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AMPEG BA115 Tweeter Removal

#22
If the tweeter and woofer are wired in parallel, and you disconnect the tweeter, haven't you changed in impedence of the cabinet and presented a higher load to the amp? If so, the rated output of 100w may now be closer to 75or 50w. Would an 8 ohm resister across the tweeter wires help?
 
#23
No, R does not = z (impedanec) because z is frequency dependent.

That's why I can't figure how Bill say there is a short at the cross over frequency.

With the tweeter removed, the speaker is still in place for a load. The current at the cross over frequency can't flow through the open circuit.
 
#24
I think it's because a true crossover isn't an open circuit with the driver removed. At the crossover frequency, current will flow through the components as if there were a short. Makes sense to me.
 
#26
john m said:
But I don't understand the theory behind the statement that
a "dead short" at crossover frequency can happen with a component missing (open circuit).
Click to expand...

Maybe this will help. Here's a simplified two way crossover, 2nd order both ways:

Link Removed

R1 is the tweeter. Remove it and now on the high pass leg you have C1 and L1 in series to ground. Do the math and tell us what you get.;)
 
#27
And the z of the inductor is relatively high @ roughly 1kHz and above, so the inductive reactance limits the current at higher frequencies. Current can't pass throug the inductor at the cross over frequency.
 
#29
john m said:
And the z of the inductor is relatively high @ roughly 1kHz and above, so the inductive reactance limits the current at higher frequencies. Current can't pass throug the inductor at the cross over frequency.
Click to expand...

The circuit shown resonates at approximately 727Hz. Please solve the series RLC impedance at that frequency. You can use Invalid Link Removed to do so.

Do you understand the phase relationships at work between the cap and inductor here?
 
#30
I calculate a z of about 2 ohms @ resonance, not a short.

And as the frequency goes up, so does the z.

There must be some amount of the current at resonance passed to the woofer (rolloff).

Many of my cabs have light bulb protection on the tweet end of the crossover. Light bulb blows, circuit opens, gig goes on with no blown amp, change bulb the next morning.
 
#31
john m said:
I calculate a z of about 2 ohms @ resonance, not a short.
Click to expand...

LTSpice and the calculator I linked to disagree, both giving ~0.1 ohms. See the yellow box labeled R min?

Link Removed

If you don't call that near a dead short at crossover frequency, OK, we're done. ;) But even two ohms would be plenty bad enough.


And as the frequency goes up, so does the z.

Yes, and Z also goes up as frequency goes down from the resonance freq, as shown in the graph. What's your point?

There must be some amount of the current at resonance passed to the woofer (rolloff).

Sure, but it's a relatively small amount given the imbalance in impedance.

Many of my cabs have light bulb protection on the tweet end of the crossover. Light bulb blows, circuit opens, gig goes on with no blown amp, change bulb the next morning.

Absolutely. There are plenty of ways to make sure that's the case, and plenty of reasons why it may not matter that much anyway (what if you're not putting out much signal at the "problem frequency?", for example). If there's an attenuator in there it can act as a resistive load and change system impedance in various ways, for example. If it's a simple first order filter none of this applies. But believe me, I have seen this failure mode in pro sound equipment several times. I'm just trying to help you understand how a near dead short can and does occur in some situations where a driver fails. ;)

Here's a link that may clarify a bit. Take a close look at Case 3 in the series RLC analysis.

Off to a gig shortly. Hope I helped a little.
 
#35
john m said:
R min is not z min.

Look at the box under the frequency for the z value at that frequency---- that's what will limit the current.
Click to expand...

There is no box for the resonant frequency, but if there were you would clearly see that you are wrong. In between 700 and 800 Hz impedance continues to go down until the res frequency, where it matches the resistive component of the RLC network. At that point, impedance = resistance. Then it starts to climb again, but with an opposite phase relationship of V to I. This is clearly explained in the tutorial I linked. ;) I could post the SPICE model, but frankly I'm tired of this. We can go to PM if you are genuinely looking to learn though.

Anyhow, the OP need not worry.
 
#36
C.Linton said:
Never mind all this technical talk (yer makin' me dizzy), what ramifications does all this have with the removal of th BA115's tweeter? Any? or none?
Click to expand...

What he said. :)

I also have a BA-115 and was thinking about installing a simple on/off-switch (cut/close circuit) for the tweeter that would be accessable on the panel, so that I wouldn't have to go through the procedure in post #1 each time I want to connect/disconnect the tweeter.
Could it be done? And if so, is there anything in perticular that I'd need to think about?
Thanks.
 
#39
JensWest said:
What he said. :)

I also have a BA-115 and was thinking about installing a simple on/off-switch (cut/close circuit) for the tweeter that would be accessable on the panel, so that I wouldn't have to go through the procedure in post #1 each time I want to connect/disconnect the tweeter.
Could it be done? And if so, is there anything in perticular that I'd need to think about?
Thanks.
Click to expand...

Your question was answered in post #4. ;)
 

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