Crossover networks are just a series a filters, if the speakers were already rolling off at a rate of 6 dB/octave then adding an inductor or capacitor would effectively create a 12 dB/octave filter. A crossover would imply there is a filter on both sides of the filter point. Which there most likely is seeing as midwoofers have moderate roll-off due to the inductance of its voice coil.Originally Posted by Gregorius
If you're crossing over your tweeters at 5kHz then you're expecting too much out of your midrange driver. If you add a 4 ohm speaker to another 4 ohm speaker in parallel you net a 2 ohm load which means the "crossover" will become useless as the crossover point on the tweeter will change. This will only get worse if you add tweeters with their own inline caps.Horse-pucky. The tweeters of the 2 ways are already wired in parallel to the woofer in DB series Polk 5 1/4" speaker, using our old friend the inline, bi-polar capacitor "filter". If you parallel another tweeter/filter onto it, the worst that will happen is that the amp will see a a drop in impedance above the "crossover" point. Since that tends to be around 5K hz, I sincerely doubt any Head Unit amp will cause A FATAL AUTOMOTIVE FIRE !!! because it has to supply an additional, say, 1-2 watts to an additional tweeter. What's much worse, and a much more grave offense, is that the treble will sound like "poo".
Also, does not anyone have issues running 2 ohm stable amps at 1 ohm? Its a theoretical problem which shouldn't be ignored. You're putting stress on a device that it wasn't designed to deal with. By your logic, its impossible to destroy an amp by clipping it at a current rating it wasn't designed to put out. And in real life, its the exact opposite. When you lower the impedance of the load you're driving the amp will also increase the number of amps it draws. So if you blow the fuse on your amp (which should never happen unless you short it out), this will be why.
What are the chances of it happening? Very small, should you be concerned? Probably not. Is there a smarter way of doing this? Very much so.
If the question is how to disable the tweeters then its simple, unplug them. Why was there 10 posts if its as simple as unplugging them? But no, the OP changed his mind between writing the Subject title and the OP. If the replacements are wired in parallel that means the replacements are 8 ohms each which means wiring the stock tweeters in wouldn't be an issue.
Show me the V=IR and 1/Rtot=1/R1+1/R2...1/Rn calculations that say you're correct.To the OP: This matt5112 seems schmuck-like. This thread was 10 posts long. He may have been drinking, didn't read it, then incorrectly decided he knew something.
Quick example: 12V going to HU which draws 10 A which means it draws 120 watts. Now lets assume its a class AB amp which limits its efficiency to 50 % this means only 50% of that 120 W is availible as power to the speakers (I'll even ignore that these things need power to run the other plethora of things jammed into them). This means you have 60 watts for four channels. If you divide that by 4 you arrive at 15 watts. All HU's I've seen are rated at 22 W RMS @ 4 ohms at 14.4V input which is still bull**** because they can only put out 18 if they're class AB.
So you've got your 4 ohm stock tweeter, 8 ohm new tweeter and 8 ohm new mid. If the tweeters are wired in parallel to each other then that circuit is wired in parallel to the mid you net a 2 Ohm load. If you wire the stock tweeter in parallel with the new parallel system then you have a 1.6 ohm nominal load.
Consider that the HU was putting out 18 watts with 14.4 V input at 4 ohms. Using P=I^2*R you get I to be 2.1 A which means you have 8.5 V output using V=IR. Since amps are voltage driven devices, they will attempt to double the current instead of lowering the voltage when attempting to drive a load of half the impedance.
Lets explore that possibility:
Again using P=I^2*R and assuming double the current and the new 2 ohm load you arrive at 35 W. 35W times four is 140 W of output. If you then multiply this by the inverse of 0.5 you arrive at 280 W of input power. This IS misleading because I assumed you were doing this on four channels instead of two, but you can still see the point: If you do this, the amp will attempt to draw twice the current. Most likely it will only attempt 1.5 its rated current as most amps are designed for that low impedance load. Even then, the heat build up will cause premature failure as the enemy of all electrical components is heat. Ever run an engine too hot for too long? Ever idle an engine with 2 ranges too hot plugs with low coolant and no cooling fans in the desert?
*Edit* Before you attempt to call bull**** on this concept merely read the following link:
http://www.crownaudio.com/pdf/amps/139718.pdf
Amps rarely "double down" (double amperage) when halving the load resistance.
