Here's a proposal for modeling the frequency effects of a transformer. Comments regarding appropriateness or accuracy are welcome. -------------------------------------------------------------------- Feedback and Compensation FAQ: Transformer Frequency Response Model: Transformers are more complex than RC networks from a frequency response view. The following formulas allow you to predict the frequency response of your circuit with transformer coupling. Definitions: Lp = Primary inductance LL = Leakage inductance Cp = Primary winding capacitance Cs = Secondary winding capacitance n = Turns Ratio f = Low Frequency pole fr = High Frequency resonant point rp = Plate resistance (also add primary winding resistance) Rl = Load resistance (also add secondary winding resistance) Q = Circuit Q of the high frequency pole pair Low Frequencies: Pole: Unterminated transformer case: f = rp/(2*pi*Lp) Terminated transformer case: The resistance in the above formula is the parallel of the primary side resistance (rp) in parallel with the load resistance (rl) referred to the primary (making it rl*n*n) so... f = Rl*rp*n*n/(2*pi*Lp*(Rp + (n*n*rl))) Zero: There is a zero at DC High Frequencies: Poles: There is a set of "complex" hi frequency poles. This leads to peaking at high frequencies as has been discussed in this newsgroup. The formulas given allow you include the effect of these on your feedback circuit. Since the pole-pair is "complex" I have to introduce the concept of "Q". Higher Q leads to more peaking, lower Q to less peaking. The following table provides an insight into how much high frequency peaking to expect: Q Peaking (dB) -3dB point W.R.T. "resonant" freq. 2 6 dB 1.5 1.41 3 dB 1.4 1 1 dB 1.25 .8 0.3 dB 1.1 .7 0 dB 1 .6 no peaking 0.8 .5 no peaking 0.65 Resonant Frequency: fr = 1/(2*pi*sqrt(LL*(Cp+(n*n*Cs))) Circuit Q: Q = (2*pi*fr*LL)/(rp paralleled with (n*n*Rl)) Example: A 1:1 transformer couples a 6SN7 plate into the grid of a class AB1 triode. The grid resistance is assumed high. (rp = 7.7k) Transformer characteristics Lp = 25 Henry LL = 0.05 Henry Cp = 100 pF Cs = 100 pF Winding resistance assumed negligible. Use "unterminated" cases: Low Frequency pole is at 7700/(2*pi*25) = 49 Hz. Hi Frequency Resonance is at 1/(2*pi*sqrt(0.05*200pf) = 50.3 kHz Q is (2*pi*50300*0.05)/7700 = 2.05 From the table, the -3 dB point is 1.5*50300 = 75.5 kHz. There will be a 6 dB peak in the response at about 50kHz.