Lecture 15: Transformer Shunt Inductance. Tuned Transformers.

Transcription

1 White, EE 3 ecture 15 Page 1 of 8 ecture 15: Tranforer Shunt nductance. Tuned Tranforer. n the lat lecture, we derived the tranforer equation V = V (6.1) and V jω (6.) where = agnetization current and t = tranforer current. An equivalent electrical circuit for uch a nonideal tranforer can be contructed fro thee two equation (Fig. 6.4a): T = deal tranforer t t V V : n articular, we have a hunt inductor that aear at the riary terinal of an ideal tranforer. With the hunt inductance in the odel, the higha nature of a hyical tranforer i roerly accounted for, ince at DC the riary terinal of T will be horted. 006 Keith W. White

2 White, EE 3 ecture 15 Page of 8 Without the hunt, the ideal tranforer aear to tranfor voltage and current equally well for all frequency, which cannot be true (by Faraday aw). Secondary nductance Equivalently, the hunt inductance can alo have been incororated fro the econdary. To do thi, we begin again with Fig. 6.: ψ V V A l ψ and (6.1): ψ = A l A l (1) Solving for ψ = A () l V ow, fro (6.10) V = jωψ or ψ = jω (3) Subtituting (3) into () leave V = (4) jω A l

3 White, EE 3 ecture 15 Page 3 of 8 or = V jω (5) t The equivalent electrical circuit for (6.1) and thi lat exreion i (Fig. 6.4b): t V V : To check the direction for the current hown in thi figure, we can aly KC at the econdary: t = or = t (6) The inu ign here agree with (5). To uarize thi work o far, whether the agnetization current effect i included on the riary ide or the econdary ide of the tranforer i iaterial: they are equivalent. Actually, we can develo thi latter econdary inductance equivalent circuit ily fro the iedance tranforation roerty of the ideal tranforer! Begin with

4 White, EE 3 ecture 15 Page 4 of 8 deal T V V : We ve een reviouly that for an ideal tranforer = (7), (6.19) Here = jω and = A l (8), (6.3) o that j ( A l ) j A l ω ω or = jω which we can odel ily a an ideal tranforer with a hunt a hown on the reviou age (Fig. 6.4b). Tuned Tranforer Achieving iedance atch between the variou ubyte in a ultitage counication circuit i very iortant. Otherwie, reciou ignal i needlely wated.

5 White, EE 3 ecture 15 Page 5 of 8 Tranforer ecifically ideal tranforer can be ued a atching network ince, a we ve already een, = Ω (6.19) We can chooe / to change (or tranfor ) to a deired value for atching. ote that (6.19) i valid only for ideal tranforer. One way to negate the effect of the agnetization current in a ractical tranforer (o that the ideal T equation aly) i to ue a tuning caacitor: deal T V C V : We can adjut C to reonate out the effect of at the deired frequency of oeration. That i, uoe the tranforer i deigned to oerate at f = f 0. For an C reonance at f0 = 1/ ( π C ), then adjut C uch that 1 C = 4π f F (9) 0 Conequently, now at thi oerating frequency f 0

6 White, EE 3 ecture 15 Page 6 of 8 c c = = c and the equivalent circuit for thi tuned tranforer circuit becoe deal T V C V no effect at reonance : which i ily an ideal tranforer. Very cool! Thi reonant ethod i only a narrowband olution, but it can be extreely ueful. Caacitive tranforer tuning effectively ake a banda filter fro a higha filter. Exale The two tuned tranforer in the orcal 40A are T (RF Filter) and T3 (atching between RF Mixer and F Filter). et conider both of thee quickly once again in the light of our exanded undertanding of tranforer.

7 White, EE 3 ecture 15 Page 7 of 8 1. T3 (between RF Mixer and F Filter): Thi tranforer wa alo conidered in the reviou lecture. T3 i ued to tranfor the outut iedance fro the RF Mixer to atch the inut iedance of the F Filter (00 Ω). Fro the data heet for the SA60A C, the outut iedance i 1500 Ω= 3000 Ω. Uing (7): 6 = = 3000 = 04. Ω 3 which i very cloe to the deired 00 Ω for the F Filter!. T (RF Filter): Conider once again the econd order Butterworth banda filter exale we dicued earlier in ecture 1: 50 Ω 1 C 1 V 86. nh 6.0 nf 50 Ω 1 and C1 are coonent that are oldered onto your PCB. Where do the 86. nh and 6.0 nf coonent coe fro? A we entioned earlier in ecture 1, they both coe fro T! 9 To ee thi exlicitly for T, = A l = H/turn 1 = 66 nh, which i cloe to the 86. nh hown above that i needed for a econdorder banda filter.

8 White, EE 3 ecture 15 Page 8 of 8 What about the C? That coe fro C and the iedance tranforing roertie of T: = = C C With = 0, = 1 and C = 15 F, then 0 C = 15 F = 6 nf 1 which i exactly the value needed for the econd order Butterworth banda filter! Very cool.