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IEEE TRANSACTIONS ON ENERGY CONVERSION 1

Simple Analytical Method for DeterminingParameters of Discharging Batteries

Tingshu Hu, Senior Member, IEEE, Brian Zanchi, and Jianping Zhao

Abstract—This paper derives simple and explicit formulas forcomputing the parameters of Thevenin’s equivalent circuit modelfor a discharging battery. The general Thevenin’s equivalent circuitmodel has n pairs of parallel resistors and capacitors (nth-ordermodel). The main idea behind the new method is to transform theproblem of solving a system of high-order polynomial equationsinto one of solving several linear equations and a single-variablenth-order polynomial equation, via some change of variables. Thecomputation can be implemented with a simple MATLAB code lessthan half-page long. Experimental and computational results areobtained for three types of batteries: Li-polymer, lead–acid, andnickel metal hydride. For all the tested batteries, the first-ordermodels are not able to generate voltage responses that closely matchthe measured responses, while second-order models can generatewell-matched responses. For some of the batteries, a third-ordermodel can do a better job matching the voltage responses.

Index Terms—Battery model, linear algebraic equations, poly-nomials, Thevenin’s equivalent circuit.

I. INTRODUCTION

DYNAMIC models for batteries are important for analy-sis, design, and simulation of battery-powered electronic

systems (e.g., see [1]–[7]). They are also important for char-acterization of battery performance, lifetime estimation, powermanagement, and efficient use of batteries [8]–[12].

In Fig. 1, a Thevenin’s equivalent circuit model is depicted,which has been widely used to model the discharging dynam-ics of various types of batteries such as lead–acid, lithium-ion(Li-ion), Li-polymer, nickel metal hydride (NiMH), and fuelcells (e.g., see [1]–[20]). Although the parameters in the modeldepend on many factors [1], [7], [11], [13], such as the state ofcharge (SOC), the load, the temperature, and even the historyof charge and discharge, under certain working conditions andover a relatively short period of time, they are assumed to beconstants.

The most commonly used model is the circuit with only onepair of parallel resistor and capacitor (RC). For this first-order

Manuscript received August 30, 2010; revised January 17, 2011; acceptedMarch 8, 2011. This work was supported in part by the National Science Foun-dation (NSF) under Grant ECS-0621651 and Grant ECCS-0925269. Paper no.TEC-00331-2010.

T. Hu and B. Zanchi are with the Department of Electrical and ComputerEngineering, University of Massachusetts, Lowell, MA 01854 USA (e-mail:[email protected]; [email protected]).

J. Zhao is with Juniper Networks, Westford, MA 01886 USA (e-mail:[email protected]).

Color versions of one or more of the figures in this paper are available onlineat http://ieeexplore.ieee.org.

Digital Object Identifier 10.1109/TEC.2011.2129594

Fig. 1. Thevenin’s equivalent model for a discharging battery.

Fig. 2. Battery connected to a load with constant current.

model, the parameters can be easily estimated from the exper-iment. However, the voltage response by the first-order modelcan be quite different from the response of the real battery, aswill be demonstrated with examples. For models with two ormore pairs of parallel RCs, there seems to be a lack of system-atic methods to identify the parameters, except for using bruteforce numerical methods, such as minimizing a cost functionof the sum (or integral) of squares of the difference betweenthe experimental data and the analytical expression [13]. Thedrawback with such kind of a numerical optimization method isthat the cost function is highly nonlinear with respect to circuitparameters and there is no guarantee to find the global mini-mum. In addition, it is a very tricky issue to find proper initialparameters for optimization.

The purpose of this paper is to derive a simple analyticalmethod to identify the parameters for the general case with twoor more pairs of parallel RCs.

II. EXPLICIT FORMULAS FOR COMPUTING THE PARAMETERS

We consider the battery model with n pairs of parallel resistorsand capacitors, (R1 , C1), (R2 , C2), . . . , (Rn,Cn ), as depictedin Fig. 1. For simplicity, this will be called an nth-order model.

A common setup to obtain the parameters is to connect thebattery to an electronic load that absorbs a constant current Ifrom the battery, see Fig. 2.

Assume that the load is connected at t = 0 and all the initialcapacitor voltages are 0. Then, the first two parameters to be

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2 IEEE TRANSACTIONS ON ENERGY CONVERSION

obtained are

E = v(0−), R0 = (v(0−) − v(0+))/I.

The other parameters (R1 , C1), . . . , (Rn,Cn ) have to be eval-uated via the time response of the voltage v(t) over a periodof time. The advantage of using a constant current load (over aresistor) is that v(t) for the model can be simply expressed as

v(t) = E − R0I −n∑

k=1

RkI(1 − e− t

R k C k ).

If only one pair of parallel resistor and capacitor (R1 , C1) isconsidered in the model, they can be estimated from the approx-imate time constant of the experimental response. However, thevoltage response of this simple model can be very different fromthat of the actual battery.

In this paper, we present explicit formulas to compute theparameters for the general nth-order model. We will first presentthe detailed results for the third-order model to explain the mainideas. Then, we describe how the method can be simplified tothe second-order model, and finally, how to extend it to higherorder models.

A. Detailed Results for the Third-Order Model

For the third-order model,

v(t) = E − R0I − R1I(1 − e−t

R 1 C 1 ) − R2I(1 − e−t

R 2 C 2 )

− R3I(1 − e−t

R 3 C 3 ). (1)

To find the six unknown parameters (Rk ,Ck ), k = 1, 2, 3, wepick six equally spaced time instants, tk = kT, k = 1, . . . , 6.Let

d1 = e−T

R 1 C 1 , d2 = e−T

R 2 C 2 , d3 = e−T

R 3 C 3 .

Then, for k = 1, 2, . . . , 6,

v(kT ) = E − R0I − R1I(1 − dk1 ) − R2I(1 − dk

2 )

− R3I(1 − dk3 ), k = 1, 2, . . . , 6. (2)

Theoretically, the six parameters (R1 , d1), (R2 , d2), (R3 , d3)can be determined from the six equations with v(kT ) obtainedfrom experiment. With these six parameters, C1 , C2 , and C3 areeasy to compute.

At first sight, it may seem that the six nonlinear equations in(2) are impossible to solve by an analytical method, since thelast equation has three seventh-order terms R1d

61 , R2d

62 , R3d

63 .

However, after further examination on the structure of the equa-tions, all the solutions can be explicitly obtained with somechange of variables.

The main idea of the new method is to define the followingvariables:

u1 = d1 + d2 + d3 , u2 = d1d2 + d2d3 + d3d1 , u3 = d1d2d3 .(3)

It turns out that these variables can be solved via a system oflinear equations. Then, d1 , d2 , d3 can be obtained by solvinga third-order single-variable polynomial equation with coeffi-cients formed with u1 , u2 , u3 . After that, the computation of

Rk ,Ck , k = 1, 2, 3, is straightforward. The main result is sum-marized as follows and the proof will be provided later.

Main result: Given E,R0 , v(kT ), k = 1, 2, . . . , 6, from theexperiment. The parameters (Rk ,Ck ), k = 1, 2, 3, can be com-puted from the following steps.

1) Compute b1 = E − R0I − v(T ). For k = 2, . . . , 6, bk =v((k − 1)T ) − v(kT ).

2) Compute

⎡

⎣u1u2u3

⎤

⎦ =

⎡

⎣b3 −b2 b1b4 −b3 b2b5 −b4 b3

⎤

⎦−1 ⎡

⎣b4b5b6

⎤

⎦.

3) Let the roots of q3 − 2u1q2 + (u2

1 + u2)q + (u3 −u1u2) = 0 to be q1 , q2 , and q3 . Then,

⎡

⎣d1d2d3

⎤

⎦ =

⎡

⎣0 1 11 0 11 1 0

⎤

⎦−1 ⎡

⎣q1q2q3

⎤

⎦ .

4) Let

⎡

⎣x1x2x3

⎤

⎦ =

⎡

⎣1 1 1d1 d2 d3d2

1 d22 d2

3

⎤

⎦−1 ⎡

⎣b1b2b3

⎤

⎦. Then, Rk =

xk

I (1−dk ) , Ck = − TRk ln(dk ) , k = 1, 2, 3.

The aforementioned algorithm can be realized with the fol-lowing MATLAB code:% Step 1: form b1,...,b6,%v(k) stands for v(kT)b1=E-R0*I- v(1); b2=v(1)- v(2);b3=v(2)- v(3); b4=v(3)- v(4);b5=v(4)- v(5); b6=v(5)- v(6);

% Step 2: Compute u=[u1;u2;u3];A=[b3 - b2 b1;b4 - b3 b2;b5 - b4 b3];u=inv(A)*[b4;b5;b6];

% Step 3: Compute d1, d2, d3c=[1 - 2*u(1) (u(1)ˆ2+u(2))];c=[c (u(3)- u(1)*u(2))];q=roots(c);dd=inv([0 1 1;1 0 1;1 1 0])*q;d1=dd(1);d2=dd(2);d3=dd(3);

% Step 4: Compute R1,C1,R2,C2,R3,C3Q=[1 1 1;d1 d2 d3;d1ˆ2 d2ˆ2 d3ˆ2];x=inv(Q)*[b1;b2;b3];R1=x(1)/I/(1-d1); C1=-T/R1/log(d1);R2=x(2)/I/(1-d2); C2=-T/R2/log(d2);R3=x(3)/I/(1-d3); C3=-T/R3/log(d3);To verify the main result and the algorithm, one may generate

v(kT ) from a model with arbitrary parameters. For instance,E=6; R0=0.02;I=2;Re1=0.02;Re2=0.014;Re3=0.05;Ce1=10;Ce2=100;Ce3=20;T=1;de1=exp(- T/Re1/Ce1);de2=exp(- T/Re2/Ce2);de3=exp(- T/Re3/Ce3);v=zeros(6,1);for i=1:6

v(i)=E- R0*I- Re1*I*(1- de1î);v(i)=v(i)- Re2*I*(1- de2î);v(i)=v(i)- Re3*I*(1- de3î);

end


HU et al.: SIMPLE ANALYTICAL METHOD FOR DETERMINING PARAMETERS OF DISCHARGING BATTERIES 3

Then, use the first code to compute (Rk ,Ck ), k = 1, 2, 3. Wewill see that they are the same as (Rek , Cek ), k = 1, 2, 3, mostlikely rearranged.

The proof for the main result is provided in the remainder ofthis section.

Proof of the main result: Define

x1 = R1I(1 − d1), x2 = R2I(1 − d2), x3 = R3I(1 − d3).

The six equations in (2) can be rewritten as follows (note that1 − dk = (1 − d)(1 + d + d2 + · · · + dk−1)):

x1 + x2 + x3 = E − R0I − v(T ) (4)

(1 + d1)x1 + (1 + d2)x2 + (1 + d3)x3 = E − R0I − v(2T )

(5)

(1 + d1 + d21)x1 + (1 + d2 + d2

2)x2 + (1 + d3 + d23)x3

= E − R0I − v(3T ) (6)

3∑

k=0

dk1 x1 +

3∑

k=0

dk2 x2 +

3∑

k=0

dk3 x3 = E − R0I − v(4T ) (7)

4∑

k=0

dk1 x1 +

4∑

k=0

dk2 x2 +

4∑

k=0

dk3 x3 = E − R0I − v(5T ) (8)

5∑

k=0

dk1 x1 +

5∑

k=0

dk2 x2 +

5∑

k=0

dk3 x3 = E − R0I − v(6T ). (9)

By keeping (4), subtracting (4) from (5), subtracting (5) from(6), and so on, we obtain

x1 + x2 + x3 = E − R0I − v(T ) =: b1 (10)

d1x1 + d2x2 + d3x3 = v(T ) − v(2T ) =: b2 (11)

d21x1 + d2

2x2 + d23x3 = v(2T ) − v(3T ) =: b3 (12)

d31x1 + d3

2x2 + d33x3 = v(3T ) − v(4T ) =: b4 (13)

d41x1 + d4

2x2 + d43x3 = v(4T ) − v(5T ) =: b5 (14)

d51x1 + d5

2x2 + d53x3 = v(5T ) − v(6T ) =: b6 . (15)

The aforementioned six equations still have high-order poly-nomials (in the variables d1 , d2 , d3 , x1 , x2 , x3) but the structureis very clear. With further change of variables, the complexitywill be reduced. Let u1 , u2 , u3 be defined as in (3).

In what follows, we provide the key step to derive threelinear equations for u1 , u2 , u3 . Then, it is straightforward to findd1 , d2 , d3 , x1 , x2 , x3 and the parameters (Rk ,Ck ), k = 1, 2, 3.

Let

A =

⎡

⎣1 1 1d1 d2 d3

d21 d2

2 d23

⎤

⎦ .

From (10)–(12), we have⎡

⎣x1x2x3

⎤

⎦ = A−1

⎡

⎣b1b2b3

⎤

⎦ . (16)

Applying this to (13), we have

d31x1 + d3

2x2 + d33x3 = [ d3

1 d32 d3

3 ]

⎡

⎣x1x2x3

⎤

⎦

= [ d31 d3

2 d33 ] A−1

⎡

⎣b1b2b3

⎤

⎦ = b4 . (17)

It can be verified via straightforward computation that

[ d31 d3

2 d33 ] = [u3 −u2 u1 ] A (18)

where u1 , u2 , u3 are defined in (3). Thus,

[ d31 d3

2 d33 ] A−1 = [u3 −u2 u1 ] . (19)

It follows from (17) that

b1u3 − b2u2 + b3u1 = b4 . (20)

Similarly, from (11)–(13), we have⎡

⎣d1x1d2x2d3x3

⎤

⎦ = A−1

⎡

⎣b2b3b4

⎤

⎦ .

Applying this to (14)

[ d31 d3

2 d33 ] A−1

⎡

⎣b2b3b4

⎤

⎦ = b5 .


b2u3 − b3u2 + b4u1 = b5 . (21)

From (12)–(14), we have⎡

⎢⎣d2

1x1

d22x2

d33x3

⎤

⎥⎦ = A−1

⎡

⎣b3b4b5

⎤

⎦ .

Applying this to (15)

[ d31 d3

2 d33 ] A−1

⎡

⎣b3b4b5

⎤

⎦ = b6 .


b3u3 − b4u2 + b5u1 = b6 . (22)

Combine (20), (21), and (22), we have⎡

⎣b3 −b2 b1b4 −b3 b2b5 −b4 b3

⎤

⎦

⎡

⎣u1u2u3

⎤

⎦ =

⎡

⎣b4b5b6

⎤

⎦ .

Thus, u1 , u2 , u3 can be solved as⎡

⎣u1u2u3

⎤

⎦ =

⎡

⎣b3 −b2 b1b4 −b3 b2b5 −b4 b3

⎤

⎦−1 ⎡

⎣b4b5b6

⎤

⎦ .

With u1 = d1 + d2 + d3 , u2 = d1d2 + d2d3 + d3d1 , andu3 = d1d2d3 computed, we can find d1 , d2 , and d3 by



the following procedure. Let p = d1d2 , q = d1 + d2 ; then,d3 = u1 − q and

u3 = p(u1 − q), u2 = p + q(u1 − q).

By substituting p = u2 − u1q + q2 into u3 = p(u1 − q), we ob-tain a third-order equation for q:

q3 − 2u1q2 + (u2

1 + u2)q + (u3 − u1u2) = 0. (23)

Let the three roots be q1 , q2 , and q3 . Then, q = d1 + d2 isone of the roots. Here, we note that the relationship between(d1 , d2 , d3) and (u1 , u2 , u3) is symmetric. This means that thevalues of u1 , u2 , and u3 are the same if d1 , d2 , and d3 areexchanged in any manner. If we let p = d2d3 , q = d2 + d3 , orlet p = d1d3 , q = d1 + d3 , we obtain the same equation (23).This implies that both q = d2 + d3 and q = d1 + d3 satisfy(23). Therefore, (23) must have three positive roots that ared1 + d2 , d2 + d3 , and d1 + d3 , respectively, i.e.,

d2 + d3 = q1 , d1 + d3 = q2 , and d1 + d2 = q3 .

From the three roots q1 , q2 , and q3 , we can solve ford1 , d2 , and d3 as

⎡

⎣d1d2d3

⎤

⎦ =

⎡

⎣0 1 11 0 11 1 0

⎤

⎦−1 ⎡

⎣q1q2q3

⎤

⎦ .

With d1 , d2 , and d3 solved, we obtain x1 , x2 , and x3 from (16).Finally,

Rk = xk/(I(1 − dk )), Ck = −T/(Rk ln(dk )), k = 1, 2, 3.

B. Algorithm for the Second-Order Model

For the case with two capacitors, we need to obtain the voltageresponse v(t) at four equally spaced time instants: v(kT ), k =1, 2, 3, 4. The new variables are defined as

u1 = d1 + d2 , u2 = d1d2 .

The parameters (Rk ,Ck ), k = 1, 2, can be computed from thefollowing steps.

1) Compute b1 = E − R0I − v(T ). For k = 2, 3, 4, bk =v((k − 1)T ) − v(kT ).

2) Compute

[u1u2

]=

[b2 −b1b3 −b2

]−1 [b3b4

].

3) The roots to the second-order equation d2 − u1d + u2 =0 will be d1 and d2 .

4) Let

[x1x2

]=

[1 1d1 d2

]−1 [b1b2

]. Then,

Rk =xk

I(1 − dk ), Ck = − T

Rk ln(dk ), k = 1, 2.

C. Algorithm for Higher Order Models

For the case with n pairs of parallel RCs, we need to obtainthe voltage response v(t) at 2n equally spaced time instants:v(kT ), k = 1, . . . , 2n. The new variables are defined as

u1 = d1 + d2 + · · · + dn (24)

u2 =∑

dk1 dk2 , k1 �= k2 , k1 , k2 ≤ n (25)

u3 =∑

dk1 dk2 dk3 , k1 , k2 , k3 are distinct, ki ≤ n (26)

...

un−1 =∑

dk1 dk2 · · · dkn −1 ,

k1 , . . . , kn−1 are distinct, ki ≤ n (27)

un = d1d2 · · · dn . (28)

The parameters (Rk ,Ck ), k = 1, 2, . . . , n, can be computedfrom the following steps.

1) Compute b1 = E − R0I − v(T ). For k = 2, . . . , 2n,bk = v((k − 1)T ) − v(kT ).

2) Compute

⎡

⎢⎢⎣

u1u2...

un

⎤

⎥⎥⎦ =

⎡

⎢⎢⎢⎣

bn −bn−1 · · · (−1)n+1b1

bn+1 −bn · · · (−1)n+1b2...

b2n−1 −b2n−2 · · · (−1)n+1bn

⎤

⎥⎥⎥⎦

−1

×

⎡

⎢⎢⎢⎣

bn+1

bn+2...

b2n

⎤

⎥⎥⎥⎦ .

3) Solve d1 , . . . , dn from u1 , . . . , un by forming an nth-orderpolynomial from (24)–(28):

f(q) = qn + c1qn−1 + · · · + cn = 0. (29)

The roots are qj = (Σnk=1dk ) − dj , j = 1, 2, . . . , n. Then,

dj ’s can be obtained from qj ’s by solving a system of linearequations. The procedure in this step can be implementedwith the following MATLAB code (where u(i) standsfor ui):c=[ 1 -u(1) u(2)]for i=3:nc=conv(c,[1 -u(1)])+[zeros(1,i) u(i)];endq=roots(c);d=inv(ones(n,n) -eye(n))*q;The first four lines form the coefficients of the polynomialf(q) with a recursive procedure, where “conv” computesthe coefficients of the product of two polynomials via theconvolution of their coefficient vectors. (More explanationwill be provided about this step.)

4) Let

⎡

⎢⎢⎣

x1x2...

xn

⎤

⎥⎥⎦ =

⎡

⎢⎢⎣

1 1 · · · 1d1 d2 · · · dn...

dn−11 dn−1

2 · · · dn−1n

⎤

⎥⎥⎦

−1 ⎡

⎢⎢⎣

b1b2...

bn

⎤

⎥⎥⎦ .

Then, for k = 1, 2, . . . , n,

Rk =xk

I(1 − dk ), Ck = − T

Rk ln(dk ).

The explanation for step 3 is given as follows.



Explanation for step 3: The purpose of this step is to solvedi’s from ui’s with the relationship given by (24)–(28). Define

v1 = d1 + d2 + · · · + dn−1 (30)

v2 =∑

dk1 dk2 , k1 �= k2 , k1 , k2 ≤ n − 1 (31)

v3 =∑

dk1 dk2 dk3 , k1 , k2 , k3 are distinct, ki ≤ n − 1

(32)

...

vn−1 = d1d2 · · · dn−1 . (33)

Then, it is easy to see that

u1 = v1 + dn ⇔ −dn = v1 − u1 (34)

u2 = v2 + v1dn ⇔ v2 = u2 + v1(−dn ) (35)

u3 = v3 + v2dn ⇔ v3 = u3 + v2(−dn ) (36)

...

un−1 = vn−1 + vn−2dn ⇔ vn−1 = un−1 + vn−2(−dn ) (37)

un = vn−1dn ⇔ 0 = un + vn−1(−dn ). (38)

Denote q = v1 . Then, −dn in (35)–(38) can be replaced withq − u1 , and v2 can be expressed as a second-order polynomialof q. By plugging the expression of v2 into (36), we express v3 asa third-order polynomial. With this iterative procedure, vn−1 isan (n − 1)th-order polynomial of q (=v1) and the last iterationyields an nth-order polynomial equation. Each iteration involvesmultiplying a polynomial with q − u1 and adding a constantterm u3 , or subsequently u4 , . . ., un . Note that the coefficientvector of the product of two polynomials is the convolution of thecoefficient vectors of the two polynomials, which can be easilycomputed with the MATLAB command “conv”. By symmetryof the dependence of di’s on ui’s, the roots of this polynomial areqj = (Σn

k=1dk ) − dj , j = 1, . . . , n (one of them is v1 = d1 +· · · + dn−1). Thus, we can use the MATLAB commandd=inv(ones(n,n)- eye(n))*q

to compute d = [d1 d2 · · · dn ]T , where q = [q1 q2 · · · qn ]T .For n = 4, the polynomial f(q) in step 3 is given as follows:

f(q) = q4 − 3u1q3 + (3u2

1 + u2)q2 + (u3 − 2u1u2 − u31)q

+ u4 − u3u1 + u2u21 . (39)

For n ≥ 5, the expression will be more complex but the co-efficients can be easily obtained via convolution.

III. EXPERIMENTAL EXAMPLES AND

COMPUTATIONAL RESULTS

We obtained experimental data for three types of batteries:lead–acid (6 V), NiMH (7.2 V, six cells) and Li-polymer (11.1 V,three cells). Terminal voltage responses under constant currentload were recorded via a 16-digit data acquisition (DAQ) device.

The whole experiment is conducted under room temperature(25 ◦C).

Fig. 3. Voltage response of a lead–acid battery to 1 A load and 60% SOC.

Fig. 4. Voltage and current around t = 0 s and t = 50 s.

A. Data Collection

The sampling period of the DAQ is Td = 0.01 s. The inputvoltage range is −10 to 10 V. Thus, the resolution is 20/216 =3.0518 × 10−4 V. We also measured the current via a 0.1 Ωresistor and an operational amplifier to see its transience afterthe load is turned on and also to see the ripples at the steady stateso that we can make proper adjustment to the terminal voltage.No digital or analog filter is used to process the data/signals.

We use the data obtained from a 6 V lead–acid battery rated13 Ah to demonstrate the procedure. Fig. 3 shows the voltageresponse of the battery to a 1 A load. The initial voltage is6.117 V (corresponding to about 60% residual capacity or SOC).Fig. 4 is plotted to show the voltage and the current aroundt = 0 and at the steady state. The plot at lower left shows thatthe electronic load generates a current that jumps from 0 to 1 Awithin one sampling period. Since the current load can be turnedon any instant within a sampling period of the DAQ, the value ofthe current at the first sampling instant (after the load is turnedon) varies between 0 and the set value. Thus, it may also taketwo sampling steps for the current to reach the set value, whichhas been observed in some of our tests. But mostly the set valueof the current is reached in one step, as depicted in Fig. 4. Thetime 0 is chosen as the sampling instant when the current firstreaches the set value.



The upper-left plot in Fig. 4 shows the initial voltage drop.From these initial values, we can obtain E = v(−Td) (=6.117 Vin this case), and R0 = (v(−Td) − v(0))/I (=0.0656 Ω in thiscase). If the set value of the current is reached in two steps, wetake E = v(−2Td) and R0 = (v(−2Td) − v(0))/I .

The two plots on the right show the ripples of the current atthe steady state and the spikes of the voltage (light colored).We see that the current oscillates around 1 A with a period offive sampling periods. Due to the relatively large time constantsRkCk for the parallel resistors and capacitors, the small currentripples mostly affect the voltage across the inner resistance R0 .Thus, we can make slight corrections to the battery voltage byreplacing v(jTd) with v(jTd) − R0 ∗ (1 − I(jTd)) for all j >0. The darker curve in the upper-right plot shows the correctedvoltage. The spikes are reduced. The corrected voltage will beused for estimating the parameters.

B. Numerical Optimization for Estimating the Parameters

We present the numerical optimization method for compar-ison with the analytical method. We arrange the parameters ofthe battery in a vector p = [R1 , . . . , Rn , C1 , . . . , Cn ]. Denotethe voltage obtained by the experiment as v(jTd) and the volt-age by the model as vm (jTd, p). To be specific

vm (jTd, p) = E − R0I −n∑

k=1

RkI(1 − e− j T d

R k C k ).

The cost function is defined as the root mean square of the error(RMSE) between the measured voltage and that by the model

J(p) =

⎛

⎝N∑

j=0

(vm (jTd, p) − v(jTd))2/(N + 1)

⎞

⎠1/2

where N + 1 is the total number of points. We see that J(p)is a complicated nonlinear function of the parameters and itgenerally has many local minima. Usually, an initial estima-tion of the parameters are needed for an optimization algo-rithm. We may use “fminsearch” in MATLAB. The commandis “p=fminsearch(@(p)cost(p),p0)”, where “cost(p)” is theMATLAB function for computing J(p) and p0 is an initialguess for the parameters.

For a first-order model, there are only two parameters(R1 , C1), so it is possible to find the global minimum. For asecond-order or high-model model, there are at least four pa-rameters and it may be very hard to find a good initial estimation.For example, we formed a cost function to match the experi-mental data for generating Fig. 3 with a third-order model. Thefollowing is a list of p0 and the resulting “minimized” costfunction J∗ by using “fminsearch:”

p0 = [0.1 0.2 0.3 1 2 3], J∗ = 0.0121

p0 = [0.01 0.02 0.03 10 20 30], J∗ = 6.9 × 10−3

p0 = [0.05 0.1 0.15 10 20 30], J∗ = 4.6 × 10−3

p0 = [0.1 0.1 0.2 100 100 100], J∗ = 3.1 × 10−3

p0 = [0.1 0.2 0.3 100 200 300], J∗ = 4.7 × 10−3 .

Fig. 5. Voltage response by a third-order model resulting from a localminimum.

Fig. 5 compares the experimental voltage (light colored) and theresponse by the model (darker curve) corresponding to the localminimum leading from the last p0 .

However, if we use the parameters obtained from the alge-braic method of this paper as initial values and use the afore-mentioned optimization algorithm, we will be able to get muchbetter estimation with a small RMSE.

C. Models for a Lead–Acid Battery

A lead–acid battery rated 6 V, 13 Ah is used for the tests.1) Comparison of the first-, second-, and third-order models:

The voltage response is collected under a load of 1 A. The opencircuit voltage is 6.117 V (corresponding to a residual capacityof about 60%) and R0 = 0.0656 Ω. The voltage response isplotted in Fig. 3.

In this section, we use the explicit formula of Section II toestimate the parameters for the model of different order. For thenth-order model, we need the voltage v(kT ), k = 1, . . . , 2n,where T = MTd for a certain integer M . Due to the noises andother nonidealities, the resulting parameters (Rk ,Ck ) will bedependent on M . We may let M vary in a proper range andchoose the one which yields the least RMSE. To further reducethe effect of noises, we may use the average around v(kT ), e.g.,the average of v(kT − mTd), . . . , v(kT ), . . . , v(kT + mTd)for a certain integer m. The best m may depend on the noisepattern of the experimental setup. For the data we have collected,the best m is between 0 and 5.

For the first-order model, the minimal RMSE obtained is4.96 × 10−3 by taking T = 31.3 s (or M = T/Td = 3130)and m = 2. The resulting parameters are R1 = 0.0703 Ω, C1 =247.3 F. The response by the model and that from the exper-iment are compared in Fig. 6, where the fuzzy light-coloredcurve is from the experiment and the smooth darker curve isby the first-order model. If we use the resulting R1 , C1 as theinitial values for the optimization algorithm in Section III-B, theRMSE can be reduced to 4.2 × 10−3 .

For the second-order model, the minimal RMSE obtained is9.14 × 10−4 , by taking T = 22.1 s, m = 0. The parameters are

R1 = 0.0334Ω, R2 = 0.044Ω, C1 = 84.4F, C2 = 834F.



Fig. 6. Matching the response with a first-order model, a lead–acid battery,1 A load, and 60% SOC.

Fig. 7. Matching the response with a second-order model, a lead–acid battery,1 A load, and 60% SOC.

The response by the second-order model (darker, smooth)and that from the experiment (light-colored, fuzzy) are com-pared in Fig. 7. The asterisk “*” symbols mark the points(ti , v(ti)), ti = T, 2T, 3T, 4T that are used to solve for the pa-rameters. At these points, the values of the two responses areexactly the same. In subsequent figures, we use the same sym-bols and notations to compare voltage responses. If we usethe resulting R1 , R2 , C1 , C2 as initial values for the optimiza-tion algorithm in Section III-B, the RMSE can be reduced to8.15 × 10−4 .

For the third-order model, the minimal RMSE obtained is4.62 × 10−4 , by taking T = 8.83 s,m = 4. The parameters are

R1 = 0.0187 Ω, R2 = 0.0260 Ω, R3 = 0.0365 Ω,

C1 = 79.7 F, C2 = 358.9 F, C3 = 1684.2 F.

The response by the third-order model and that from the exper-iment are plotted in Fig. 8 for comparison.

Again, we may use the parameters obtained from the algebraicmethod as initial values for the optimization algorithm in SectionIII-B. The RMSE is further reduced to 2.78 × 10−4 .

From these three figures and the RMSE values, we see thatthe first-order model matches the experimental response poorly.The second-order model shows significant improvement. How-ever, there are still visible difference in the first interval (betweent = 0 and the first “*”). The third-order model matches the ex-perimental response almost perfectly. Recall that the resolution

Fig. 8. Matching the response with a third-order model, a lead–acid battery,1 A load, and 60% SOC.

Fig. 9. Parameters for the second-order model versus SOC, 1 A load, alead–acid battery.

of the DAQ is 3.0518−4 V. The RMSE by the third-order modelcan be less than this.

2) Estimated Parameters Versus State of Charge, LoadCurrent: Here, we examine how the parameters of the modelsvary with the residual capacity (or SOC) and the load current.The experimental/computational results and figures presentedlater are only intended to demonstrate the method and to givesome rough ideas of the relationship.

Fig. 9 plots the parameters for the second-order model versusthe SOC, where the load current is fixed at 1 A. In the figure,R0 , (R1 , C1), and (R2 ,C2) are represented by “◦”, “*,” and“�”, respectively. Same symbols are used in subsequent figures.

Fig. 10 plots the parameters for the third-order model versusSOC, where (R3 , C3) are denoted with “�′′.

From these two figures, we see that the parameters do notvary significantly for the SOC between 10% and 80%. Above80%, R0 vary drastically and R2 of the third-order model visiblyincreases.

Fig. 11 plots the parameters for the second-order model versusthe load current, where the residual capacity is between 55% and60%.

Fig. 12 plots the parameters for the third-order model versusthe load current. From these figures, we see that R2 of thesecond-order model and R3 of the third-order model vary mostsignificantly with the load current. Other parameters do notchange much.



Fig. 10. Parameters for third-order model versus SOC, 1 A load, a lead–acidbattery.

Fig. 11. Parameters for second-order model versus load current, 55–60%SOC, a lead–acid battery.

Fig. 12. Parameters for third-order model versus load current, 55–60% SOC,a lead–acid battery.

To completely characterize the properties of a particular typeof batteries, more extensive tests should be carried out by vary-ing the load current, residual capacity, even temperature, overa 3-D grid. For each parameter, a nonlinear function can benumerically derived to match the numbers computed from ex-perimental data. Alternatively, for simplified modeling, we maydetermine a nominal value for each parameter and a correspond-

Fig. 13. Matching the response with a second-order model, an NiMH battery,1 A load, and 55% SOC.

ing range around this nominal value. This will result in a nom-inal model with uncertain parameters, which may be useful inrobustness analysis and design of a system involving batteries.

D. Models for an NiMH Battery

The battery under test is a 7.2 V (six cell) NiMH battery rated5000 mAh.

1) Comparison of the First-, Second-, and Third-OrderModels: The voltage response is measured under a load of1 A. The open circuit voltage is E = 7.87 V (about 55%residual capacity). From the initial voltage drop, we obtainedR0 = 0.0588 Ω. We use the same procedure as that for the lead–acid battery to obtain the first-, second-, and third-order models.

For the first-order model, the minimal RMSE obtained is5.1 × 10−3 , with R1 = 0.083 Ω and C1 = 503 F. The responseby the model also differs significantly from the experimentalresponse, similarly to that for the lead–acid battery.

For the second-order model, the minimal RMSE obtained is9.57 × 10−4 , by taking T = 21.2 s, and m = 3. The parametersare

R1 = 0.022Ω, R2 = 0.104Ω, C1 = 95.7F, C2 = 1290F.

The response by the second-order model and that from the ex-periment are plotted in Fig. 13.

For the third-order model, the minimal RMSE obtained is2.88 × 10−4 , by taking T = 12.17 s and m = 2. The parametersare

R1 = 0.0129 Ω, R2 = 0.0099 Ω, R3 = 0.1117 Ω,

C1 = 8.6 F, C2 = 663 F, C3 = 1350.7 F.

The response by the third-order model and that from theexperiment are plotted in Fig. 14.

As with the lead–acid battery, the first-order model does avery poor job matching the terminal voltage, the second-ordermodel shows significant improvement, while the third-ordermodel matches the experimental response almost perfectly.

2) Estimated Parameters Versus State of Charge, LoadCurrent: Fig. 15 plots R0 and the parameters for the second-order model versus the residual capacity, where the load current



Fig. 14. Matching the response with a third-order model, an NiMH battery,1 A load, and 55% SOC.

Fig. 15. Parameters for the second-order model versus SOC, 1 A load, anNiMH battery.

Fig. 16. Parameters for third-order model versus SOC, 1 A load, an NiMHbattery.

is fixed at 1 A. Only R2 shows significant changes. Other pa-rameters do not change much.

Fig. 16 plots the parameters for the third-order model versusthe residual capacity. R3 is plotted separately since it is muchgreater than R1 and R2 . Among the six parameters, R3 showsmost visible changes.

Fig. 17 plots the parameters for the second-order model versusthe load current, where the residual capacity is about 50–55%.

Fig. 17. Parameters for second-order model versus load current, 50–55%SOC, an NiMH battery.

Fig. 18. Parameters for third-order model versus load current, 50–55% SOC,an NiMH battery.

Fig. 18 plots the parameters for the third-order model versusthe load current. From these figures, we see that R2 of thesecond-order model and R3 of the third-order model vary mostsignificantly with the current. Other parameters do not changemuch.

E. Models for a Li-Polymer Battery

The battery under test is a 11.1 V (three cell) Li-polymerbattery rated 5000 mAh.

1) Comparison of the First-, Second-, and Third-orderModels: The voltage response was obtained under a load of 2 A.The open circuit voltage was 11.51 V (about 55% residual capac-ity). From the initial voltage drop, we obtained R0 = 0.0483 Ω.

For the first-order model, the minimal RMSE obtained is2.8 × 10−3 , with R1 = 0.0245 Ω and C1 = 1124.2 F. The re-sponse by the model also differs significantly from the experi-mental response, as with the other two batteries.

For the second-order model, the minimal RMSE obtained is5.04 × 10−4 , by taking T = 25 s and m = 4. The parametersare

R1 = 0.0097 Ω, R2 = 0.0220 Ω, C1 = 571.6 F,

C2 = 4051.1 F.



Fig. 19. Matching the response with a second-order model, 2 A load, 55%SOC, and a Li-polymer battery.

Fig. 20. Matching the response with a third-order model, 2 A load, 55% SOC,and a Li-polymer battery.

The response by the second-order model and that from experi-ment are plotted in Fig. 19.

For the third-order model, the minimal RMSE obtained is4.11 × 10−4 by taking T = 17.13 s and m = 5. The parametersare

R1 = 0.0056 Ω, R2 = 0.0083 Ω, R3 = 0.0332 Ω,

C1 = 638.3 F, C2 = 2124.8 F, C3 = 7866.5 F.

The response by the third-order model and that from the exper-iment are plotted in Fig. 20.

The parameters for the third-order model of the Li-polymerbattery seem to be very sensitive to the selection of T and m. Weobtained another third-order model that yields a slightly largerRMSE, 4.22 × 10−4 , by taking T = 11.72 s and m = 4. Butthe parameters are visibly different

R1 = 0.0044 Ω, R2 = 0.0097 Ω, R3 = 0.0439 Ω

C1 = 419.9 F, C2 = 1569.5 F, C3 = 8554.5 F.

The responses are compared in Fig. 21.Actually, several other third-order models have been obtained

to produce responses that match the experimental data quitewell, by varying T and m. The resulting parameters for themodels are quite different. After all, the third-order models do

Fig. 21. Response by another third-order model.

Fig. 22. Parameters for the second-order model versus SOC, 2 A load, aLi-polymer battery.

not seem to yield much better RMSE than the second-ordermodel. These results may indicate that there is some redundancyin the third-order model and that a second-order model may besufficient. This is somewhat unexpected for a three-cell battery.

2) Estimated Parameters Versus State of Charge, LoadCurrent: Fig. 22 plots R0 and the parameters for the second-order model versus the residual capacity, where the load currentis fixed at 2 A. Among these parameters, R2 and C2 have mostsignificant changes.

Fig. 23 plots the parameters for the third-order model versusthe residual capacity. We see that all the parameters have visiblechanges. This may be an indication that the parameters for thethird-order models are also sensitive to the SOC, which is notdesirable for the analysis and design of dynamical systems.Since the third-order models are also sensitive to other factors,maybe the second-order models are better choices.

Fig. 24 plots the parameters for the second-order model versusthe load current, where the residual capacity is between 58% and66%. We see that R2 changes most visibly.

Fig. 25 plots the parameters for the third-order model versusthe load current. From these figures, we also see that the pa-rameters for the third-order models are more sensitive to loadcurrent than those for the second-order models.



Fig. 23. Parameters for the third-order model versus SOC, 2 A load, a Li-polymer battery.

Fig. 24. Parameters for the second-order model versus load current, 58–66%SOC, a Li-polymer battery.

Fig. 25. Parameters for third-order model versus load current, 58–66% SOC,a Li-polymer battery.

IV. DISCUSSIONS AND FUTURE WORK

We derived simple analytical algorithms to compute the pa-rameters for Thevenin’s equivalent circuit for batteries. Experi-ment and computation are implemented on three different typesof batteries. For all these batteries, the computation shows thatthe first-order model does a very poor job matching the exper-imental responses, the second-order model shows significant

improvement and the third-order model can produce a responsethat matches the experimental data almost perfectly. However,parameters for the third-order model of some battery can be verysensitive to operating conditions and noises. If the improvementis not significant, a second-order model may be a better choice.

The battery model obtained from one test is valid under spe-cific work conditions when the load current and the SOC varyin a certain range. To completely characterize the properties ofa particular type of batteries, more extensive tests should becarried out by varying the load current, residual capacity, eventemperature, over a 3-D grid. For each parameter, a nonlinearfunction can be numerically derived to match the numbers com-puted from experimental data. For a simplified model, we maychoose a nominal work condition and determine a range for eachparameter. These ranges will be useful for robustness analysisof battery-driven electronic systems.

One limitation with the battery model considered in this paperis that it is not able to account for the decrease of the opencircuit voltage during discharge. A possible solution to dealwith this limitation is to replace the ideal voltage source in thecircuit in Fig. 1 with a capacitor (possibly in parallel with aresistor with a very large resistance). This revised model willbe considered in our future work. Our first task is to derive ananalytical method to extract the parameters for this new modelfrom experimental responses. We will see if the method in thiswork can be modified/extended for the new model. However, wemay face numerical challenges since the capacitor replacing theideal voltage source is expected to have very large capacitance.We may have to measure the response for a much longer time,further reduce the measurement noises, and/or use a DAQ withbetter resolution.

Another significant problem about batteries is to model theircharging behaviors. We may also use a similar Thevenin’s equiv-alent circuit but the ideal voltage source in Fig. 1 has to be re-placed with a capacitor. It is possible that the analytical methodin this paper can also be modified for this purpose. It will beintriguing to find out if the charging and discharging dynamicscan be unified in the same new model, and how the parametersfor the charging dynamics differ from those for the dischargingdynamics.

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Tingshu Hu (SM’01) received the B.S. and M.S. de-grees in electrical engineering from Shanghai JiaoTong University, Shanghai, China, in 1985 and1988, respectively, and the Ph.D. degree in elec-trical engineering from the University of Virginia,Charlottesville, in 2001.

She was a Postdoctoral Researcher at the Uni-versity of Virginia and the University of California,Santa Barbara. In January 2005, she joined the Fac-ulty of Electrical and Computer Engineering, Univer-sity of Massachusetts, Lowell, where she is currently

an Associate Professor. Her research interests include nonlinear systems theory,optimization, robust control theory, power electronics, and energy systems.

Brian Zanchi is currently working toward the B.S.degree in electrical engineering in the Department ofElectrical and Computer Engineering, University ofMassachusetts, Lowell.

He has been working in the control systems lab-oratory with Dr. T. Hu since January 2010. He hasalso been working part time with Maxim IntegratedProducts, North Chelmsford, MA, since May 2010and will be a full time employee after graduation.His research interests include power electronics andbattery evaluation.

Jianping Zhao received the B.S. degree in electricalengineering and the M.S. degree in system engineer-ing from Shanghai Jiao Tong University, Shanghai,China, in 1985 and 1988, respectively.

He has been at Juniper Networks, Westford, MA,since 2001. His research interests include test au-tomation and methodologies, and battery evaluation.

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