Resistor/LPAD wattage question

[DerrickM]

Hi,

Everything I read says that the resistors in an LPAD must add up to the wattage the driver could see. (Say, 50W for a 50W tweeter). What I'm wondering is why in all the car audio crossovers I've seen they always use 5W resistors (on speakers that are claimed to be 100W).

Reason I'm asking is I'm building another set of crossovers, and I had planned to use 50W LPADs, but I'm considering just tuning the tweeter level and using resistors instead.

Also, I feel like an idiot for asking, but why are LPADs offered in impedances to match drivers (4/8/16 ohm). I mean isn't the point to leave the impedance the crossovers sees unchanged?

Thanks everybody,

Derrick


PS - For anybody that read my previous posts, the Dynaudio drivers worked so well after I crossed them over correctly, that I went out and bought another set of them for the fronts (of my car).

[Inductor]

Originally posted by DerrickM
What I'm wondering is why in all the car audio crossovers I've seen they always use 5W resistors (on speakers that are claimed to be 100W

Did you try to see if they are hot?

[DerrickM]

Nothing to check--I just built new crossovers, and they don't have tweeter attenuation. I'm planning to add an LPAD (pot or two resistors), and wondering if I really need 50 watts worth of resistors.

As for the variable LPAD question, it became obvious when I thought about it some more. (i.e. if it was turned all the way down, you'd want it to present the same load across the inputs that the driver would have had).

[Svante]

Originally posted by DerrickM
What I'm wondering is why in all the car audio crossovers I've seen they always use 5W resistors (on speakers that are claimed to be 100W).

The power rating on the components does not have to match the rating on the speaker. It tells how much power the resistor can take, and that power does typically not need to be as large as the power that goes into the driver.

Imagine the following example, where I have simplified the driver's impedance to a 8 ohm resistor and the filter to a simple series resistor, just to make the point.

If the amplifier delivers 9 volts, 1 volt will end up over the resistor, and 8 volts over the driver. The power in the resistor will be U²/R=1²/1=1 watt. The draver will take U²/R=8²/8=8 watts.

In crossovers, similar effects occur. Most of the power ends up in the drivers, luckily.

[Inductor]

Svante,
do you know what happens if you de-rate L-pad resistors? you may have an escalating failure. Also depends on the power they are using...

[Svante]

Originally posted by Inductor
Svante,
do you know what happens if you de-rate L-pad resistors? you may have an escalating failure. Also depends on the power they are using...

"de-rate"? I don't think I understand what you mean by that.

[Inductor]

Originally posted by Svante

"de-rate"? I don't think I understand what you mean by that.

Svante,
if you change a 50W by a 5W.
Cheers.

[DerrickM]

Here's what I came up with last night. It's a selectable -6dB / -9dB LPAD for a 4ohm driver. All the resistors are 10W (Dayton from Parts Express). I doubled up the 5's to make 2.5 and the 8's to make 4 ohms to increase the wattage. It'll be 20 or 30 watts of resistors on each leg of the lpad, depending on the switch postion.

[Pit Hinder]

Derrick,
throw your headache pills into the trashcan. Unless you try to attenuate a PA bass, or try to bake Idaho potatoes on your resistors, 10W rating is perfectly OK. If you are a "belt and braces" man, two 10W of twice the value in parallel will last a lot longer than your eardrums.



Pit

[Svante]

Originally posted by Inductor



Svante,
if you change a 50W by a 5W.
Cheers.

Ah, well given that the original design was sensible, ie that the 50 W resistor was indeed needed, a 5 W resistor would burn, yes.