Hello, new here with a question regarding Lpad resistors.

Question

First of I'm new so Hi,

I have been messing about with Lpads (I have a few horns I need to tame in P.A enclosures). I have created a crossover in LTspice, added the appropriate Lpad resistor values then I ran a simulation to check everything over. First the voltage supply was set to 56V (level required to drive 8 ohm speakers to their rated 400Watts) and set it as a sine wave at 8Khz. The crossover and Lpad are definitely working as designed (tested with pink noise source and checked fourier as well), however the the first Resistor in the Lpad (R1) is dissipating 120Watts. Which seems at odds with the 30Watt resistors I've seen in similar rated speaker cabs. Our overlord google.com has told me they don't need to be high wattage resistors, but the circuit and simulation here (designed using commonly available calculators) tells me otherwise.

R1 and R2 are the Lpad, R3 is the horn.

The rest being a 4th order for the horn and 2nd order for the woofer.

Any Ideas what I am doing wrong?

Many thanks.

P.S still trying to work out how this site all works, forgive me if I've posted in the wrong group or something.

shabaz · Answer

Hi Matthew, I didn't have a chance to read/examine this in detail, but what stuck out was that you simulated with a sine wave. That may explain why the real-life resistors are of a lower power rating, since ordinarily the average power from speech or music will be a lot less than the peak. Most resistors can easily withstand higher power provided it is not sustained for a long time.

michaelkellett · Answer

Good point Shabaz - in real full range prgramme material the amount of energy at high frequencies will only be a small proprtion of the total. @ mrsouth Looking at the circuit, at high frequencies we can approximate that the inductors are infinite impedance (open circuit) and the capacitors zero impedance (short cicruit). The horn driving arm of the filter becomes: And half the delivered power will go into R1 and a quarter into each of R2 and R3. (remember this is an approximation !) So your results look sensible. How much power goes into R1 if you use noise rather than a sine wave excitation ? MK

jc2048 · Answer

I don't understand anything about crossover design. What was the rationale for adding R2 to the circuit and lowering R1 to 4 Ohms to match? Is it because the horn is starting to look somewhat inductive as you go up towards 20kHz? It looks to me like the filter was probably designed for 8 Ohms (with the assumption that the horn load was resistive). At 8 ohms it has a smooth Butterworth response. As you've got it here, with the resistors either end changed to 4 Ohms, it's halfway to a Chebyshev, with ripple in the passband. (As I said, I don't know anything about crossover design, so maybe it's all very, very clever and compensates for the horn response, or something.) This is the response as you've got the circuit (R1=4, R2=8) (red=woofer, green=horn) This is how it would be at 8 Ohms (R1=8, R2=infinity) One thing to notice from both those is that the 4-pole horn filter has a very abrupt phase change at cut-off, where the woofer is still sending out a modest amount of sound. Wouldn't that then acoustically result in some cancellation between the horn and the woofer? If so, you may find a bit of a notch in the response around 2kHz when you physically test it.

mrsouth · Answer

Thanks Jon, effectively R1 and R2 make up a voltage divider network for the horn driver to better match it with the woofer (there is a 6db SPL difference between them). It's probably not immediately obvious as a divider because R1 is on the other side of the filter. You are right in that it can work with only R1 (and look better on paper), however my reading of late has lead me to believe it is better to run with the divider instead of a one large resistor on grounds that the it somehow results a more even response up higher. I believe if the horn is a piezo tweeter they are almost purely capacitive changing the rules again. In any result, what I am trying to avoid is having to spend time correcting the response with excessive EQ, and thus am open to having my mind changed. Interference at the crossover likely won't be as noticeable as it looks on paper, a typical gig see's way too much change in the room (not just between venues but in the number of people moving in and out of a venue) to make that the biggest concern. At the moment I really need better matched and clear high frequency sound. Which is why I am incorporating an Lpad and spending a bit more time on the horn side of things. I really shouldn't be testing them with a straight up resistor because the drivers have a freq. dependent inductance, and exactly as you say, that and the difference in slopes will have an effect on the final response. Having said that, there is a method to my madness on that front and the bit I don't understand is the power side of it. Especially as I have already overlooked a few fundamentals of signal power. It is also the first time I am using LTspice (in fact it's the first time since 1994 that I have used any simulator), so It's kinda like I am trying to learn how to ride a bike again, only now I am almost 50 and bikes have changed so much I need youtube lessons. Thank you for your work on those graphs too, I appreciate that. Just to add some background information to what's going on here, I have four speaker sets, one I made years ago that barely have a crossover (they are still going strong but only used for Small gigs),I have an old dodgy pair that rarely get used anymore, I have a commercial pair that are starting to die and I currently have set half complete but will need more than a basic crossover if they are to sound half decent. The later three sets are all for rock band use so they do/will be pushed hard. I can usually compensate for excessive high end with the EQ but you know what they say about lipstick on pigs.

jc2048 · Answer

Thanks for the explanation. It sounds interesting, though I can imagine there's a danger of 'fiddling around' for ever on these kind of things, what with all the electrical, electromechanical, and acoustic variables involved. I did my simulation with Tina-TI (Texas Instrument's free simulator), but that's simply because I'm a bit more familiar with it than LTSpice. LTSpice will be able to do you a swept Bode plot without any difficulty at all. With Tini-TI, I can also do things like this (I imagine the same goes for LTSpice, or there will be another way to achieve the same end): Here I'm measuring the current through the resistor and the voltage across it and then using the post-processor to calculate the product of the two. This is what the result shows for an 8kHz 56V pk sinewave from the generator. That's showing a peak of 132W, so an average of about 66W in the resistor. Be wary if you decided to give Tina-TI a try. There are some bugs with displaying the postprocessor results (at least, with the version I've got there are). With more than one calculated curve, it will only display one of them correctly, the rest being wrongly shown as a duplicate of the one that it gets right.

michaelkellett · Answer

I think you have R1 in the wrong place - the idea of the Lpad is to make two resistors look like the same impedance to the drive source.

R1 in orignal position, filter sees 4R load and 4R source

R2 in correct position, filter sees 8R load and 0R source

I expect that you know that you will get much better results with an active crossover and two power amps.

What drivers are you using ?

You can buy a 100 W Lpad at a very reasonable price:

https://cpc.farnell.com/monacor/at-62h/high-power-l-pad-mono/dp/LS00547

I'm not that confident about the power rating but I think it would be OK for the HF driver.

might be worth just tuning it by ear.

MK

michaelkellett · Answer

If you need more attenuation then just alter the LPAD resistors.

The rules are:

Rn = R3 = 8 (impedance of horn driver), Rn||R2 is the impedance of the horn in parallel with R2

Rn||R2 = 1/(1/R2 + 0.125)

R1 = 8 - Rn||R2

Attenuation = Rn||R2 / 8

You can solve the equations (not before breakfast) so I did a little Excel spreadsheet:

(all these tools make us lazy !)

MK

Hello, new here with a question regarding Lpad resistors.

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