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Chapter 5

Simulation and Results

In this chapter the simulations of the Braking step-down and of the motoring step-up will be discussed. As said, simulations are run by Simulink using the equations and the diagrams of chapter 3, and using Simplorer. The energy dissipated from the switches during the of f − state of are neglected because they are small compared to the other losses.

In figure 5.1 the converter topology is depicted. The inductance L

in

is a small inductance and represent the parasitic inductance of the cables. In the simulation L

in

is needed to reach a ripple at the voltage capacitor, which would otherwise be a constant value, because a constant voltage is directly applied across it.

V

b

C

1

L

ind

C

2

M

bd

D

bd

D

mu

M

mu

D

bu

M

bu

D

md

M

md

V

m

L

in

L

in

Figure 5.1: Topology of the bidirectional buck-boost converter

Simulations are run using the following data:

• Capacitor battery side C

1

modeled by a capacitance C

1

= 8800µF in series with a

34

(2)

resistance R

c1

= 12.75mΩ

• Main inductance L

ind

modeled by an inductance L

ind

= 30µH in series with a resis- tance R

ind

= 5.17mΩ

• Capacitor motor side C

2

modeled by a capacitance C

2

= 5000µF in series with a resistance R

c1

= 0.6mΩ

• Automotive MOSFETS IRF model IRF P 2907

5.1 MOSFET

IRF P 2907 is a MOSFET specifically designed for automotive applications. Its main char- acteristics are:

- Drain-to-source breakdown Voltage V

DSS

= 75V with V

GS

= 0V and I

D

= 250µA - Static drain-to-source On-Resistance (max)R

DS(on)

= 4.5mΩ with V

GS

= 10V and

I

D

= 125A

- Continuous drain current I

D

= 209A with case temperature T

c

= 25

and V

GS

= 10V As said in paragraph 2.1 the MOSFET model used in the simulink simulation is resistive.

The value of the on resistance is found using Simplorer. In Simplorer there is a library called Manuf actures, in which there are several components for the most important companies.

The IRF P 2907 MOSFET model is in the library and several simulations are run for find its on resistance R

mon

and the reversal diode resistance R

don

and forward voltage diode V f d. The value of the resistances are:

• R

mon

= 4.1225mΩ

• R

don

= 3mΩ and V

f d

= 0.5V

5.2 Energy Balance of the Converter

The Objective of the simulation is to find an efficiency map at different voltage and current

values, at the steady-state. It means that the initial value of the current in the inductor and

the initial values of the voltage in the capacitors, must be the same as the final values of a

switching period. Hence the energy stored in every storage element in a complete switching

period must be zero. The initial values found by the formulae of chapter 4, which have used

(3)

for the simulation, are more accurate , but again approximate, it follows that the initial state of the storage elements is not exactly the same as the final. Hence the energy stored in every storage element in a complete switching period is not zero. Thus the variation of the energy in the storage elements in the balance of the energies must be considered.

Equation 5.1 is the energy balance of the converter, where E

in

is the input energy, E

out

the output energy, E

r

the leakage, ∆E

l

is the energy stored in the inductor and ∆E

c

the energy stored in the capacitor.

E

in

= E

out

+ E

r

+ ∆E

l

+ ∆E

c

(5.1)

∆E

c

and ∆E

l

are calculated as:

∆E

c

= C

c

·(V

end2

− V

ini2

)

2 (5.2)

∆E

l

= L

ind

·(I

end2

− I

ini2

)

2 (5.3)

I

ini

and I

end

are the currents at the beginning and at the end of a switching period respec- tively, V

ini

and V

end

are the voltages at the beginning and at the end of a switching period respectively.

5.2.1 Calculation of the Efficiency

Efficiency is calculated with respect to just one switching period. For the calculation of the efficiency, the initial values of the currents and the voltages in the storage elements are assumed to be correct. As mentioned above, the final values are not the same as the initial values, thus the stored energies must be considered for the efficiency calculation. The efficiency formula is:

η = E

out

E

in

(5.4) Energy stored in the storage components must be considered in the equation 5.4 as an addendum to add or subtract at the output or at the input energy. A typical behavior of the current in the inductor is depicted in figure 5.2, where the dashed line represents the current progress, if its value had been exact. For the calculation of the efficiency, it is assumed that the initial value is exact. The challenge at this point is understand if the

E

must be added or subtracted at the input or at the output energy. It is impossible to make correct choice, but it is possible to say what is the impact of ∆E on E

in

or E

out

. For example, considering i

in

, its variation in one switching period, has a big impact on E

in

, because E

in

= R

t

V

in

·i

ind

dt and the same is for v

C2

with respect to E

out

.

(4)

T

t

on

t

off

t t

I

ini

I

end

Figure 5.2: typical current progress in the inductance

In general, if ∆

2

is the energy to subtract from E

out

and ∆

1

the energy to subtract from E

in

, it is possible to write:

η = E

out

− ∆

2

E

in

− ∆

1

(5.5)

and from the equation 5.1: E

out

= E

in

− E

r

− ∆

1

− ∆

2

, and substituting in equation 5.5:

η = 1 − E

r

+ 2·∆

2

E

in

− ∆

1

(5.6)

In the simulation, the value of efficiency was calculated by both formulae 5.6 and 5.5,

5.3 Simulation using Simulink

5.3.1 Braking Step-Down

Figure 5.3 shows the Brake step-down mode circuit. Figure 5.4 shows the circuit with the components model. For comparison and for a first step, a further simplification is made: voltage at the points B is considered constant and equal to the supply voltage. It is possible do that because of the capacitor, which keeps the voltage relatively constant.

Consequently, the voltage at the point A is a square wave that, during the on-state, is equal

to V

in

subtracted from the voltage drop across the Mosfet, and during the off-state from the

voltage drop across the diode. By the above simplification, the circuit in fig. 5.4 become

the circuit in fig. 5.5. It follows that v

A

is a square wave with the following features:

(5)

V

b

L

ind

V

in

= V

m

C

2

D

bd

M

bd

D

bu

C

1

Figure 5.3: Brake step-down circuit

T

s

= 100 kHz = (switchingperiod); voltage at turn-on ˆ V

in

= V

in

− i

m

· R

onm

; voltage at turn-off ˆ V

d

= −(i

d

·R

ond

);

For simulating the circuit in every point of operation, Matlabscript was used. A f or − loop starts the simulation of the circuit automatically, varying the values of I

load

, of V

load

and of V

in

for each loop. Hence for each loop the initial conditions of the storage elements are also calculated using the formulae in chapter 4. In the block diagram, I

load

is varied changing the value of the load resistance R. V

load

is varied, changing the value of the duty cycle d.

V

in

= V

m

C

1

R

c1

C

2

R

c2

R

ind

R

load

i

in

L

in

i

ind

L

ind

i

c2

i

m

R

onm

R

ond

i

load

A

v

load

B

V

A

i

c1

i

d

R

ond

Figure 5.4: Most simple model of the Braking step down circuit mode of operation

(6)

i

ind

R

load

i

c1

t

on

t

off

V

in

V

d

R

ind

L

ind

v

load

i

load

R

ond

C

1

R

c1

v

A

Figure 5.5: Brake step-down supplied by a square wave

The relation between d and V

load

is given by equation 4.10, and for clarity is rewritten as:

V

load

= V

in

·d − V

f d

·(1 − d) b·d + a·(1 − d) d = V

load

·a + V

f d

V

in

− V

load

·(b − a) + V

f d

The initial value of the voltage across the capacitor C

1

(V

C01

), is given by equation 4.32 and the initial value of the current through the main inductance (I

ind0

) from equation 4.29.

A simulation for a I

load

range of [10A, 200A] each 3A, for a V

in

range of [26A, 52A] each

2V was run, and for a V

load

range of [10V, 14V ] each 1V also simuated. The simulation

results are shown in the next page, each of them is coloured for different value of V

load

. The

contour lines are coloured for the same values of efficiency, calculated using the formula 5.6

with ∆

2

= ∆E

C1

and ∆

1

= ∆L

ind

(7)

20 40 60 80 100 120 140 160 180 200 30

35 40 45 50

0.8

0.810.82

0.83

0.84

0.85

0.86

0.87

0.88

0.89

0.90.91

0.92 0.93 0.95 0.94 0.96

η =EEoutin−∆E−∆Elc (Vload= 10V )

Vin

Iload

Figure 5.6: Contour lines for V

load

= 10V

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.82

0.83

0.840.85

0.86

0.87 0.88 0.89 0.9

0.91

0.93 0.92

0.94 0.95 0.96

η =EEoutin−∆E−∆Elc (Vload= 11V )

Vin

Iload

Figure 5.7: Contour lines for V

load

= 11V

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.83

0.84

0.85

0.86

0.87

0.88 0.9 0.89 0.91 0.92

0.94 0.93 0.95

0.96

0.97

η =EEout−∆Ec

in−∆El (Vload= 12V )

Vin

Iload

Figure 5.8: Contour lines for V

load

= 12V

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.84

0.85

0.86

0.87

0.88

0.89

0.9

0.91 0.92 0.93 0.94

0.95

0.96 0.97

η =EEout−∆Ec

in−∆El (Vload= 13V )

Vin

Iload

Figure 5.9: Contour lines for V

load

= 13V

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.850.86

0.87 0.88

0.89

0.90.91

0.92 0.93 0.94

0.95

0.96

0.97

η =EEoutin−∆E−∆Elc (Vload= 14V )

Vin

Iload

Figure 5.10: Contour lines for V

load

= 14V

The results shows that current is the most important factor.

(8)

The converter is now considered with the input capacitor (C

C1

) and input inductance (L

in

). Efficiency is calculated using the formula 5.6, considering ∆

2

= ∆E

c2

and ∆

1

=

∆E

Lind

+ ∆E

c1

+ ∆E

Lin

. The results of the simulation are shown in the figures in the next

page, where each is coloured for a different value of V

load

. The contour lines are coloured

for the same values of efficiency, calculated by the formula 5.6. The efficiency is smaller

than the efficiency of the device supplied by a square wave, which is due to the presence of

a further resistance (R

C1

).

(9)

20 40 60 80 100 120 140 160 180 200 30

35 40 45

50 0.8

0.810.82

0.83

0.84

0.85

0.86

0.87

0.88

0.89

0.9

0.91

0.92 0.93

0.94

0.96 0.95

η =Ein−(∆EELout−∆Ec2

ind+∆Ec1+∆ELin)

Vin

Iload

Figure 5.11: Contour lines for V

load

= 10V

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.82

0.83

0.84

0.85

0.86

0.87

0.88

0.890.9

0.91

0.92

0.93

0.94

0.95

0.96

η =Ein−(∆EELout−∆Ec2

ind+∆Ec1+∆ELin)

Vin

Iload

Figure 5.12: Contour lines for V

load

= 11V

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.83 0.85 0.84

0.86

0.87 0.88

0.890.9

0.91 0.92 0.93 0.94

0.95

0.96

η =Ein−(∆EELout−∆Ec2

ind+∆Ec1+∆ELin)

Vin

Iload

Figure 5.13: Contour lines for V

load

= 12V

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.84

0.85 0.86

0.87

0.88

0.89

0.90.91

0.92

0.93

0.94

0.95 0.96 0.97

η =Ein−(∆EELout−∆Ec2

ind+∆Ec1+∆ELin)

Vin

Iload

Figure 5.14: Contour lines for V

load

= 13V

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.85

0.86

0.87

0.88

0.890.9

0.91 0.92 0.93

0.94 0.95 0.96 0.97

η =Ein−(∆EELout−∆Ec2

ind+∆Ec1+∆ELin)

Vin

Iload

Figure 5.15: Contour lines for V

load

= 14V

(10)

5.3.1.1 Calculation of theoretical efficiency

In this section the efficiency is calculated by using theory. The currents and the voltages

in the device are calculated by the formulae of chapter 4. Results are shown in the next

page, and are similar to the results obtained by the simulation.

(11)

20 40 60 80 100 120 140 160 180 200 30

35 40 45 50

0.84

0.85

0.86 0.87 0.88 0.89

0.9 0.91 0.92

0.93 0.94 0.95

η =EEoutin (Vload= 10V )

Vin

Iload

Figure 5.16: contour lines for V

load

= 10V (theoretic)

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.85

0.86

0.87

0.88 0.9 0.89

0.91 0.92 0.93 0.94 0.95

0.96

η = 1 −EEoutin (Vload= 11V )

Vin

Iload

Figure 5.17: contour lines for V

load

= 11V (theoretic)

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.86

0.87 0.88 0.89

0.91 0.9 0.92 0.94 0.93

0.95 0.96

η =EEout

in (Vload= 12V )

Vin

Iload

Figure 5.18: contour lines for V

load

= 12V (theoretic)

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.87

0.88 0.89

0.9

0.91 0.92 0.93

0.94 0.95

0.96

0.97

η =EEout

in (Vload= 13V )

Vin

Iload

Figure 5.19: contour lines for V

load

= 13V (theoretic)

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.88 0.89

0.91 0.9 0.92 0.93 0.94

0.95 0.97 0.96

η =EEout

in (Vload= 14V )

Vin

Iload

Figure 5.20: contour lines for V

load

= 14V

(theoretic)

(12)

5.3.2 Motoring Step-Up

As with the braking step-down converter, the evaluation of the losses was made. The circuit is depicted in figure 5.21. As explained in chapter 2, M

md

is constantly on in this mode

C

1

L

ind

C

2

M

md

D

mu

V

in

= V

m

M

mu

V

b

Figure 5.21: Motor step up circuit

of operation, hence its model is a resistance. The circuit with the model components is drawn in figure 5.22. As with the step-down, a f or − loop starts the simulation of the

vload Vin= Vb

C2 Rind

iind Lind

Ronm

iload

id

im

Rond

Vfd

+

- Rc2

Rload

ic2

- +

iin Lin

C1

Rc1 c i1

Ronm

Figure 5.22: Motor step-up supply by a square wave

circuit automatically, varying the values of I

load

, of V

load

and of V

in

for each loop, but in

this case the load is at the motor side of the converter. Hence for each loop the initial

conditions of the storage elements are also calculated using the formulas of chapter 4. In

the block diagram, I

load

is varied changing the value of the load resistance R, and V

load

is

(13)

varied changing the value of the duty cycle d. The relation between d and V

load

is given from equation 4.46, and for clarity is rewritten as:

V

load

= V

in

− V

f d

·(1 − d)

(1 − d)

2

+ (c − e)·d + e ·(1 − d)

(5.7)

d1 =

−Vload c+2 Vload −Vin+Vload e+2 Vfd

2 (Vload +Vfd)

+

+

(2 Vload cVin−2 Vin Vload e−4 Vload cVfd+

2 (Vload +Vfd)

−2 Vload2ce−4 Vload2cVin2+Vload2e2+Vload2c2)12

2 (Vload +Vfd)

Looking at picture 4.9, it is composed of two parts, one for d < ¯ d and one for d > ¯ d. For the simulation, only the part for d < ¯ d is considered. The initial value of the voltage across the capacitor C

1

(V

C01

), is given from equation 4.65 and the initial value of the current through the main inductance (I

ind0

) from equation ??. A simulation for a I

load

range of [10A, 100A]

each 3A, for a V

in

range of [26A, 52A] each 2V and for a V

load

range of [10V, 14V ] each 1V

was run. The simulation results are shown in the next page, each of them is coloured for

different value of V

load

. The contour lines are coloured for the same values of the efficiency,

calculated using the formula 5.6 with ∆

2

= ∆E

c2

and ∆

1

= ∆E

Lind

+ ∆E

c1

+ ∆E

Lin

.

(14)

10 20 30 40 50 60 70 80 90 100 30

35 40 45

50 0.5

0.6 0.55 0.65 0.720.7 0.74 0.76 0.78

0.8 0.82 0.84 0.86 0.87 0.88

0.90.89

0.91

0.92 0.93 0.94

η =Ein−(∆EELout−∆Ec2

ind+∆Ec1+∆ELin)

Vload

Iload

Figure 5.23: contour lines for V

in

= 10V

10 20 30 40 50 60 70 80 90 100

30 35 40 45 50

0.5

0.55 0.65 0.6

0.720.7 0.760.74 0.78 0.8

0.84 0.82

0.86 0.87

0.890.88 0.9

0.92 0.91 0.93 0.94

0.95

η =Ein−(∆EELout−∆Ec2

ind+∆Ec1+∆ELin)

Vload

Iload

Figure 5.24: contour lines for V

in

= 11V

10 20 30 40 50 60 70 80 90 100

30 35 40 45 50

0.5

0.55 0.6 0.65 0.7 0.72

0.760.74 0.78

0.8 0.84 0.82 0.86 0.880.87

0.89 0.9 0.91 0.92

0.94 0.93

0.95

η =E Eout−∆Ec2

in−(∆ELind+∆Ec1+∆ELin)

Vload

Iload

Figure 5.25: contour lines for V

in

= 12V

10 20 30 40 50 60 70 80 90 100

30 35 40 45 50

0.5 0.55

0.6 0.65

0.7 0.72 0.74

0.78 0.76 0.8 0.84 0.82

0.86

0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95

0.96

η =E Eout−∆Ec2

in−(∆ELind+∆Ec1+∆ELin)

Vload

Iload

Figure 5.26: contour lines for V

in

= 13V

10 20 30 40 50 60 70 80 90 100

30 35 40 45 50

0.6 0.7 0.65

0.72 0.74

0.78 0.76 0.8 0.84 0.82

0.86 0.88 0.87 0.89 0.91 0.9

0.92 0.93 0.94

0.95 0.96

η =E Eout−∆Ec2

in−(∆ELind+∆Ec1+∆ELin)

Vload

Iload

Figure 5.27: contour lines for V

in

= 14V

As with the brake step-down, the current is the most important factor. In this case the

values of V

in

and V

load

are closer, the efficiency is higher, but the reason is different. In this

(15)

case in fact is that the closer values of the voltage are, the smaller the duty cycle is. For the same value of the load current, the larger d is, the larger is the current in the inductance (I

ind

= I

load

/(1 − d)). Consequently the currents, and hence the losses in the Mosfet and in the diode are larger too.

5.3.2.1 Efficiency Calculated by Theoretic Consideration

In the same way as with the brake step-down, the efficiency is calculated by theoretical

consideration. The currents and the voltages in the device are calculated by the formulae

of Chapter 4. The results are shown in the next page, and are similar to the results obtained

by the simulation.

(16)

10 20 30 40 50 60 70 80 90 100 30

35 40 45

50 0.5

0.55 0.650.6

0.72 0.7 0.74

0.780.76 0.8 0.82 0.84 0.86 0.87 0.88

0.89 0.9 0.91

0.92 0.94 0.93 0.95

η =EEoutin (Vin= 10V )

Vload

Iload

Figure 5.28: Contour lines for V

in

= 10V (theoretical)

10 20 30 40 50 60 70 80 90 100

30 35 40 45 50

0.5 0.55 0.650.6

0.7 0.72

0.74 0.76 0.78 0.8

0.82 0.84

0.86

0.87

0.88 0.9 0.89 0.91 0.92 0.94 0.93

0.95

η = 1 −EEoutin (Vin= 11V )

Vload

Iload

Figure 5.29: Contour lines for V

in

= 11V (theoretical)

10 20 30 40 50 60 70 80 90 100

30 35 40 45 50

0.550.5 0.6 0.65 0.7

0.74 0.72 0.780.76

0.8

0.82 0.86 0.84

0.87 0.88 0.89 0.9 0.91

0.92 0.93 0.95 0.94 0.96

η =EEoutin (Vin= 12V )

Vload

Iload

Figure 5.30: Contour lines for V

in

= 12V (theoretical)

10 20 30 40 50 60 70 80 90 100

30 35 40 45

50 0.60.550.5

0.65 0.720.7 0.74 0.76 0.78

0.8 0.84 0.82

0.86 0.880.87 0.89 0.91 0.9

0.92 0.93 0.94 0.95

0.96

η =EEoutin (Vin= 13V )

Vload

Iload

Figure 5.31: Contour lines for V

in

= 13V (theoretical)

10 20 30 40 50 60 70 80 90 100

30 35 40 45

50 0.7

0.72 0.74 0.76

0.78 0.8 0.82 0.86 0.84

0.87 0.89 0.88 0.9 0.91

0.93 0.92 0.95 0.94

0.96

η =EEoutin (Vin= 14V )

Vload

Iload

Figure 5.32: Contour lines for V

in

= 14V (theoretical)

5.4 Simulation using Simplorer

As with simulink, a f or − loop starts the simulation of the circuit automatically, varying

the values of I

load

, of V

load

and of V

in

for each loop. Hence for each loop the initial condi-

(17)

tions of the storage elements are also calculated using the formulae of the chapter 4. The only differece among the models are the MOSFETS, which in this case are modeled more accurately.

5.4.1 Braking Step-down

Simulations were run for V

load

= 10V and V

load

= 12V and the efficiency maps are shown in figures 5.33, 5.34

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.8

0.8 0.81

0.81 0.82

0.82 0.83

0.83

0.84

0.84

0.85

0.85 0.86

0.86 0.87

0.87 0.88

0.88 0.89

0.89

0.89 0.9

0.9

0.9 0.91

0.91 0.92 0.93

η =E Eout−∆Ec2

in−(∆ELind+∆Ec1+∆ELin)

Vin

Iload

Figure 5.33: Contour lines for V

load

= 10V

20 40 60 80 100 120 140 160 180 200

30 35 40 45 50

0.8

0.8 0.81

0.81 0.82

0.82 0.83

0.83 0.84

0.84 0.85

0.85 0.86

0.86 0.87

0.87 0.88

0.88 0.89

0.9 0.9

0.9 0.91

0.91

0.91 0.92

0.92 0.93

η =E Eout−∆Ec2

in−(∆ELind+∆Ec1+∆ELin)

Vin

Iload

Figure 5.34: Contour lines for V

load

= 12V

5.4.2 Motoring step-up

Simulations were run only for V

in

= 12V and the efficiency maps are shown in figures 5.35.

10 20 30 40 50 60 70 80 90 100

30 35 40 45 50

0.5

0.5

0.5 0.55

0.55

0.55 0.6

0.6

0.6 0.65

0.65

0.65 0.7

0.7

0.7 0.72

0.72 0.74

0.74 0.76

0.76 0.78

0.78 0.8

0.8 0.82

0.82 0.84

0.84 0.86

0.86 0.88

0.88 0.89

0.89 0.9

0.9 0.91

0.91 0.92

0.92

0.93

0.93

η =Ein−(∆EELindout−∆E+∆Ec2

c1+∆ELin)

Vload

Iload

Figure 5.35: Contour lines of the motoring step-up for V

in

= 12V

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