
#1
May313, 04:28 PM


#2
May313, 05:19 PM

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I got a error message when I tried to follow the link to that image. Added in Edit: OK. It's visible now. 



#3
May313, 06:34 PM

P: 26

Ok I think that should work




#4
May313, 07:45 PM

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Equivalent Mutual Inductor CircuitsFor the "T" circuit, consider the unloaded case where there is no secondary load resistance. If M=1, the secondary voltage equals the primary excitation voltage. So what does that imply about the bottom inductor value? And when M=0.5, you get half of the primary voltage making it to the secondary, so what does that imply about the ratio of the upper inductors to the lower inductor? I haven't gotten very far yet on the case where there are source and load impedances, and including the turns ratio part. But I think approaching it from the ideal cases is helpful to start to figure out the method to use... 



#5
May313, 07:55 PM

P: 26

Thanks for your contribution. I will work on what you have provided :) 



#6
May313, 10:21 PM

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If you write the portion of the KCL loop equations which include the inductances in Laplace form for the original and "Y" (or "T") cases, you can equate the coefficients. The result is quite nifty.




#7
May413, 11:42 AM

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I have my doubts about this. How can you get a voltage gain > 1 from either of those circuits, or can one or more inductances be negative? 



#8
May413, 01:40 PM

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$$K = \frac{M}{\sqrt{L_1 L_2}}$$ and 0 ≤ K ≤ 1. 



#9
May413, 02:25 PM

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Take the L1L2L3 network, ascribe any values you like to those inductors, and show me how the output can exceed the input? 



#10
May413, 02:28 PM

P: 26

whoa whoa whoa guys I'm still stumped on how I can do KCL on this. I'm a mechanical engineer not EE.




#11
May413, 02:50 PM

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What's preventing you? Not that I especially recommend it, especially with no output load element. Then it's just ac voltage dividers ...
Anyway, these "equivalent" circuits can't handle the case where k*sqrt(Ls/Lp) > 1, k = M/sqrt(Lp*Ls), 0< k< 1. In other words, turns ratio > 1. 



#12
May413, 02:55 PM

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We have this nifty "Quote" feature to resolve this issue  especially when more than two people are involved in the discussion. After all, it is anol1258's thread. 



#13
May413, 04:07 PM

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#14
May413, 05:35 PM

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#15
May413, 05:42 PM

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These ckts are really, really pointless. They don't include winding resistance, not to mention stepup capability. They do include leakage inductance, ya gotta give 'em credit for that i suppose. 



#16
May413, 10:34 PM

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A negative inductance is really a capacitance. jωL looks just like 1/(jωC) for a suitable choice of C. So the circuit becomes second order and resonance effects obtain. For a sinusoidal stimulation, this can appear as amplification (stepup). The relationship that I found using term matching does yield a negative value for one of the "T" inductors when the mutual inductance exceeds the value of either winding's inductance. 



#17
May513, 11:16 AM

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If the equiv ckt is only good at one frequency then I wouldn't want to put a step function into it ... use it as a pulse transformer ... 


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