I got the following question:
Thank you for your responce stephan.
Let input voltage v = V sin(wt) ;
effective resistance seen by the driver circuit
= resistance in the primary ( assumed to be very small)
+ load resistance in the secondary
= r + R = R' (say)
now at resonance (after tunning the whole circuit ),
current in the pimry circuit i ~ v/R'
the voltage across the inductor = quality factor times the inputvoltage
= Q * v
Therfore simalar voltage Q*v and current I appears across the secondary coil of 1:1 transformer as well as across load resistance R .
Now ,
power supplied by the source P = ( V * I ) / 2 (product r.m.s values )
( also it is known that at resonance power factor = 1 (approximatly) )
power received by the load resistance P' = (Q* V*I)/2
.
. . efficiency of the circuit = (P' / P ) = Q
which can be made well overunity.
Regards
NAGARJUNA.NALLAM
Hi,
I will post it in my forum,
in my opinion normally the energy stored in coils
is limited and the Q will fall to a level under 1
if you draw the power fromm the secondary ?!
Is this true ?
But stefan,
when simulate the circuit using software (multisim) interestingly i observed both voltge & current in the secondary much close to those values in the primary . During simulation i observed that the Q is inversly proposional to resistance in the primary but directy proposional to the resistance in the secondary ,i.e. as load resistance increases Q of the circuit increses ( here even though voltage increses to a high value , current in both cicuits falls drastically , but still efficiency ( = Q) is much grater than one.).
Can you please post the output of the simulation as pictures over here ?
Many thanks.
Also the circuit diagramm please.
Pleaseattach it to a reply. Thanks.
Thanks.
Regards, Stefan.
sorry for the delay stefan. I am busy with my semester work in my college.
you are asked to send the simulation results & circuit diagram.But i don't have my own computer.I am doing all this work at internet cafe, this is also one of the resons for responding slowly .
stefan ,i hope u understood my circuit . once again i am going to discribe it. I am using the primary coil of a 1:1 transformer as the inductor of a series resonant circuit and across the secondary coil i am putting my load .Previously i forget simple thing that even though the current in a series resonant circuit is in phase with the source voltage , it is 90 degrees out of phase with respect to voltage across the inductor/ capacitor.Therefore same voltage & currents as in the primry coil will appear in the secondary ( which are 90 degrees out of phase) and therfore in the load.
Now let us concentrate on the secondary circuit.since secondary coil and load resistance are in series same currents should flow through both of them but what about the voltages? since load resistance and the secondary coil are the only elements in the secondary circuit and more over load resistance is across the secondary coil their volteges should also same.Here is the contradaction . voltage and current in pure inductor are 90deg out of phase but they should be in phase in the load resistance.Suppose if we assume voltages are equal ,how the phase of current changes when it pases from inductor to resistor so that it is in phase with the volage in the resistor. OR riversly If we assume current is same through both inductor and resistor how voltage across each element of a simple parellel combination will chage.
I hope you understand what i am asking. And finally stephan tommorow or day after tommorow i will keep very intersting results & observations my simulation, because currently i didn't brought them.
thank you for your pationce.
regards,
nagarjuna