From: ruffrecords@yahoo.com   
      
   flipper wrote:   
   > On Sat, 02 Oct 2010 11:03:10 +0100, Ian Bell   
   > wrote:   
   >   
   >> Ian Iveson wrote:   
   >>>>>> One thing I was wondering was if I use a transformer   
   >>>>>> with   
   >>>>>> a nominal 16 ohm secondary but use it with 32 ohm   
   >>>>>> headphones for example, then ra reflected into the   
   >>>>>> secondary would be so low relative to the load that  the   
   >>>>>> output could be considered essentially a voltage source.   
   >>>>>> Equally the load reflected to the plate would be four   
   >>>>>> times higher so distortion would be lower?   
   >>>>>>   
   >>>   
   >>> Language. Lower load = higher resistance.   
   >>>   
   >>> If you follow my logic, my point is clear. The load   
   >>> presented by the primary of a transformer comprises:   
   >>>   
   >>> 1. Primary winding resistance   
   >>> 2. Leakage inductance in series with 1   
   >>> 3. Reflected secondary winding resistance in series with 1   
   >>> and 2   
   >>> 4. Reflected secondary load in series with 1, 2 and 3   
   >>> 5. Primary inductance in shunt with 3 and 4   
   >>>   
   >>> Think.   
   >>>   
   >>   
   >> Nope, maybe I am just thick. You said "10H is in shunt with load. Lower   
   load makes shunt more   
   >> significant."   
   >>   
   >> Now you are saying "Language. Lower load = higher resistance."   
   >>   
   >> Which means "Higher resistance makes shunt more significant"   
   >>   
   >> That means to drive a higher resistance load I need MORE inductance in the   
   primary.   
   >>   
   >> Does not make sense to me.   
   >   
   > It's because the inductance is in parallel with the load impedance.   
   >   
   > With a 5k (or any other value) OPT you'd like it to be that same   
   > impedance at all frequencies but it won't because the inductance   
   > impedance, which is in parallel with the load impedance, falls in   
   > value as the frequency decreases, so the parallel value also falls.   
   >   
   > Let's pick a convenient reference point: effective impedance half of   
   > nominal. That's going to occur at a frequency where the 10 Henry   
   > impedance equals the OPT nominal impedance, I.E. 5k. If the nominal   
   > OPT impedance is 10k then that (parallel impedance half of nominal) is   
   > going to happen at a higher frequency.   
   >   
   > Now, that was presuming the EL84 in pentode mode where the tube's   
   > anode impedance is high and, so, not much of a factor. See Patrick's   
   > post for triode mode.   
   >   
      
   Yes, you are of course right. I have a stinking headache today and I just   
   cannot think straight.   
   Bigger L means higher Zl or lower load - duh??   
      
   Cheers   
      
   ian   
      
   >> Cheers   
   >>   
   >> Ian   
   >>   
   >>> Ian   
   >>>   
   >>>   
      
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