Type A replacement caps slightly off spec

[mboxler]

4 hours ago, glens said:

 

Think about what you just stated...

 

2.83 minus 2 equals 0.83, minus another 2 ("2 volts each") equals -1.17!  Pretty hard to get into a hole ;^)

Although the K-77 isn't a true resistor, for the sake of this explanation let's say it is...

 

The impedance of a 2uf capacitor at 9947 Hz is 8 ohms at -90 degrees.  That impedance added to the driver's 8 ohm 0 degree impedance = 11.38 ohm at -45 degrees.

 

2.83 volts through a 11.38 ohm load = 0.248682 amps.

 

.248682 amps through 8 ohms = 1.99 volts, or -3db.  There is a 2 volt drop across the capacitor and a 2 volt drop across the driver.

 

Result. the current through the driver is 45 degrees ahead of the voltage at the crossover frequency.

If there is a comparable low pass filter to, say, a woofer, the current through it will be 45 degrees behind the voltage at the crossover frequency.

This is what creates the 90 degree difference between drivers in a first order filter.

 

Mike

[glens]

Oh,  now you're bringing phase angle into the equation...  I didn't anticipate that.  You go!  (CIVIL)

[JohnA]

20 hours ago, mboxler said:

That's an 8 ohm Lpad.  It will work for tweeter, but not the squawker, which is 14 ohms (?).   

 

If you make your own 14 ohm Lpad, then a 26uf cap will be needed.  With a 16 ohm Lpad, it will be a little off, but now by much.

 

Mike

 

It's waay more complicated than that.  A K-55-V is nominally 16 ohms.  The impedance at a La Scala's crossover point on a plane wave tube is about 12 ohms.  Robbie, you need to change the caps based on the L-pad you design. 

http://ccs.exl.info/installation/crossovers-installation-tweaking/crossover-calculators/

It is also possible to build an L-pad that attenuates output as well as reducing the apparent impedance to the amplifier.  You could make the K-55 look like it is 8 Ohms to the crossover as well as the amp.  Great for some tube amps. 

https://www.electronics-tutorials.ws/attenuators/l-pad-attenuator.html

 

https://www.atlasied.com/f/681/ATS004136 PD Series Datasheet RevA.pdf

OTOH, 1st order crossovers are very tolerant of changes.  +/-10% on the values is fine as long as the part's tolerance is not to the high side; a +20% cap would not be suitable.  A 6uF and a 7uF in parallel makes 13. 

 

You won't need an L-pad for the tweeter, if you use a Klipsch design. 

[glens]

19 hours ago, mboxler said:

The impedance of a 2uf capacitor at 9947 Hz is 8 ohms at -90 degrees.  That impedance added to the driver's 8 ohm 0 degree impedance = 11.38 ohm at -45 degrees.

 

2.83 volts through a 11.38 ohm load = 0.248682 amps.

 

.248682 amps through 8 ohms = 1.99 volts, or -3db.  There is a 2 volt drop across the capacitor and a 2 volt drop across the driver.

 

As an aside, this sort of shenanigans is why high-electricity-usage industry invests heavily in power factor correction equipment.  The utilities don't like to have to make and send 1.2 megawatts only to be able to bill for 1, so they bill for what they're sending, but the customer doesn't like paying the penalty...

[mboxler]

Likewise, the amplifier needs to be designed for this.  2.83 volts into an 11.38 ohm load = .7 watts.  However, the only "work" being done is the 2 volt signal across the 8 ohm tweeter, or .5 watts.  A capacitor is an open circuit, it does no work.  The output transistors of the amplifier need to be able to dissipate the excess .2 watts as heat.  In class d amps with single ended outputs, this excess work is "pumped" back into the power supply as increased voltage.

 

At least I think that's right.  Sorry, getting way off topic.