28. 선형 변압기

28. 선형 변압기

반사 임피던스

$\displaystyle\mathbf{V}_{s}=(R_{1}+j\omega L_{1})\mathbf{I}_{1}-j\omega M\mathbf{I}_{2}$

$\displaystyle0=-j\omega M\mathbf{I}_{1}+(R_{2}+j\omega L_{2}+\mathbf{Z}_{L})\mathbf{I}_{2}$

$\mathbf{Z}_{11}=R_{1}+j\omega L_{1}$, $\mathbf{Z}_{22}=R_{2}+j\omega L_{2}+\mathbf{Z}_{L}$이라 하면 $\displaystyle\mathbf{V}_{s}=\mathbf{Z}_{11}\mathbf{I}_{1}-j\omega M\mathbf{I}_{2}$, $0=-j\omega M\mathbf{I}_{1}+\mathbf{Z}_{22}\mathbf{I}_{2}$이고 입력 임피던스는 $\displaystyle\mathbf{Z}_{in}=\frac{\mathbf{V}_{s}}{\mathbf{I}_{1}}=\mathbf{Z}_{11}-\frac{(j\omega)^{2}M^{2}}{\mathbf{Z}_{22}}$

(입력 임피던스를 반사 임피던스라고 한다.)

$R_{22}=\text{Re}\mathbf{Z}_{22}$, $X_{22}=\text{Im}\mathbf{Z}_{22}$라 하면 $\displaystyle\mathbf{Z}_{in}=\mathbf{Z}_{11}+\frac{\omega^{2}M^{2}}{R_{22}+jX_{22}}=\mathbf{Z}_{11}+\frac{\omega^{2}M^{2}R_{22}}{R_{22}^{2}+X_{22}^{2}}-j\frac{\omega^{2}M^{2}X}{R_{22}^{2}+X_{22}^{2}}$

T와 $\Pi$ 등가회로망

(1): $\displaystyle\frac{di_{2}}{dt}=\frac{v_{2}}{L_{2}}-\frac{M}{L_{2}}\frac{di_{1}}{dt},\,v_{1}=L_{1}\frac{di_{1}}{dt}+M\frac{v_{2}}{L_{2}}-\frac{M^{2}}{L_{2}}\frac{di_{1}}{dt}\,\Rightarrow\,\frac{di_{1}}{dt}=\frac{L_{2}}{L_{1}L_{2}-M^{2}}v_{1}-\frac{M}{L_{1}L_{2}-M^{2}}v_{2}$

0부터 $t$까지 적분하면 $\displaystyle i_{1}-i_{1}(0)u(t)=\frac{L_{2}}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{v_{1}dt'}-\frac{M}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{v_{2}dt'}$

(2): $\displaystyle\frac{di_{1}}{dt}=\frac{v_{1}}{L_{1}}-\frac{M}{L_{1}}\frac{di_{2}}{dt},\,v_{2}=L_{2}\frac{di_{2}}{dt}+M\frac{v_{1}}{L_{1}}-\frac{M^{2}}{L_{1}}\frac{di_{2}}{dt}\,\Rightarrow\,\frac{di_{2}}{dt}=-\frac{M}{L_{1}L_{2}-M^{2}}v_{1}+\frac{L_{1}}{L_{1}L_{2}-M^{2}}v_{2}$

0에서 $t$까지 적분하면 $\displaystyle i_{2}-i_{2}(0)u(t)=\frac{-M}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{v_{1}dt'}+\frac{L_{1}}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{v_{2}dt'}$

$\displaystyle i_{1}-i_{1}(0)u(t)=\frac{L_{2}}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{v_{1}dt'}-\frac{M}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{v_{2}dt'}=\frac{L_{2}-M}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{v_{1}dt'}+\frac{M}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{(v_{1}-v_{2})dt'}$

$\displaystyle i_{2}-i_{2}(0)u(t)=\frac{-M}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{v_{1}dt'}+\frac{L_{1}}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{v_{2}dt'}=\frac{L_{1}-M}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{v_{2}dt'}+\frac{M}{L_{1}L_{2}-M^{2}}\int_{0}^{t}{(v_{2}-v_{1})dt'}$

이때 $\displaystyle\frac{1}{L_{B}}=\frac{M}{L_{1}L_{2}-M^{2}},\,\frac{1}{L_{A}}=\frac{L_{2}-M}{L_{1}L_{2}-M^{2}},\,\frac{1}{L_{C}}=\frac{L_{1}-M}{L_{1}L_{2}-M^{2}}$이고 $\displaystyle L_{A}=\frac{L_{1}L_{2}-M^{2}}{L_{2}-M},\,L_{B}=\frac{L_{1}L_{2}-M^{2}}{M},\,L_{C}=\frac{L_{1}L_{2}-M^{2}}{L_{1}-M}$(참고: $M\leq\sqrt{L_{1}L_{2}}$이므로 $L_{1}L_{2}-M^{2}\geq0$)

참고자료:

Engineering Circuit Analysis 8th edition, Hayt, Kemmerly, Durbin, McGraw-Hill