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In Reply to: RE: You're right, I'm wrong. posted by Tre' on July 10, 2015 at 07:50:07
I think that is a backwards way to approach a design. I prefer to consider the worst case first and then move onward. in this case the load at 20 hz is 5k. if we deem that as acceptable, and we are driving a tube grid, do we then add additional loading in an attempt to keep the load unchanged?
i much prefer the results of the lighter loadat higher frequencies
dave
Follow Ups:
"i much prefer the results of the lighter load at higher frequencies"
Me too but you've missed my point.
If 20Hz is present in the signal the load for the "higher frequencies" will, in effect, be 5k not 10k (9800).
With the low frequencies filtered out the load for the higher frequencies will be much closer to the value (and shape) of the reflected impedance.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
If 20Hz is present in the signal the load for the "higher frequencies" will, in effect, be 5k not 10k (9800).
I don't see it that way. The higher frequencies will still see the value of the inductance as a load. here are two loadlines of a small signal 1000Hz signal plus a large signal 20Hz signal for a fixed amount of inductance. I do not see the width of the loadline as anywhere near the problem as the overall range of current.
Can you share the asc file pease?
Naz
It is just a twist on a file Stephie Bench did and is attached below.
I am still trying to get my head around what is happening particularly in a dynamic sense.
dave
Naz
Tre', you are correct regarding the shunt inductance of the transformer. The required inductance is a function of the *operating impedance* of the circuit, and that in turn is determined by the impedance reflected back onto the primary from the load. This operating impedance can be represented by △E/△I, and that term is controlled by load Z and turns ratio. It is relatively unaffected by the source Z, which generally has more effect on output power and efficiency than anything else.
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