Background informations:
I have one pair of 2 ways speakers with a 12 inches woofer and a 5 inches mid-tweeter. All drivers are 8 ohms.
One of the woofer suffers from a intermittent breakdown...it plays then it stops for a few seconds/minutes and start to play again (without touching anything).
I have interchanged the faulty enclosure with the good one and the problem still occurs on the faulty one.
Wiring and connections have been redone and the problem still occurs.
I'm convinced it's the woofer that is at fault.
Since I'm working with vintage speakers, Wharfedale from the early sixties, it' pretty difficult to find a replacement driver.
I happen to have an identical woofer BUT it's a 15 ohms (measures 12 ohms on ohmmeter)
I applied the OHM's Law:
R (equivalent) = (R1 ie: 12 ohms x R2 ie: 24 ohms) divided by (R1 ie: 12 ohms + R2 ie: 24 ohms).
Maths calculation gives 8 ohms.
Question:
If I add a 24 ohms 50 watts resistor in parallel with the woofer I should be matching the other speaker at 8 ohms.
I'm aware that we are talking impedance ... stop me if I'm wrong and please explain!
Do you think the load seen by the amplifier will have an impact on it's performance ?
Is the 50 watts resistor big enough wattage to match with a woofer rated at 30 watts RMS maximum and driven by a 50 watts amplifier ?
As somebody here ever done this type of arrangement ?
If the answer is yes can you comment of the performance.
I will be trying this tooday but I want to have your opinion.
Thanks.
I have one pair of 2 ways speakers with a 12 inches woofer and a 5 inches mid-tweeter. All drivers are 8 ohms.
One of the woofer suffers from a intermittent breakdown...it plays then it stops for a few seconds/minutes and start to play again (without touching anything).
I have interchanged the faulty enclosure with the good one and the problem still occurs on the faulty one.
Wiring and connections have been redone and the problem still occurs.
I'm convinced it's the woofer that is at fault.
Since I'm working with vintage speakers, Wharfedale from the early sixties, it' pretty difficult to find a replacement driver.
I happen to have an identical woofer BUT it's a 15 ohms (measures 12 ohms on ohmmeter)
I applied the OHM's Law:
R (equivalent) = (R1 ie: 12 ohms x R2 ie: 24 ohms) divided by (R1 ie: 12 ohms + R2 ie: 24 ohms).
Maths calculation gives 8 ohms.
Question:
If I add a 24 ohms 50 watts resistor in parallel with the woofer I should be matching the other speaker at 8 ohms.
I'm aware that we are talking impedance ... stop me if I'm wrong and please explain!
Do you think the load seen by the amplifier will have an impact on it's performance ?
Is the 50 watts resistor big enough wattage to match with a woofer rated at 30 watts RMS maximum and driven by a 50 watts amplifier ?
As somebody here ever done this type of arrangement ?
If the answer is yes can you comment of the performance.
I will be trying this tooday but I want to have your opinion.
Thanks.
Well it might and might not, it's really a simple Q but difficult to answer without all the data ie matching sensitivities and crossover circuit. I would investigate the fault more thoroughly. have you taken the suspect driver out of the enclosure yet? Some times internal connectors come loose, solder connections fail, crossover parts age and become intermittent, open or shorted driver tinsel leads, driver suspensions sag allowing voice coils to rub and short. My bet > if you localize the fault and examine and tighten up everything, reflow any solder joints in the path, and rotate the woofer 180 degrees you will have joy once again.
Yes, I have taken the woofer out of the box.
This afternoon I replaced the suspect woofer with the 15 ohms version and I didn't like the sound so I put the original back in. I replaced the crossover so it's presently on my workbench getting tested again...
Your suggestion of turning the woofer 180 degrees is far fetch but I will definitely try it. I have nothing to loose and evrything to gain.
Thanks.
Ahem, not quite.
1. Since the nominal impedance of the driver is 15 ohms that's the value to plug in to the equation, which yields a 17 ohm resistor.
2. The power lost is inversely proportional to the resistance, so that would have been 1/3 in the example.
3. It may work after a fashion, but the bass will be unbalanced between the sides du to different power to the drivers, and general frequency response will be different between the sides.
Now, speaker drivers are really very simple in a wiring sense. There's a coil which is wound from a continous copper (most of the time) wire. At the bottom of the cone it's connected via solder joints to two soft copper braids. These in turn connect to the connection lugs via another pair of solder joints.
The problem as described is usually due to either of these solder joints working loose, and a resolder will fix the problem. Unfortunately it's not unusual for the coil-to-braid joints to be hidden under the dust cap, but there's not really much to lose, so If I were you I'd consider using a scalpel to remove the dust cap, do the soldering and then glue the dust cap back.
yes replacing the woofer is not so straight forward as matching DC ohms. S/b the last resort and usually means alot more work even if initially sounding OK.
Test the suspected woofer by itself on the bench, playing a 400Hz tone and moving the cone in and out as part of examining things ie tinsel leads.
2) most amps are constant voltage, so no appreciable power lost to the driver. The amp may deliver more power in this case.
3) I would expect most differences due to mismatches from primarily both sensitivity, T/S, and secondarily cone materials/profiles.
2. You're missing the point. With the replacement driver being 15 ohms, even though paralleling it with a resistor might bring the total impedance down to 8 ohms, at a given input voltage to this parallel combination, only part of the power will be seen by the driver (about half if the driver were paralleled by a 17 ohm resistor) especially because most amps are constant voltage sources. The other speaker whose 8-ohm driver is still okay, would be receiving ~twice the power as this 15-ohm driver at the same time. Unless the 15-ohm driver happened to be twice as sensitive as the 8-ohm driver, there will be an unbalanced output between the two.
Paul
Paul
My choice would be to repair the old driver.
As infinia says it is usually a break in the coil winding close to the cone terminations. That is the first place to look, and does not require any dis-assembly.
Any other fault requires dis-assembly.With many drivers from the sixties, the magnet is bolted on, and the cone, coil and basket can be removed complete, exposing the coil. If you do this, the basket and cone will need recentering, which requires care and a modium of skill, but is not overly difficult Removing the dustcap is usually not required, but makes it easier.Ask if you don't know how.
The old glues are prone to failure with age, and the so next most likely point of failure is at the top end of the coil. Sometimes it is possible to wind off one turn, (this does not affect performance much) to allow a joint to be made outside the gap. Take care as the wire is fine and easy to break. Note Wharfdale used aluminium wire in some of there drivers, (usually the full-range, rather than the woofers) and joining this is difficult to say the least.
If the fault is anywhere else a rewind of the coil is required.
As infinia says it is usually a break in the coil winding close to the cone terminations. That is the first place to look, and does not require any dis-assembly.
Any other fault requires dis-assembly.With many drivers from the sixties, the magnet is bolted on, and the cone, coil and basket can be removed complete, exposing the coil. If you do this, the basket and cone will need recentering, which requires care and a modium of skill, but is not overly difficult Removing the dustcap is usually not required, but makes it easier.Ask if you don't know how.
The old glues are prone to failure with age, and the so next most likely point of failure is at the top end of the coil. Sometimes it is possible to wind off one turn, (this does not affect performance much) to allow a joint to be made outside the gap. Take care as the wire is fine and easy to break. Note Wharfdale used aluminium wire in some of there drivers, (usually the full-range, rather than the woofers) and joining this is difficult to say the least.
If the fault is anywhere else a rewind of the coil is required.
And you will only have 1/3 the audio output from the driver. In other words, that woofer will be much less loud than its other channel cabinet.
Keep looking for the right woofer.
You're still missing the point. You said there would be no power lost to the driver (with a resistor in parallel with the 15-ohm driver to create a nominal 8-ohm load), but that's incorrect because the power, from a constant-voltage source with a specific output voltage, will be divided, part to the driver and part to the resistor based on their impedance/resistance values, resulting in the 15-ohm driver receiving less power (about half) than the 8-ohm driver in the other speaker. You are correct in that from a constant voltage source (low output impedance) with a given output, the 15-ohm driver will receive the same power whether or not there's a resistor in parallel with it (as long as the amplifier has enough output capability).
Paul
Paul
A simpler way to look at it is if the resistor gets warm, then some of the power must be dissipated as heat and obviously, not all of the power can be going to the driver.
The long and short of it is, the resistor is a bad idea and the wrong approach.
Is that what Paul is saying in that sentence?
I previously said or meant to say "adding a parallel resistor does NOT change the drivers initial sensitivity appreciably given a normal amp(ie voltage source) and crossover with reasonable insertion loss. Maybe he is pre- supposing a 16 ohm driver might have about 3 dB less sensitivity compared to an equivalent 8 ohm?
Using a parallel resistor is not efficient but it might work I don't know enough yet.
I doubt it. I suspect that the 15Ω transducer's efficiency is the same as the 8Ω transducer.
If you parallel a resistor and drive the circuit with 50 Watts, 50 Watts gets divided between the resistor and the transducer pretty evenly. That would yield a 3dB loss of power to the transducer and the need for a pretty large resistor.
That extra heat will also alter the crossover slightly as well as the transducer, but it probably won't be really that critical with old speakers like that.
Keep looking for the right driver.
Think about what you are saying.
A zobel will provide a somewhat constant impedance at the input for a varying impedance at the driver.
In this case you have an impedance mismatch (8Ω to 16Ω). A zobel will not do the job. If you work out a circuit to give you a relatively constant 8Ω impedance at the input you still need to dissipate 50% of your power through one or more resistors.
You are really paddling upstream here.
If both woofers have the same sensitivity (unlikely) it becomes a matter of matching the second woofers impedance just before the crossover frequency. say an 1/2 octave before not across the whole band. It's not difficult using a parallel zobel (series cap and resistor). Depending if the original Xover design uses a zobel, the added resistor will be mostly be higher than 16 ohms and thus only Pdiss comes to play around the cutoff and even much less beyond.
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