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Driver Impedence


eth2

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I am confused about driver impedance. I understand the concept of impedance, but what are the practical implications for a system. Specifically, if I go with a 6 ohm driver, how does that effect the type and build of the crossover. In a K'horn, what other factors need to be considered if I go to a 6 ohm driver rather than the 16 ohm K-55?

Thanks again

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I have just bought Eliptrac horns and tweeters. I now know I cannot use the standard Klipshorn crossovers. I was considering buying a set of AP-12 350's. A friend has offered to sell me a pair of Electro Voice EV DH1A's (6 ohm) for the Eliptrac's. I am using a 40 watt tube amp at present with the Khorns.

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I believe the AP-12 350's have a 10 ohm "swamping" resistor across taps 0 and 5 of the autoformer.  A 16 ohm driver attached to the -6 db

taps actually increases the impedance the amplifier "sees" by a factor of 4, in this case the driver's impedance is now 64 ohms.

 

A 10 ohm resistor in parallel with a "64" ohm driver looks like 8.6 ohms of resistance.

 

Replacing the 16 ohm driver with a 6 ohm driver, using the same taps, would be 6 X 4 or 24 ohms.

 

A 10 ohm resistor in parallel with a "24" ohm driver looks like 7 ohms of resistance.

 

That 1.6 ohm difference probably won't make much of a differance.  If you replaced the swamping resistor with, say, a 14 ohm

resistor, the difference would be even less.

 

An e-mail to Al at alkeng.com might ease your mind :)

 

Hope this helps.

 

Mike

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I am confused about driver impedance. I understand the concept of impedance, but what are the practical implications for a system. Specifically, if I go with a 6 ohm driver, how does that effect the type and build of the crossover. In a K'horn, what other factors need to be considered if I go to a 6 ohm driver rather than the 16 ohm K-55?

Thanks again

 

Are you talking about impedance or resistance?

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When using an auto-former / auto-transformer, I believe the impedance changes with the turns ratio within the transformer.   As djk said, you can change the tap to change the impact on the crossover point to get a driver that is not the correct impedance back closer to the original drivers impedance.  The impact on crossover points will be determined by the end result of both components put together impedance.

You can look at the DIY audio's site for "crossover designer" to see various impacts of an impedance change on the component values.

Impact on sound, you won't know until you try it.  Does the driver have the same efficiency and usable frequency range?

 

Since you already know what drivers you are trying to switch to, you can follow what has been everyone has said here.  If you find out it is not adequate for your ears, the above information I put in will be helpful in you in your tweaking quest.  Generic solutions are good but not optimal.

 

Edited due to brain fade  :wacko:

Edited by pzannucci
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When using an auto-former / auto-transformer, I believe the impedance changes with the turns ratio within the transformer.  So if the is a 2 to 1 turn ratio, impedance should double.  

 

Actually, The impedance changes by the square of the turns ratio

 

https://drive.google.com/viewerng/viewer?url=http://www.critesspeakers.com/3636atz.pdf

 

Again, unless someone disagrees, the 10 ohm resistor in parallel with the varying impedance changes will eliminate

the need for crossover mods.  Without that resistor, every tap change would require modification of the squawker filter.

 

If this were, say, a stock AA crossover, the 13uf cap to the K-55 would need to be replaced with a different

value for each tap change.  Otherwise, the crossover frequency would change.  The 13uf cap's impedance

is around 31 ohms at 400 Hz, matching the K-55 impedance doubling that the 0 - 4 output taps create.

 

Cheers

 

Mike

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When using an auto-former / auto-transformer, I believe the impedance changes with the turns ratio within the transformer.  So if the is a 2 to 1 turn ratio, impedance should double.  

 

Actually, The impedance changes by the square of the turns ratio

 

https://drive.google.com/viewerng/viewer?url=http://www.critesspeakers.com/3636atz.pdf

 

Again, unless someone disagrees, the 10 ohm resistor in parallel with the varying impedance changes will eliminate

the need for crossover mods.  Without that resistor, every tap change would require modification of the squawker filter.

 

If this were, say, a stock AA crossover, the 13uf cap to the K-55 would need to be replaced with a different

value for each tap change.  Otherwise, the crossover frequency would change.  The 13uf cap's impedance

is around 31 ohms at 400 Hz, matching the K-55 impedance doubling that the 0 - 4 output taps create.

 

Cheers

 

Mike

 

Mike,

Thanks for that correction.  I was looking at current vs impedance.

As for the other, your solution might work but if their are efficiency and response differences in the drivers, you may want to know a little more about the interaction of the components.  I find the calculator very handy along with other items referenced off that web page on DIY.

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When using an auto-former / auto-transformer, I believe the impedance changes with the turns ratio within the transformer.  So if the is a 2 to 1 turn ratio, impedance should double.  

 

Actually, The impedance changes by the square of the turns ratio

 

https://drive.google.com/viewerng/viewer?url=http://www.critesspeakers.com/3636atz.pdf

 

Again, unless someone disagrees, the 10 ohm resistor in parallel with the varying impedance changes will eliminate

the need for crossover mods.  Without that resistor, every tap change would require modification of the squawker filter.

 

If this were, say, a stock AA crossover, the 13uf cap to the K-55 would need to be replaced with a different

value for each tap change.  Otherwise, the crossover frequency would change.  The 13uf cap's impedance

is around 31 ohms at 400 Hz, matching the K-55 impedance doubling that the 0 - 4 output taps create.

 

Cheers

 

Mike

 

Mike,

Thanks for that correction.  I was looking at current vs impedance.

As for the other, your solution might work but if their are efficiency and response differences in the drivers, you may want to know a little more about the interaction of the components.  I find the calculator very handy along with other items referenced off that web page on DIY.

 

I'm just learning this stuff, and the terms still confuse me  :)

 

I agree that there are many factors to designing a passive crossover.  The OP was referring to driver nominal impedance only.  A drivers 

impedance curve varies so much by frequency, that crossover calculators are guidelines at best.

 

I'm leaning more and more towards active crossovers.   They seem to make more sense to me.

 

Mike

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"Again, unless someone disagrees, the 10 ohm resistor in parallel with the varying impedance changes will eliminate

the need for crossover mods.  "

 

Prove it mathematically (you can't), it reduces variations, it does not eliminate them.

 

The EV is 110dB/W on a horn with a Q=7, so is the K55V.

 

If you move the 8Ω EV down a tap it should play the same level as the 16Ω K55V and have the same crossover point.

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Guess I have a lot to learn  :unsure:

 

I'd appreciate it if someone would correct my math...

 

The 8 ohm and 16 ohm drivers have equal sensitivities...110 db @ 1 watt.

 

The 8 ohm driver would need a 2.83 volt signal to generate 110 db.

The 16 ohm driver would need a 4 volt signal to generate 110 db.

 

Putting the 16 ohm driver on the 0 - 4 taps would double its impedance to 32 ohms.

4 volts into 32 ohms equals 1/2 watt, which equals 107 db.

 

Putting the 8 ohm driver on the 0 - 3 taps would quadruple its impedance to 32 ohms.

2.83 volts into 32 ohms equals 1/4 watt, which equals 104 db. 

 

Yes, the "crossover frequency" is unchanged,  but the wouldn't the amplitude would be different?

 

djk is correct.  If I have this math wrong, I have everything wrong!

 

Thanks, Mike

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Not to go to far off topic, but if one is running a active crossover with amps that have different ohm taps should you go with the correct ohm tap for the driver. I guess my question would picking the same corresponding ohm tap to match the driver be the same as what eth2 is talking about with passive networks. 

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"but the wouldn't the amplitude would be different?"

 

What are you on about?

 

My head hurts, this is so simple.

 

Feed 5.6V into tap 5, you get 4V on tap 4, and 2.8V on tap 3.

 

4V is 1W/16Ω, 2.8V is 1W/8Ω

 

Both drivers are 110dB/W, so what is it you don't understand?

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"but the wouldn't the amplitude would be different?"

 

What are you on about?

 

My head hurts, this is so simple.

 

Feed 5.6V into tap 5, you get 4V on tap 4, and 2.8V on tap 3.

 

4V is 1W/16Ω, 2.8V is 1W/8Ω

 

Both drivers are 110dB/W, so what is it you don't understand?

 

Thanks for the clarification.  And sorry about the headache.

I knew I had something wrong...just couldn't see it.

 

The turns ratio on tap 4 is 1.414.

In your example, 5.6 volts / 1.414 = 4 volts.

 

The turns ratio on tap 3 is 2.000

5.6 / 2 is 2.8.

 

My apologies to the OP.  

 

Mike

 

 

 

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