Introduction to designing crossovers without measurement

My amp can do 4 ohms . Since I am bi-amping I can get away with no resistor .I'm guessing the crossover will add resistance also . I'm guessing the crossover values for capacitance and inductance are doubled ? The tweeter sits at 92 db and the midrange is 89 db . They will run off their own channel . The cutoffs I picked are 1000 hz and 3500 hz . my amp AVR-X4300H | Powerful 9 channel Network AV receiver with HEOS and 3D Sound | Denon .
 
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Such a resistor would not be in a convenient place anyway as you'd have to consider it against a complex impedance, so it's good your amp is OK with 4 ohms as is.

A series inductor will add some resistance but not much at all. The inductor will add reactance (somewhat similar to resistance), but only at the top end. The resonance impedance will be higher than 4 ohms. This just leaves a small region in the low hundreds of Hz that might be close to 4.

Double capacitors and halve inductors and resistors.
 
Such a resistor would not be in a convenient place anyway as you'd have to consider it against a complex impedance, so it's good your amp is OK with 4 ohms as is.

A series inductor will add some resistance but not much at all. The inductor will add reactance (somewhat similar to resistance), but only at the top end. The resonance impedance will be higher than 4 ohms. This just leaves a small region in the low hundreds of Hz that might be close to 4.

Double capacitors and halve inductors and resistors.

A couple of woofers I'm using for a centre channel are nominal 8 ohm woofers.
Their minimum resistance is 5.2 ohms, and I've a 5.6 ohm series resistor on each.
For the amp's sake, is this okay, or am I better off just running these woofers in series.
The minimum resistance is in the middle of the vocal range. There's power to spare, even in series.
 
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Dropping the resistors and putting the drivers in series, they should remain at around the same level. Your amp should draw around half the current. You might find that the response changes.

Having the drivers against a series resistor, one impedance varies with frequency and the other doesn't.. so the Voltage doesn't divide equally at some frequencies. When you use two of the same driver in series they each vary the same way at the same frequencies, so they share fairly (assuming they are well matched).

If you look at the impedance plot you can see the areas that would be affected. These should be a peak around the bass resonance, and a rising response toward the higher frequencies using a resistor compared to not.
 
About flattening the impedance of the woofer:

We'll need to use one resistor and one capacitor (per woofer).

Excuse me AllenB, what do you mean with "(per woofer)"? I'm designing a creossover for a speaker with 2 woofers and 1 tweeter, so do you mean that i have to consider 2 resistors and 2 capacitors? If so, how? Because i'm treating the 2 woofers as one (in parallel).

Thanks for your article, it has helped me so much!!
Nicola
 
About flattening the impedance of the woofer:



Excuse me AllenB, what do you mean with "(per woofer)"? I'm designing a creossover for a speaker with 2 woofers and 1 tweeter, so do you mean that i have to consider 2 resistors and 2 capacitors? If so, how? Because i'm treating the 2 woofers as one (in parallel).

Thanks for your article, it has helped me so much!!
Nicola
Put two 8ohms drivers in parallel and the crossover sees them as a 4ohms load.
You design the crossover for that 4ohms load.
 
Put two 8ohms drivers in parallel and the crossover sees them as a 4ohms load.
You design the crossover for that 4ohms load.

Thank you AndrewT this is how i worked:
I have 2 woofer (8ohm) in parallel with

RE = 5,7 ohm
Le = 1,06mH

So
Resistance is (5,7 * 1,25)=7,125 ohm
Capacitor is (0,00106 / (7,125*7,125))=20 uF circa

Is this correct?
So the impedance of the woofers is not important in this step right?

Thanks
 
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what do you mean with "(per woofer)"?
Yes, I meant per side. If you use two woofers together you can do it a few different ways. Andrew has shown the most common. If you do it that way then just halve all the resistance values compared to using one driver... The driver resistance will be halved, the resistor to use in the crossover should come out at half the value that it would be for just one driver. The capacitor should come out at double the value for one driver.
 
AllenB...... Thanks for the educative material. Its a fantastic read for beginners like me.

A couple of questions with regard to my build which is a take on Paul Carmody's Sprite with a couple of twists. I am planning on using the ND90-8 full range drivers as separates using the same filter as the Sprite (0.9mH coil and 20uF resistor in parallel). I have already made a stock Sprite build last year which was quite good in the bass department, but I found the treble to be not very open (BTW, its quite shouty without the filter).

So, it got me thinking on making separates with the ND90-8's which would cross over 1st order to an 89dB AMT-8 tweeter around 9000Hz. Each separate will be in a 5W x 8H x 10D box (outer dimensions using 1/2" MDF). I had another set of of the Sprite filter lying with me and planned on using them in the build. My confusion was on where to add the low pass filter on the full range i.e. before or after the BSC.

After reading your article, I am considering on going with a 2nd order crossover on the AMT-8 and the ND90-8 and crossover at 7000Hz to take care of the slight rise in response of the ND90-8 at this region. The components that i'll be using for the crossover will be a 0.25mH coil in series and 2uF cap in parallel on the full range and a 2uF cap in series and 0.25mH coil in parallel on the tweeter. I'll be adding a 9dB L-Pad on the tweeter after the high pass filter using a 5.1uf resistor in series and a 4.3uF resistor in parallel.

Now the issue would be to ensure that the speaker doesn't sound shouty and adding the BSC components. Going by your article, the BSC can be added to the woofer after the crossover i.e by adding the inductor in series and the resistor in parallel. The inductor value shows 0.9mH (570Hz region) if i consider the 8"H or 0.5mH (900Hz region) if I consider the 5"W of the box. As I mentioned earlier, I already have a pair of 0.9mH coils with me.

Please let me know if I am on the right path or there is a different way to go about this. My confusion was how to incorporate the BSC in the full range crossover. The BSC point and crossover point are quite far off from one another.
 
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In the original filter for the 8 ohm version, the reactance of the 0m9H inductor gets to 20 ohms around 3k5Hz. This means that by 7kHz you are more or less dealing with just a 20 ohm resistor in series with the driver and you can use this when calculating your values. (Since the driver impedance will be partially inductive, the total will be slightly less than the sum of the driver impedance and the 20 ohms.)

The driver impedance will be rising at this frequency. If you were to use the RC impedance flattening method you will also affect the original filter. In this case I think it is the intention to try to avoid that. (not to say that you couldn't try it or there aren't a number of different ways this could all be done.) However if you use a 2nd order low pass filter you will gain some immunity from the variations.

It is unfortunate that inductor values tend to come out large when you are working into a high impedance.
 
Thanks Allen.....

So, my understanding from your response:
1. If I add the Sprite filter, the amp will see around 26-28 ohm past the 5000Hz mark (the 20 ohm resistor in series with the 8 ohm ND90-8).

2. I need to calculate the 2nd order low pass filter at 7000Hz based on a 26-28 ohm impedance for the ND90-8. The inductor value would be around 1.2mH based for a 2nd order LR.

If my above understanding is correct, I needed some more clarifications:
1. By adding the filter, have I changed the total impedance/load which the amp will see per channel. If so, then with the tweeter being 8 ohm and the ND90-8 now presenting a higher load and the wiring between the two being parallel, will this affect the total load being seen by the amp. Or is the calculations for parallel speaker connection load only applicable for two completely separate speakers on each channel and not for 2 or 3 way crossover networks where multiple drivers are used per channel.

2. Will I need to tweak the L-Pad on the tweeter in light of the above. I had planned a 9db L-Pad based on the 89db AMT-8 tweeter and the 81.4db ND90-8.

3. Will the BSC be before the low pass filter i.e. Amp---BSC---LPF---Zobel network (if any)---Driver.

Sorry for asking such basic questions. You are one of the very few pros on this forum who answer questions of newbies without being condescending or admonishing us.

Really appreciate your patience and time.
 
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If I add the Sprite filter, the amp will see around 26-28 ohm past the 5000Hz mark (the 20 ohm resistor in series with the 8 ohm ND90-8).
According to this plot of impedance, the driver is at 28ohms at 7kHz. Although this adds up to 48 ohms, the impedance phase shows it is not entirely resistive. I would guess 40 ohms is close enough. (but more on this below..)
By adding the filter, have I changed the total impedance/load which the amp will see per channel. If so, then with the tweeter being 8 ohm and the ND90-8 now presenting a higher load and the wiring between the two being parallel, will this affect the total load being seen by the amp. Or is the calculations for parallel speaker connection load only applicable for two completely separate speakers on each channel and not for 2 or 3 way crossover networks where multiple drivers are used per channel.
The short answer is that the amp will see the woofer below the crossover frequency and the tweeter above. This will not be a problem for most amps.
Will I need to tweak the L-Pad on the tweeter in light of the above.
This is one of the first things often done after constructing a crossover
and you can assume you'll want to do it.. but not because of the difference in impedance. Most amplifiers that you buy have a near zero output impedance and this means they provide the same Voltage regardless of the impedance of the load they are driving even if it is different at different frequencies.
Will the BSC be before the low pass filter i.e. Amp---BSC---LPF---Zobel network (if any)---Driver.
When you say BSC, I assume you are talking about the resistor and inductor in the original plan.

Here is where it really comes down to choice. If you want to avoid altering the existing sound and also keep it simple, then use Amp-LPF(L and C)-BSC-Driver. If this gives you a reasonable value of components to work with then this should give you a start.

The typical way to do this would be the way you have suggested above. If you do this, the BSC will have a greater effect on the filter. This will make it more difficult to predict the result if you aren't measuring. That's the only reason for making this suggestion.
 
When you say BSC, I assume you are talking about the resistor and inductor in the original plan.

Here is where it really comes down to choice. If you want to avoid altering the existing sound and also keep it simple, then use Amp-LPF(L and C)-BSC-Driver. If this gives you a reasonable value of components to work with then this should give you a start.

[/QUOTE]

The picture is now becoming more clear.

A couple of questions:

1. Would this be valid for a LPF 1st order too? i.e. Amp-LPF(L)-BSC-Zobel-Driver.

2. Would the Zobel stabilise the impedance to around upto the crossover point i.e will the amp see 8 ohms if I use an 8 ohm Zobel network?
 
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If you go first order, you will be putting a capacitor (should be “lnductor”-Ed.) in series with the RL circuit. It can go on either side of the RL, it makes no difference because each part has the same current running through it.

Please tell me what the Zobel is for and why you want it. Is it to make the amp see 8 ohms? Is it to make the crossover see 8 ohms? Is it to make the BSC see 8 ohms?
 
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