Calculation of inductances for a coupling network

[brainbaby]

Hi guys

As per text...
To match 10k with 500Ω a coil is used as an autotransformer tapped at point 3...now for a certain bandwidth and centre frequency what should be the value of Lp and Ls...?? How do I calculate it..??

and
if the value of Ls and Lp is found out to be low then how could they be increased by using tapping to yield a high value of Q..??

Thanks in advance

 

[tech99]

Hi guys

As per text...
To match 10k with 500Ω a coil is used as an autotransformer tapped at point 3...now for a certain bandwidth and centre frequency what should be the value of Lp and Ls...?? How do I calculate it..??

and
if the value of Ls and Lp is found out to be low then how could they be increased by using tapping to yield a high value of Q..??

Thanks in advance

View attachment 110803

My understanding is that if an autotransformer is tuned then it may be regarded as a "perfect" transformer, because leakage inductance has been tuned out.
Between point 2 and ground there is 10k, and when the 500 Ohms of the secondary is also transformed up to 10k, this will result in a total resistance between point 2 and ground of 5k. Now decide what loaded Q you want, from bandwidth considerations. For instance, 10kHz B/W at 500 kHz is a Q of 50. Now assume for a moment that the capacitor will be placed from point 2 to ground. Its required reactance will be decided by the requirement to have a Q of 50. Q = Rp/Xc, so that Xc = Rp/Q = 5000 / 50 = 100 ohms. And the inductive reactance between point 2 and ground must also be 100 ohms.
But you actually want to use smaller capacitor, located at point 1, so assume a desired X value for it, say 200 ohms, then choose turns ratio between points 2 and 1 to give this step up in X. Turns ratio will be SQRT 200/100 = 1.4.
Having fixed the capacitor X you now make the inductance from point 1 to ground an equal value.
The inductive reactance between point 2 and ground has already been calculated, X=100 Ohms, so you can find inductance from
L = X / 2 pi f.
The tapping point 3 for the secondary is now done by just taking the SQRT of the turns ratio. So to obtain 500 Ohms, with 10k on point 2, the resistance ratio is 10000/500 = 20, and the turns ratio is SQRT 20 = 4.5.
The inductance from point 3 to ground is the square of the turns ratio, so it is again 1/20 of that from point 2 to ground.
In summary,
Assume a perfect auto transformer.
Relate everything to anyone tapping point.
Use SQRT (turns ratio) for stepping impedances up and down.
Find inductive reactance from your chosen value for Xc.
Find other inductance from the turns ratio squared.

 

[brainbaby]

My understanding is that if an autotransformer is tuned then it may be regarded as a "perfect" transformer, because leakage inductance has been tuned out.
Between point 2 and ground there is 10k, and when the 500 Ohms of the secondary is also transformed up to 10k, this will result in a total resistance between point 2 and ground of 5k. Now decide what loaded Q you want, from bandwidth considerations. For instance, 10kHz B/W at 500 kHz is a Q of 50. Now assume for a moment that the capacitor will be placed from point 2 to ground. Its required reactance will be decided by the requirement to have a Q of 50. Q = Rp/Xc, so that Xc = Rp/Q = 5000 / 50 = 100 ohms. And the inductive reactance between point 2 and ground must also be 100 ohms.
But you actually want to use smaller capacitor, located at point 1, so assume a desired X value for it, say 200 ohms, then choose turns ratio between points 2 and 1 to give this step up in X. Turns ratio will be SQRT 200/100 = 1.4.
Having fixed the capacitor X you now make the inductance from point 1 to ground an equal value.
The inductive reactance between point 2 and ground has already been calculated, X=100 Ohms, so you can find inductance from
L = X / 2 pi f.
The tapping point 3 for the secondary is now done by just taking the SQRT of the turns ratio. So to obtain 500 Ohms, with 10k on point 2, the resistance ratio is 10000/500 = 20, and the turns ratio is SQRT 20 = 4.5.
The inductance from point 3 to ground is the square of the turns ratio, so it is again 1/20 of that from point 2 to ground.
In summary,
Assume a perfect auto transformer.
Relate everything to anyone tapping point.
Use SQRT (turns ratio) for stepping impedances up and down.
Find inductive reactance from your chosen value for Xc.
Find other inductance from the turns ratio squared.

Thanks for your analysis...I will be back once I have
done