Charge Recovery - Aalborg Universitet Design of a Fuel Cell Hybrid Electric Vehicle Drive Syste

5. ULTRACAPACITOR MODELING

6 8 10 12 14 16 18

0 0.5 1 1.5 2 2.5x 10⁷

VoltagevU C [V]

Selfdischargetimeconstantτsd[s]

Calculation Curve ﬁt

Figure 5.5: Discharge time constant.

The self discharge resistance can now be calculated by solving Equation (5.8), i.e.

R_sd = τ_sd

C_eq [Ω] (5.10)

5.4. Charge Recovery

In order to calculate the first capacitorC₁in Figure 5.6 is it assumed that the current i₂in Figure 5.6 is constant just before and after the inverse step at timeT_B. Due to the fast rise of the voltage the self discharge is neglected. Therefore the ultracapacitor can be modeled as the circuit in Figure 5.7 just before and after the inverse current step.

From the two circuit diagrams the capacitorC₁can be calculated, i.e.

i₂(TB−) =I₂ =i₂(T_B+) [A] (5.11)

i_{U C}(T_B₋)−C₁v˙₁(T_B₋) = −C₁v˙₁(T_B+) [A] (5.12)

C₁ = i_{U C}(T_B₋)

v₁(TB−)−v˙₁(T_B+) [F] (5.13) This means that the capacitorC₁ can be calculated from the derivative of the voltage v₁ just before and after the inverse step.

- C1

iC1

iUC +

- C1

iC1

(a) (b)

Figure 5.7: Equivalent circuit model at inverse step. (a) Just before the inverse step at timeT_B₋. (b) Just after the inverse step at timeT_B+.

Self discharge compensation

In Figure 5.8 the voltage of Figure 5.1 is shown after the inverse step at timet =T_B. It is seen that the voltage increases for many hours. The self discharge mechanism will therefore affect the voltage. The self discharge currentisd is given by

i_sd = v₁ Rsd

[A] (5.14)

The charge dissipated in the self discharge resistor R_sd during the charge recovery period is therefore

q_sd =

_T_C

TB+

i_sddt [C] (5.15)

If this charge level is not removed from the equivalent capacitance C_eq the com-pensated voltage of the capacitorC₁ will be

v_1,comp=v₁+ q_sd

C_eq [V] (5.16)

The compensated voltagev_1,compis also shown in Figure 5.8. It is seen that the charge recovery circuits charge the capacitorC₁ with more than30 mV.

5. ULTRACAPACITOR MODELING

Figure 5.8: Compensated and non-compensated voltage of capacitor C₁ during the charge recovery period.

Time Constant Calculation

The voltage rises withktime constants during the charge recovery time. It is assumed that for eachR_iC_i-circuit with time constantτ_i,τ_i is bigger than the time constantτ_i₋₁ ofR_i₋₁C_i₋₁-circuit, i.e.τ₁ < τ₂· · ·< τ_k. It is therefore assumed that for theR_iC_i-circuit in Figure 5.6 the current i_i+1 is constant, when the R_iC_i-circuit is dominating. The equivalent circuit used for calculating time constantτ₂ is shown in Figure 5.9(a). It is seen that the current contribution of the higher RC-circuits is modeled as a constant current source with currentI₃. In Figure 5.9(b) the equivalent circuit model for time constant τ₃ is shown. It is seen that this circuit is obtained by replacing the current source I₃ with a RC-circuit and another current source I₄. However, in order to sim-plify it is assumed that the currenti₂ is almost constant in the time period where the time constant τ₃ is dominating. This means that the voltage slope of capacitor C₂ is equal to the voltage slope of capacitorC₁, i.e.

v₂ =v_1,comp,3−R₂I₂ [V] (5.17)

⇓

v₂ = ˙v_1,comp,3 [V/s] (5.18)

where v_1,comp,3 [V] Compensated voltage of capacitorC₁in the interval where time constant 3τ₃is dominating

The capacitorsC₁andC₂ therefore behave as they were in parallel. In Figure 5.9(c) it is shown an equivalent circuit diagram for time constant τ₃ when they are in pa-rallel. It is seen that the resistor R₃ is in series with an equivalent resistance R_2,cha. The equivalent resistanceR_2,cha is inserted in order to compensate for the moving of capacitorC₂. The equivalent resistanceR_2,chacan be calculated by using the constraint

5.4. Charge Recovery

C1 +

- C2

iC2 +

v1,comp,2

-i2

iC1

(a)

C1 C2 +

v1,comp,3

-i3

C3 +

-iC3 +

-i2

iC2

iC1

R2,cha

C1 C2 +

v1,comp,3

-i3

C3 +

-iC3 +

-(b) (c)

R2,cha

C1 C2 +

v1,comp,4

-I5

C3 +

-iC3

i4 R4

iC4

(d)

C1 C2 +

v1,comp,4

-I5

C3 +

-i4 R4

iC4

R3,cha

-(e)

Ri-1,cha

C1 C2 +

v1,comp,i

-ii

Ii+1

Ci +

-iCi +

vi-1

Ci-1

-(f)

Figure 5.9: Equivalent ultracapacitor circuit modeling during the charge recovery pe-riod. (a) Equivalent circuit for time constantτ₂. (b) Equivalent circuit for time constant τ₃. (c) Modified equivalent circuit for time constantτ₃. (d) Equivalent circuit for time constantτ₄. (e) Modified equivalent circuit for time constantτ₄. (f) General structure of equivalent circuit diagrams for time constantτ_i.

that the voltagev₂ of Figure 5.9(b) and (c) should be identical. Therefore:

v₂ =v_1,comp,3+R₂C₁v˙_1,comp,3

=v_1,comp,3+R_2,cha(C₁+C₂) ˙v_1,comp,3 [V] (5.19)

R_2,cha =R₂ C₁

C₁+C₂ [Ω] (5.20)

If the current sourceI₄ in Figure 5.9(c) is replaced by a RC-circuit and another cur-rent source I₅ the equivalent circuit in Figure 5.9(d) is obtained. In order to simplify this circuit is also modified. The modified circuit is seen in Figure 5.9(e). It is seen that it has the same structure as the circuit for the previous time constant in Figure 5.9(c).

The equivalent resistanceR_3,chais calculated in the same way as the previous

equiva-5. ULTRACAPACITOR MODELING

lent resistanceR_2,cha, i.e.

R_3,cha = (R_2,cha+R₃) C₁+C₂

C₁+C₂+C₃ [Ω] (5.21)

For each time constant τ_i this principle can be repeated, i.e. replacing the current source of the previous τ_i₋₁-circuit by a RC-circuit and another current source, and afterwards modifying the circuit by putting all the capacitors to the left of thei^th ca-pacitors. In Figure 5.9(f) a general equivalent circuit structure of thei^thtime constant can be seen. For thei^thtime constant the equivalent resistorR_i_−1,cha is therefore

R_i_−1,cha =

⎧⎪

⎨

⎪⎩

0 , i= 2

(R_i_−2,cha+R_i₋₁)

_l=i−2

l=1 C_l

_l=i−1

l=1 C_l , i∈[3;k] [Ω] (5.22) From the general equivalent circuit in Figure 5.9(f) the currents are given by

i_i = −^l=i⁻¹

l=1

C_lv˙_1,comp,i [A] (5.23)

i_C_i =C_iv˙_i =i_i −I_i+1 =−^l=i⁻¹

l=1

C_lv˙_1,comp,i−I_i+1 [A] (5.24) From Kirchhoff’s voltage law one obtains

v_i =v_1,comp,i−(R_i_−1,cha+R_i)i_i

=v_1,comp,i+ (R_i_−1,cha+R_i)

l=i−1 l=1

C_lv˙_1,comp,i [V] (5.25)

v_i = ˙v_1,comp,i+ (R_i_−1,cha+R_i)

l=i−1 l=1

C_l¨v_1,comp,i [V/s] (5.26)

When inserting Equation (5.26) in (5.24) the following second order differential equa-tion is obtained:

− I_i+1

_l=i

l=1C_l = (R_i_−1,cha+R_i)^l=i−1_l=1 C_lC_i

_l=i

l=1C_l v¨_1,comp,i+ ˙v_1,comp,i [V/s] (5.27) The second order differential equation has a homogeneous solution v_1,comp,h and particular solutionv_1,comp,p. The homogeneous solution has the form [38]

v_1,comp,h,i =a_i +b_ie⁻^τi¹^(t⁻^t^B⁾ [V] (5.28) The time constantτ_i can be determined by solving the characteristic equation, i.e.

0 = (R_i_−1,cha+R_i)^l=i_l=1⁻¹C_lC_i

_l=i

l=1C_l s²+s

s = 0 ∨ −s = 1 τ_i =

_l=i

l=1C_l

(R_i_−1,cha+R_i)^l=i_l=1⁻¹C_lC_i (5.29) 58

5.4. Charge Recovery

The particular solution is given by

v_1,comp,p,i=ci(t−tB) [V] (5.30)

ci = − I_i+1

_l=i

l=1Cl

[V/s] (5.31)

The system is therefore given by

v_1,comp,i =v_1,comp,h,i+v_1,comp,p,i

=a_i +b_ie⁻^τi¹^(t⁻^t^B⁾+c_i(t−t_B) [V] (5.32)

v_1,comp,i = − 1

τ_ibie⁻^τi¹^(t⁻^t^B⁾+ci [V/s] (5.33)

v_1,comp,i = 1

τ_i²b_ie⁻^τi¹^(t⁻^t^B⁾ V/s² (5.34)

ci =τiv¨_1,comp,i+ ˙v_1,comp,i [V/s] (5.35)

For each n sample it is assumed that the voltage is approximated to be a linear function, i.e. v_1,comp,i[n] = v_1,comp,i(T_B)[n] + ˙v_1,comp,i(t−T_B) [n]. The derivative of the voltagev_1,comp,i is calculated for each samplen by using least squares in the interval n−N_ls,iton+N_ls,i, i.e.

⎡

⎢⎢

⎣

v_1,comp,i[n−N_ls,i] ...

v_1,comp,i[n+N_ls,i]

⎤

⎥⎥

⎦

⎡

⎢⎢

⎣

1 (t[n−N_ls,i]−T_B)

... ...

1 (t[n+N_ls,i]−T_B)

⎤

⎥⎥

⎦

v_1,comp,i(TB)[n]

v_1,comp,i[n]

X = A^TA⁻¹A^TY (5.36)

v_1,comp,i[n] = X[2] (5.37)

The double derivative is calculated in the same way, i.e.

⎡

⎢⎢

⎣

v_1,comp,i[n−N_ls,i] ...

v_1,comp,i[n+N_ls,i]

⎤

⎥⎥

⎦

⎡

⎢⎢

⎣

1 t[n−N_ls,i]−T_B ... ...

1 t[n+N_ls,i]−T_B

⎤

⎥⎥

⎦

v˙_1,comp,i(T_B)[n]

v_1,comp,i[n]

X = A^TA⁻¹A^TY (5.38)

v_1,comp,i[n] = X[2] (5.39)

The voltage is sampled withf_s = 5 Hz. The sampling frequency should be fast enough to capture the fast transients of the ultracapacitor module, but on the other hand it should not be too fast; otherwise the logged data files will be too big, if data is logged

5. ULTRACAPACITOR MODELING

for several hours. The selection of the intervalN_ls,i used for eachR_iC_i-circuit is also a compromise. The interval should be big enough to suppress the influence of the previousR_i₋₁C_i₋₁-circuit. However, it should not suppress the response due to the actual R_iC_i-circuit. It has therefore been decided to let N_ls,i have the length of two time constants of the previousR_i₋₁C_i₋₁-circuit, i.e.

N_ls,i = 2τ_i₋₁f_s, i= 2. . . k (5.40) The derivative and double derivative ofv_1,compare now calculated. The constants ci and τi can now be estimated by using least squares method on Equation (5.35).

However, the result depends of which part of the data set that is used. That part generated from smaller or higher time constants should therefore not be taken into account. Only data from time t = Tstart to t = Tstop will be included. Therefore, in order to neglect the contribution of smaller time constants, data before two time constants of the previous time constant will not be included. The start timeT_start,i is therefore

T_start,i=T_B+ 2τ_i₋₁ [s] (5.41)

The number of RC-circuitskis therefore given by thatR_iC_i-circuit that makesT_start,i = TB+ 2τi≥TC. In order to neglect the contribution from higher time constants the stop time T_stop is chosen to be the time when the derivative of the compensated voltage

v_comp,ihas dropped to15 %of its initial value at timet=T_B, i.e.

0.15·v˙_1,comp,i(T_B) = − 1

τ_ib_ie⁻^τi¹^(T^stop,i⁻^T^B⁾+c_i [V/s] (5.42) Therefore, from Equation (5.35):

⎡

⎢⎢

⎣

v_1,comp,i(T_start,i) ...

v_1,comp,i(T_stop,i)

⎤

⎥⎥

⎦

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

⎡

⎢⎢

⎣

1 −v¨_1,comp,i(T_start,i)

... ...

1 −v¨_1,comp,i(T_stop,i)

⎤

⎥⎥

⎦

c_i τ_i

, i < k

⎡

⎢⎢

⎣

−v¨_1,comp,i(T_start,i) ...

−v¨_1,comp,i(Tstop,i)

⎤

⎥⎥

⎦

τ_i , i=k

(5.43)

c_i τ_i

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

A^TA⁻¹A^TY , i < k

⎡

⎣ 0

A^TA⁻¹A^TY

⎤

⎦ , i=k (5.44)

For the last time constantτ_k no current source is contributing to the charge recovery.

Therefore is the constantci equal to zero fori=k.

In document Aalborg Universitet Design of a Fuel Cell Hybrid Electric Vehicle Drive System Schaltz, Erik (Sider 67-74)