Results don't match the theory

[Jean Clochet]

I'm confused what you mean about another 15k resistor across the secondary though, did you mean to put 150k? At the moment it is just the 15k output of the transformer with the 150k load across it, no 15k resistors :S

No, not another 150K.  But DO leave the 150K in place. 
You are expecting the impedance to be reflected up to about 15K
If this reflected voltage is loaded by it's same value (15K), we have a 15K into 15K divider.  This is a 6dB attenuator.  Or half of the initial voltage.
So:

do you mean use a pot as a means of finding out what the output Z is? So when the pot is applying 6dB attenuation, take it out the circuit and measure its resistance, and whatever that is should be the output Z?

Yes! exactly

With regards to taking the resistance across the shunt, shouldn't it be the 599Ω from all the input resistors and shunt in parallel, also in parallel with the 150k divided by 25 as it is reflected back to the primary?

150k/25=6k
1/ (1/599 + 1/6000) = 544.6Ω

No.  You can't measure directly an impedance so the 150K load R /25 doesn't come into it.  This merely reflects an input "impedance" (which, again, you can't measure directly ) of circa 6K to the primary.  The secondary dcr does come into it but not enough to bother about here.
If it's convenient,  disconnect the transformer primary connections from all this stuff and measure the shunt value.  Then measure just the transformer dcr of the primary.  I think you'll have a Eureka moment

 

[ej_whyte]

So i did the thing with the pot, came out at 19.1k, in the right ball park i guess but not entirely as expected. Im guessing if i measured the DCR of the transformer primary it would be something low, which is why when i measure the Ω across the shunt it comes out at ~50Ω yeh?

Cheers

 

[Jean Clochet]

Im guessing if i measured the DCR of the transformer primary it would be something low, which is why when i measure the Ω across the shunt it comes out at ~50Ω yeh?

Yes exactly, that's why you got a low reading.

19K1 is a bit higher than our ideal transformation calc. but it's not an unreasonable number.  If you add in the dcr of the primary in series with the shunt resistance (we didn't in our quick calc.) and also the dcr of the secondary to the reflected up impedance, then we are about where it would be. 

It would be excellent if more folks would feel driven to do what you did and compare theory and actual. 
Well done.  A nice first layout that worked without issues.  And you understood what is going on.

Jean.

 

[ej_whyte]

Yeh I guessed that the primary DCR etc would be what was bringing it up to 19.1k from ~15k.

Checking that everything matches the theory and works as it should gives me more confidence for future designs and understand all the small details that are often overlooked. Also, the project that this is part of is going to be my dissertation when i get back to uni next year, so its important i keep all the graphs etc else i'll forget it all  Roll on next month when i get paid and can afford some more components!

Thanks for all the help.

 

[Jean Clochet]

Good deal.
Here's a white paper in case you need references for this stuff:

http://www.jensen-transformers.com/apps_wp.html

AN002 is the one I mean.

As was brought up in another thread, the result of 225 ohms that the writer  Bill Whitlock gets for the impedance presented to the amplifier is a little off, it should be 229ohms.
Roll on the end of the month then...  see ya,

Jean.