Gordon W can easily shed more light for you than can i. The above mentioned fellas are very very well versed in these things.

 

[Zilch]

We've shown how to do this several times here now:

1) Define the drivers,

2) Determine the crossover components, and

3) Draw the circuit.

Filter theory will then tell us how it works.... :thmbsp:

 

[johnnm]

I guess I don't know what you mean by "define the drivers". There's a large 12" woofer, a smaller 5" woofer, and a small tweeter.

Here is what the back of the speaker looks like:

And here is what the crossover assembly looks like:

Here is my first attempt at drawing a crossover schematic. I also clearly do not know how to properly draw a pot (or L-pad) in a schematic. I looked up the symbol, but don't know which "tab" goes to the arrow or resistor-looking-part of the symbol.

I may have the small and mid driver mixed up. That is, the resistor may be in series with the tweeter and not the midrange. If I get the energy I'll get to the other speaker and open it up to confirm.

I'm not sure it's a pot and not an L-pad. What exactly is an L-pad? Google Images shows objects that are very similar in appearance to a potentiometer.

 

[geoff727]

AFAIK, the pot shouldn't affect frequency at all, it should only change the voltage passing thru that circuit.

Depending on where the series resistance is placed in the circuit, it can affect reactance on several levels, thus affecting frequency response.

Ref: Vance Dickason's LDC, the chapter on Crossovers.

 

[Zilch]

Correct?

 

[johnnm]

Opened up other speaker. Resistor is in series with the HF driver (actually a horn).
So more something like this:

My mistake, however.
Can I ask what program you used to create that image?

I'm in the process of making my way through LDC. I'll skip ahead to the crossover chapter.

 

[Zilch]

1) "Define the drivers" means what are they each, their sizes, and their impedances. Crossover's going to be different for a piezo horn vs. a compression driver, for example.

2) L-pad has two resistive elements, wirewound, typically, with a common wiper, which behave reciprocally, the opposite ends of each element terminating at the terminals. One element varies from the nominal resistance to zero, while the other from zero to ~36 Ohms, then infinity at the max setting. Dickason tells you how they are wired, and how they work to "normalize" circuit impedance.

3) Program is CircuitMaker.