Jump to content

Crossovers with a cap and coil on either side of autoformer: second order?


Recommended Posts

2 hours ago, JohnA said:

Yes.  Putting a component on one side or the other can change the size and probably the cost of that component due to the impedance it operates in to.

Thanks John. I understand why Klipsch put caps on the upstream side of the autoformer! :) 

Link to comment
Share on other sites

I'm gonna nay-say here.  I don't see much change in cap value being required either side of the transformer.  The impedance is the same enough either side, or at least should be.  If anything, it might could cut down on the size of the parallel inductor required if you count the inductance of the transformer, but I believe a coil doesn't act like much of an inductor (the way the inductor proper is being used here) when it's being used as a transformer...

Link to comment
Share on other sites

The turns ratio squared of the 3507 is around 6.4, so the series load on the 3uf cap will always be 6.4 times the impedance of the driver in parallel with the inductor.

If the cap were on the other side, it would need to be 6.4 times larger to get the same results.  

Link to comment
Share on other sites

On 5/20/2019 at 7:38 AM, glens said:

I'......... The impedance is the same enough either side, or at least should be.  If ..

 

No, not accurate.  In a TypeA/AA, the impedance is twice as high before the autoformer as the driver, resulting in a cap that is half as large. 

Link to comment
Share on other sites

I was unaware that impedance factored through a transformer.  I wonder why they were used with amps having output transformers?  I always figured output transformers were just for voltage-matching.  Got a link to any of PWK's notes on the subject?

Link to comment
Share on other sites

As with all transformers, watts in equals watts out, so let's use Ohm's Law.

 

As an example, let's use a T2A autoformer, and apply 2.83 volts to taps 0-5.

Taps 0-4 will connect to a 14 ohm K-55 driver at 2.00 volts (-3db).

 

2 volts into a 14 ohm load equals .2857 watts  (watts = volts squared/resistance, or 4/14).

 

Since there's .2857 watts output, there's got to be .2857 watts input.

 

2.83 volts at .2857 watts equals 28 ohms (ohms = volts squared / watts, or 8 / .2857).

 

As you can see, taps 0-5 appear to be a 28 ohms load.

 

Hope that helps!

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...