CLLC谐振变换器_基波分析
目录
- CLLC谐振变换器_基波分析
- 目录
- 电路图
- FHA分析
- 基于FHA的电路增益特性分析
电路图
FHA分析
- 输入电压FHA分析
谐振输入假设为理想方波
V i ( t ) = 4 V i n π ∑ n = 1 , 3 , 5... ∞ 1 n s i n ( 2 π n f s t ) V_{i}(t)=\frac{4Vin}{\pi}\sum_{n=1,3,5...}^{\infin}\frac{1}{n}sin(2\pi nf_st) Vi(t)=π4Vin∑n=1,3,5...∞n1sin(2πnfst) （方波的傅里叶变换）
基波提取
V i ( t ) = 4 V i n π s i n ( 2 π f s t ) V_i(t)=\frac{4Vin}{\pi}sin(2\pi f_st) Vi(t)=π4Vinsin(2πfst)
有效值提取
V i , F H A = 2 2 V i n π V_{i,FHA}=\frac{2\sqrt{2}Vin}{\pi} Vi,FHA=π22Vin - 输出电压FHA分析
谐振输出也设为理想方波
V o ( t ) = 4 V o π ∑ n − 1 , 3 , 5... ∞ 1 n s i n ( 2 π n f s t − ϕ ) V_{o}(t)=\frac{4Vo}{\pi}\sum_{n-1,3,5...}^{\infty}\frac{1}{n}sin(2\pi nf_st-\phi) Vo(t)=π4Vo∑n−1,3,5...∞n1sin(2πnfst−ϕ)（ ϕ 为相对输入方波的相移 \phi 为相对输入方波的相移 ϕ为相对输入方波的相移）
基波提取
V o ( t ) = 4 V o π s i n ( 2 π f s − ϕ ) V_{o}(t)=\frac{4V_o}{\pi} sin(2\pi f_s-\phi) Vo(t)=π4Vosin(2πfs−ϕ)
有效值提取
V o , F H A = 2 2 V o π V_{o,FHA}=\frac{2\sqrt{2}V_o}{\pi} Vo,FHA=π22Vo - 输出电流FHA分析
i ( t ) = 2 I r c t , F H A s i n ( 2 π f s t − ϕ ) i(t)=\sqrt{2}I_{rct,FHA}sin(2\pi f_st-\phi) i(t)=2Irct,FHAsin(2πfst−ϕ)
平均输出电流
I o = 2 T s ∫ 0 T s 2 ∣ i r c t , F H A ( t ) ∣ d t = 2 2 π I r c t , F H A I_o=\frac{2}{T_s}\int_0^{\frac{T_s}{2}}|i_{rct,FHA}(t)|dt=\frac{2\sqrt{2}}{\pi}I_{rct,FHA} Io=Ts2∫02Ts∣irct,FHA(t)∣dt=π22Irct,FHA - 输出负载FHA分析
I r c t , F H A = π 2 2 I o I_{rct,FHA}=\frac{\pi}{2\sqrt{2}}I_o Irct,FHA=22πIo
V o , F H A = 2 2 π V o V_{o,FHA}=\frac{2\sqrt{2}}{\pi}V_o Vo,FHA=π22Vo
R o , e q = V o , F H A I r c t , F H A = 8 π 2 R o R_{o,eq}=\frac{V_{o,FHA}}{I_{rct,FHA}}=\frac{8}{\pi^2}R_o Ro,eq=Irct,FHAVo,FHA=π28Ro
基于FHA的电路增益特性分析
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H_{r}(s)=\frac{Z_o(s)}{Z_{in}(s)}\frac{R_{eq}}{Z_2(s)+R_{eq}}
Hr(s)=Zin(s)Zo(s)Z2(s)+ReqReq
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\begin{cases} Z_{in}(s)=Z_1(s)+Z_o(s)\\ Z_o(s)=\frac{Z_m(s)(Z_2(s)+R_{eq})}{Z_m(s)+Z_2(s)+R_{eq}}\\ Z_1(s)=sL_1+\frac{1}{sC_1}\\ Z_2(s)=sL_2+\frac{1}{sC_2}\\ Z_m(s)=sL_m \end{cases}
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⎧Zin(s)=Z1(s)+Zo(s)Zo(s)=Zm(s)+Z2(s)+ReqZm(s)(Z2(s)+Req)Z1(s)=sL1+sC11Z2(s)=sL2+sC21Zm(s)=sLm
得
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H_r(s)=\frac{{C_1}{C_2}{L_m} R s^3}{{C_1} {C_2} {L_1} {L_2}s^4+{C_1} {C_2} {L_1} {L_m} s^4+{C_1} {C_2} {L_1} Rs^3+{C_1} {C_2} {L_2} {L_m} s^4+{C_1} {C_2} {L_m} Rs^3+{C_1} {L_1} s^2+{C_1} {L_m} s^2+{C_2} {L_2}s^2+{C_2} {L_m} s^2+{C_2} R s+1}
Hr(s)=C1C2L1L2s4+C1C2L1Lms4+C1C2L1Rs3+C1C2L2Lms4+C1C2LmRs3+C1L1s2+C1Lms2+C2L2s2+C2Lms2+C2Rs+1C1C2LmRs3
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H_r(jw)=\frac{{C_1} {C_2} j^3 {L_m} R w^3}{{C_2} j w \left(j w\left({C_1} j^2 w^2 ({L_m} ({L_1}+{L_2})+{L_1}{L_2})+{L_2}+{L_m}\right)+{C_1} j^2 R w^2({L_1}+{L_m})+R\right)+{C_1} j^2 w^2 ({L_1}+{L_m})+1}
Hr(jw)=C2jw(jw(C1j2w2(Lm(L1+L2)+L1L2)+L2+Lm)+C1j2Rw2(L1+Lm)+R)+C1j2w2(L1+Lm)+1C1C2j3LmRw3
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H_r(jw)=-\frac{{L_m} R w}{j \left(\frac{{C_1} {C_2} w^4 ({L_m} ({L_1}+{L_2})+{L_1} {L_2})-{C_2} w^2 ({L_2}+{L_m})}{{C_1}{C_2} w^2}-\frac{{C_1} w^2 ({L_1}+{L_m})-1}{{C_1} {C_2} w^2}\right)+\frac{R \left({C_1} w^2({L_1}+{L_m})-1\right)}{{C_1} w}}
Hr(jw)=−j(C1C2w2C1C2w4(Lm(L1+L2)+L1L2)−C2w2(L2+Lm)−C1C2w2C1w2(L1+Lm)−1)+C1wR(C1w2(L1+Lm)−1)LmRw
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H_r(jw)=\frac{kwC_1R}{j \left(\frac{\frac{1}{{C_1} {L_1} w^2}-k-1}{g}+{C_1} {L_1} w^2 (h k+h+k)-h-k\right)+R \left({C_1} (k+1)w-\frac{1}{{L_1} w}\right)}
Hr(jw)=j(gC1L1w21−k−1+C1L1w2(hk+h+k)−h−k)+R(C1(k+1)w−L1w1)kwC1R
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H_r(jw)=\frac{kwC_1R}{R \left({C_1} (k+1) w-\frac{1}{{L_1} w}\right)+j \left(\frac{-k+\frac{{w_1}^2}{w^2}-1}{g}+\frac{w^2 (hk+h+k)}{{w_1}^2}-h-k\right)}
Hr(jw)=R(C1(k+1)w−L1w1)+j(g−k+w2w12−1+w12w2(hk+h+k)−h−k)kwC1R
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上下同除以kwC1R并且代入R=\frac{Z_1}{Q},Z_1=\sqrt{\frac{L_1}{C_1}},\sqrt{L_1C_1}=w_1,\frac{w_1}{w_s}=w_n,得
上下同除以kwC1R并且代入R=QZ1,Z1=C1L1,L1C1=w1,wsw1=wn,得
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H_r(jw)=-\frac{1}{\frac{j Q (g h+g k+k+1)}{{C_1} g k {w_s}\sqrt{\frac{{L_1}}{{C_1}}}}+\frac{j Q {w_1}^2}{{C_1} g k{w_s}^3 \sqrt{\frac{{L_1}}{{C_1}}}}+\frac{j Q {w_s} (hk+h+k)}{{C_1} k {w_1}^2\sqrt{\frac{{L_1}}{{C_1}}}}-\frac{1}{{C_1} k {L_1}{w_s}^2}+\frac{k+1}{k}}
Hr(jw)=−C1gkwsC1L1jQ(gh+gk+k+1)+C1gkws3C1L1jQw12+C1kw12C1L1jQws(hk+h+k)−C1kL1ws21+kk+11
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H_r(jw_n)=-\frac{1}{-\frac{j Q (g h+g k+k+1)}{g k {w_n}}+\frac{j Q}{g k {w_n}^3}+\frac{j Q{w_n} (h k+h+k)}{k}-\frac{1}{k {w_n}^2}+\frac{k+1}{k}}
Hr(jwn)=−−gkwnjQ(gh+gk+k+1)+gkwn3jQ+kjQwn(hk+h+k)−kwn21+kk+11
整理得
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H_r(jw_n)=\frac{1}{\frac{j Q}{k} \left(-\frac{g h+g k+k+1}{g {w_n}}+\frac{1}{g {w_n}^3}+{w_n} (hk+h+k)\right)-\frac{1}{k {w_n}^2}+\frac{k+1}{k}}
Hr(jwn)=kjQ(−gwngh+gk+k+1+gwn31+wn(hk+h+k))−kwn21+kk+11
增益为
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M(w_n)=||H_r(jw_n)||=\frac{1}{\sqrt{(\frac{j Q}{k})^2 \left(-\frac{g h+g k+k+1}{g {w_n}}+\frac{1}{g {w_n}^3}+{w_n} (hk+h+k)\right)^2+(-\frac{1}{k {w_n}^2}+\frac{k+1}{k})^2}}
M(wn)=∣∣Hr(jwn)∣∣=(kjQ)2(−gwngh+gk+k+1+gwn31+wn(hk+h+k))2+(−kwn21+kk+1)21
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\begin{cases} a=k+h+kh \\ b=k+\frac{k}{g}+h+\frac{1}{g}\\ c=\frac{1}{g} \end{cases}
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⎧a=k+h+khb=k+gk+h+g1c=g1
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M(w_n)=||H_r(jw_n)||=\frac{1}{\sqrt{(\frac{j Q}{k})^2 \left(-\frac{b}{ {w_n}}+\frac{c}{ {w_n}^3}+ a{w_n}\right)^2+(-\frac{1}{k {w_n}^2}+\frac{k+1}{k})^2}}
M(wn)=∣∣Hr(jwn)∣∣=(kjQ)2(−wnb+wn3c+awn)2+(−kwn21+kk+1)21
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