🎯要点
- 特性测试评估卸载电池性能以及不同温度下电池容量和电阻。
- 使用两种电气模型评估和模拟卸载电池可利用性能。
- 从宝马 i3模块中提取三星三元锂方形电池作为评估电池：容量测量、电阻测量，对比新电池性能。
- 使用的等效电路模型以及用于校准和验证这些模型的程序。
🍁汽车电池制造和卸载再利用分析
🍪语言内容分比
🍇MATLAB降压转换器
当 MOSFET 关闭时，电路方程可以写成
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\left\{\begin{array}{c} \left(r_{i n}+r_{d s}+r_L\right) i_L+L \frac{d i_L}{d t}+R\left(i_L-C \frac{d v_C}{d t}\right)=v_{i n} \\ r_C C \frac{d v_C}{d t}+v_C=R\left(i_L-C \frac{d v_C}{d t}\right) \end{array}\right.
{(rin+rds+rL)iL+LdtdiL+R(iL−CdtdvC)=vinrCCdtdvC+vC=R(iL−CdtdvC)
负载电压可以写成
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v_o=R\left(i_L-C \frac{d v_C}{d t}\right)=R\left(\frac{r_C}{r_C+R} i_L+\frac{1}{R+r_C} v_C\right)
vo=R(iL−CdtdvC)=R(rC+RrCiL+R+rC1vC)
当 MOSFET 开路时，电路方程可以写成
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\left\{\begin{array}{c} V_D+\left(r_D+r_L\right) i_L+L \frac{d i_L(t)}{d t}+r_C C \frac{d v_C}{d t}+v_C=0 \\ r_C C \frac{d v_C}{d t}+v_C=R\left(i_L-C \frac{d v_C}{d t}\right) \end{array}\right.
{VD+(rD+rL)iL+LdtdiL(t)+rCCdtdvC+vC=0rCCdtdvC+vC=R(iL−CdtdvC)
输出方程可以写成
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v_o=R\left(i_L-C \frac{d v_C}{d t}\right)=R\left(\frac{r_C}{r_C+R} i_L+\frac{1}{R+r_C} v_C\right)
vo=R(iL−CdtdvC)=R(rC+RrCiL+R+rC1vC)
电感电流方程为
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\left\{\begin{array}{l} L \frac{d i_L}{d t}=-\left(r_{\text {in }}+r_{d s}+r_L+R-\frac{R^2}{R+r_C}\right) i_L-\frac{R}{R+r_C} v_C+v_{\text {in }}, n \cdot T<t<n \cdot T+d \cdot T \\ L \frac{d i_L}{d t}=-\left(r_D+r_L+\frac{R \cdot r_C}{R+r_C}\right) i_L-\frac{R}{R+r_C} v_C-V_D, \quad n \cdot T+d . T<t<(n+1) T \end{array}\right.
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⎧LdtdiL=−(rin +rds+rL+R−R+rCR2)iL−R+rCRvC+vin ,n⋅T<t<n⋅T+d⋅TLdtdiL=−(rD+rL+R+rCR⋅rC)iL−R+rCRvC−VD,n⋅T+d.T<t<(n+1)T
电容电压方程为
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\left\{\begin{array}{lr} C \frac{d v_C}{d t}=\frac{R}{R+r_C} i_L-\frac{1}{R+r_C} v_C & n \cdot T<t<n \cdot T+d . T \\ C \frac{d v_C}{d t}=\frac{R}{R+r_C} i_L-\frac{1}{R+r_C} v_C & n \cdot T+d . T<t<(n+1) T \end{array}\right.
{CdtdvC=R+rCRiL−R+rC1vCCdtdvC=R+rCRiL−R+rC1vCn⋅T<t<n⋅T+d.Tn⋅T+d.T<t<(n+1)T
最终稳态值
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\left\{\begin{array}{c} I_L=\frac{\left(( R + r _ { C } ) \left(D V_{I N}-(1-D) V_D\right.\right.}{\left(R+r_C\right) R_2+R^2+D\left(R+r_C\right)\left(R_1-R_2\right)} \\ V_C=\frac{\left(( R + r _ { C } ) \left(D V_{I N}-(1-D) V_D\right.\right.}{\left(R+r_C\right) R_2+(1-2 D) R^2+D\left(R+r_C\right)\left(R_1-R_2\right)} \times R \end{array}\right.
{IL=(R+rC)R2+R2+D(R+rC)(R1−R2)((R+rC)(DVIN−(1−D)VDVC=(R+rC)R2+(1−2D)R2+D(R+rC)(R1−R2)((R+rC)(DVIN−(1−D)VD×R
clc
clear all
syms R1 R2 R D IL VC rC rL VD vIN
eq1=-D*R1*IL-(1-D)*R2*IL-R/(R+rC)*VC-(1-D)*VD+D*vIN;
eq2=R/(R+rC)*IL-1/(R+rC)*VC;
DC_operatingPoint=solve(eq1,eq2,[IL VC]);
disp('IL=')
pretty(simplify(DC_operatingPoint.IL))
disp('VC=')
pretty(simplify(DC_operatingPoint.VC))
降压转换器的小信号传递函数计算
R=5;
VIN=50;
rin=.1;
L=400e-6;
rL=.1;
C=100e-6;
rC=.05;
rD=.01;
VD=.7;
rds=.1;
D=.41;
R1=rin+rds+rL+R*rC/(R+rC);
R2=rD+rL+R*rC/(R+rC);
IL=(R+rC)*(D*VIN-(1-D)*VD)/((R+rC)*R2+R^2+D*(R+rC)*(R1-R2));
A=[(R2*(D-1)-R1*D)/L -R/(R+rC)/L;R/(R+rC)/C -1/(R+rC)/C];
B=[(VIN+VD+(R2-R1)*IL)/L D/L;0 0];
CC=[R*rC/(rC+R) R/(R+rC)];
used for matrix
H=tf(ss(A,B,CC,0));
vO_d=H(1)
duty ratio
vO_vin=H(2)
input source
figure(1)
bode(vO_d), grid on
figure(2)
bode(vO_vin), grid on
该程序提取小信号传递函数
clc
clear all;
R=5;
VIN=50;
rin=.1;
L=400e-6;
rL=.1;
C=100e-6;
rC=.05;
rD=.01;
VD=.7;
rds=.1;
D=.41;
syms iL vC vin vD d
M1=(-(rin+rds+rL+(R*rC/(R+rC)))*iL-R/(R+rC)*vC+vin)/L;
for closed MOSFET
M2=(R/(R+rC)*iL-1/(R+rC)*vC)/C; %d(vC)/dt for
vO1=R*(rC/(rC+R)*iL+1/(R+rC)*vC);
M3=(-(rD+rL+R*rC/(R+rC))*iL-R/(R+rC)*vC-vD)/L;
M4=(R/(R+rC)*iL-1/(R+rC)*vC)/C;
vO2=R*(rC/(rC+R)*iL+1/(R+rC)*vC);
MA1= simplify(d*M1+(1-d)*M3);
MA2= simplify(d*M2+(1-d)*M4);
vO= simplify(d*vO1+(1-d)*vO2);
MA_DC_1=subs(MA1,[vin vD d],[VIN VD D]);
MA_DC_2=subs(MA2,[vin vD d],[VIN VD D]);
DC_SOL= solve(MA_DC_1==0,MA_DC_2==0,iL,vC);