vtb9046 insertion loss probelm

[skal1]

Ok lads, i have wired up the i/p tx as the document suggest for 2k4 pri in parallel and 2k4 sec in series , so this should have no drp in voltage right? .

so d/a is balanced output impedance ca .1k ohms and max. output level 2dbv, this is driving the transformer and the out of tx is going to monitor .


supposedly their should be no insertion lose with this set  but have a lot of lose, have i wired incorrectly?  am i missing some thing ?


cheers

skal1

 

[joaquins]

How much is lor of lose? What Z is the input of your monitor?

For insertion loss should be counted DC resistance of windings, driving Z and loading Z... DC of windings are there, the rest how much is?

JS

 

[skal1]

driving Z =1k ,loading Z 10k and dc resistance of primary is  87.5 ohms and secondary is 112 .

so how do i find the loss do i have to add or subtract these figures .?


skal

 

[joaquins]

this will give you about 1dB, just a voltage divider... input Z divided by total equivalent series Z, if tx is not 1:1 should be converted by impedance ratio. Assuming everything R--> 20*Log(10K/11K2)* gives your 1dB lose. Shouldn't be a problem, but I don't know, how sure you are your input is 10K? did you measure it?
*11K2 = 10K+1K+87.5+112

Could you measure how much lose you are having now? If you hear it when A/B comparison is a thing, if you connect everything and don't listen to it for some minutes and still makes you itch are diferent things...

JS