KennyB
Nov18-08, 04:28 AM
I am trying to work out the BW of my transformer.
My source resistance, Rs, is 50R. My load resistance, RL=50R. The windings ratio ,N, is 1:1. The magnetising current, Lmag=7.96uH. The data i have suggests that to find the lower -3dB point, you refer the secondary load to the primary. This gives RL'=50R. In order to find the cut off point, the resistance of the primary is equated to the reactance of Lmag. The combined resistance is defined as Rs//RL'= 50R//50R = 25R. The lower cut-off is then evaluated as 500KHz. A little confused by this because although the leakage inductances are essentially shorts at this frequency, Lmag appears in // with the primary referred load, RL'...and these two components combined are in // with Rs. Unsure.
Now for the upper -3dBpoint...the data now says that the total resistance is Rs + RL'. Why is Thevenin's not applicable here?! The primary winding leakage inductance equals the secondary winding leakage inductance = (1-K)*Lmag where K = 0.998333. The upper -3dB point was evaluated as 600MHz.
Can somebody please clear this up for me?
PS: The reason why i'm interested in the maths of this is that i want to input this data into SIMetrix for analysis with additional analogue front-end circuitry.
My source resistance, Rs, is 50R. My load resistance, RL=50R. The windings ratio ,N, is 1:1. The magnetising current, Lmag=7.96uH. The data i have suggests that to find the lower -3dB point, you refer the secondary load to the primary. This gives RL'=50R. In order to find the cut off point, the resistance of the primary is equated to the reactance of Lmag. The combined resistance is defined as Rs//RL'= 50R//50R = 25R. The lower cut-off is then evaluated as 500KHz. A little confused by this because although the leakage inductances are essentially shorts at this frequency, Lmag appears in // with the primary referred load, RL'...and these two components combined are in // with Rs. Unsure.
Now for the upper -3dBpoint...the data now says that the total resistance is Rs + RL'. Why is Thevenin's not applicable here?! The primary winding leakage inductance equals the secondary winding leakage inductance = (1-K)*Lmag where K = 0.998333. The upper -3dB point was evaluated as 600MHz.
Can somebody please clear this up for me?
PS: The reason why i'm interested in the maths of this is that i want to input this data into SIMetrix for analysis with additional analogue front-end circuitry.