Introduction to designing crossovers without measurement

[JRT256]

Working out the values
Looking at the specs (image 3) you'll see one called either 're' or 'Rdc'. When working out the value for the resistor in this circuit you should multiply this value by 1.25. For our example this is 5.5 x 1.25 Although this equals 6.875, there is no easy to find resistor of this value, but 6.8 ohms will do nicely.

For the capacitor, look for a spec called "Le" which for our example is 0.4mH (that is 0.4 milli, also known as 0.0004). Take your value of Le and divide it by R squared. For example, our resistor was 6.8 ohms, and 6.8 squared is 46, so 0.0004 divided by 46 gives us a value for our capacitor of 8.7uF, which is close enough to the easy to find value of 8.2uF.

Why do you use a 25% larger resistance than the DC resistance of the driver?

 

[witwald]

I think that the 1.25 factor is applied because the impedance we are trying to equalise is on the right-hand side of the impedance peak, and the impedance minimum there is typically a bit greater than the DC resistance of the driver.

 

[AllenB]

It's interesting that 1.25 is around eight ohms divided by the typical coil resistance.

No. What you'll usually find is that the existing impedance minimum isn't as low as the resistance, and at the top end it doesn't reach significantly high enough to need a resistor equal to Re. Higher would generally be better if it can give a good result, however when you design these on a simulator it becomes clear that the ideal value can be different for various reasons. For example how low you need or want to flatten, such as with a 3-way vs a 2-way.