「spiceの活用方法」セミナー資料(28jan2011) ppt

Copyright (C) Bee Technologies Inc. 2011 1

SPICE の活用方法

株式会社ビー・テクノロジーhttp://www.bee-tech.com/[email protected]

2011 年 1 月 28 日 ( 金曜日 )

1.PWM Buck Converter Average Model←[DEMO]　フィードバック制御におけるアベレージモデルを活用した　位相余裕度のシミュレーションの活用方法を解説していきます。

2. ステッピングモータのコンセプトキット　 [ 事例紹介 ]2.1 ユニポーラ・ステッピングモーター制御回路2.2 バイポーラ・ステッピングモーター制御回路

「コンセプトキット」でパラメータベース・シミュレーション

http://www.bee-tech.com/

コンセプトキットの位置付け


回路解析シミュレータSpice 系回路解析シミュレータ

PSpice,LTspice,MultiSim,MicroCap,HSPICE,SmartSPICE,Simplorer, and so on

[ デザインキット ] 回路方式のテンプレートコンセプトキット (NEW)←(概念設計のテンプレート )

デザインキット ( 各回路方式のテンプレート )

[ スパイスモデル ]デバイスモデリングサービス (58 種類のデバイスモデリング )

スパイス・パーク www.spicepark.comシンプルモデル (NEW)← ブロックベースのスパイスモデル

http://beetech.s287.xrea.com/spicepark/index.html

コンセプトキットとは


製品価格 ( 円 ) PSpice 版 LTspice 版ユニポーラステッピングモータ制御回路

42,000 2011 年 2 月初旬 2011 年 2 月中旬

バイポーラステッピングモータ制御回路

42,000 2011 年 2 月初旬 2011 年 2 月中旬

アベレージモデルの降圧コンバータ 84,000 2011 年 2 月中旬 2011 年 2 月下旬

過渡解析モデルの降圧コンバータ未定 2011 年 2 月中旬 2011 年 2 月下旬

コンセプトキット

[ 概念設計 ]

デザインキット[ 具体化設計 ] 実設計

デザインキット


要望が多いインバータ回路方式を中心に 20 種類の新製品を開発中。

製品分野FCC 回路電源回路RCC 回路電源回路低損失リニアレギュレータ電源回路高精度リニアレギュレータ電源回路D 級アンプアンプ回路擬似共振電源回路電源回路マイクロコントローラ電源回路ステッピングモータドライブ回路モーター制御回路PWM IC による電源回路電源回路バッテリー回路 ( リチウムイオン電池 ) バッテリーアプリケーション回路バッテリー回路 ( ニッケル水素電池 ) バッテリーアプリケーション回路バッテリー回路 ( 鉛蓄電池 ) バッテリーアプリケーション回路DCDC コンバータ電源回路DC モータ制御回路モーター制御回路

Concept Kit:PWM Buck Converter Average Model


Power Switches Filter & LoadPWM Controller (Voltage Mode Control)

VREF

VOUT

REF

PWM

1 / V p

-

+

U ?P W M _ C TR L

V P = 2 . 5V R E F = 1 . 2 3

D

U ?B U C K _ S W

L1 2

C

R lo a d

V o

E S R

Contents

• Concept of Simulation• Buck Converter Circuit• Averaged Buck Switch Model• Buck Regulator Design Workflow

1. Setting PWM Controller’s Parameters.

2. Programming Output Voltage: Rupper, Rlower

3. Inductor Selection: L

4. Capacitor Selection: C, ESR

5. Stabilizing the Converter (Example)

• Load Transient Response Simulation (Example)Appendix

A. Type 2 Compensation Calculation using Excel

B. Feedback Loop Compensators

C. Simulation Index



Power Switches

Averaged Buck Switch Model

Filter & Load

Parameter:• L• C• ESR• Rload

PWM Controller (Voltage Mode Control)

Parameter:• VP

• VREF

Models:

Block Diagram:

Concept of Simulation

VREF

VOUT

D

U ?B U C K _ S W

REF

PWM

1 / V p

-

+

U ?P W M _ C TR L

V P = 2 . 5V R E F = 1 . 2 3

L1 2

C

R lo a d

V o

E S R

L1 2

C

R lo a d

0

C o m p

C 2

R 2 C 1

F B

Type 2 Compensator

R u p p e r

R lo we r

0

d

V inD

U 2B U C K _ S W

REF

PWM

1 / V p

-

+

U 3P W M _ C TR L

V P = 2 . 5V R E F = 1 . 2 3

V o

E S R

Buck Converter Circuit


Filter & Load

PWM Controller

Power Switches

Averaged Buck Switch Model

• The Averaged Buck Switch Model represents relation between input and output of the switch that is controlled by duty cycle – d (value between 0 and 1).

• Transfer function of the model is

vout = d vin

• The current flow into the switch is

iin = d iout


D

U 2B U C K _ S W

vin

+

-

vout

+

-D

iin iout

Buck Regulator Design Workflow


Setting PWM Controller’s Parameters: VREF, VP1

Setting Output Voltage: Rupper, Rlower2

Inductor Selection: L3

Capacitor Selection: C, ESR4

Stabilizing the Converter: R2, C1, C2

• Step1: Open the loop with LoL=1kH and CoL=1kF then inject an AC signal to generate Bode plot. (always default)

• Step2: Set C1=1kF, C2=1fF, (always keep the default value) and R2= calculated value (Rupper//Rlower) as the initial values.

• Step3: Select a crossover frequency (about 10kHz or fc < fosc/4). Then complete the table.

• Step4: Read the Gain and Phase value at the crossover frequency (10kHz) from the Bode plot, Then put the values to the table

• Step5: Select the phase margin at the fc ( > 45 ). Then change the K value until it gives the satisfied phase margin, for this example K=6 is chosen for Phase margin = 46.

• Remark: If K-factor fail to gives the satisfied phase margin, Increase the output capacitor C then try Step1 to Step5 again.

Load Transient Response Simulation

5

6

Buck Regulator Design Workflow


1

2

3

4

5

L1 2

C

R lo a d

0

C o m p

C 2

R 2 C 1

F B

Type 2 Compensator

R u p p e r

R lo we r

0

d

V inD

U 2B U C K _ S W

REF

PWM

1 / V p

-

+

U 3P W M _ C TR L

V P = 2 . 5V R E F = 1 . 2 3

V o

E S R

• VREF, feedback reference voltage, value is given by the datasheet

• VP = (Error Amp. Gain vFB ) / d• vFB = vFBH – vFBL

• d = dMAX – dMIN

• Error Amp. Gain is 100 (approximated)

where

VP is the sawtooth peak voltage.

vFBH is maximum FB voltage where d = 0

vFBL is minimum FB voltage where d =1(100%)

dMAX is maximum duty cycle, e.g. d = 0(0%)

dMIN is minimum duty cycle, e.g. d =1(100%)

Setting PWM Controller’s Parameters


REF

PWM

1 / V p

-

+

U ?P W M _ C TR L

V P = 2 . 5V R E F = 1 . 2 3

vcomp

d

Error Amp.

FB

The PWM block is used to transfer the error voltage (between FB and REF) to be the duty cycle.

If vFBH and vFBL are not provided, the default value, VP=2.5 could be used.

1

TimeV(PWM)

V(osc) V(comp)0V

2.0V

3.0V

SEL>> VP

Duty cycle (d) is a value from 0 to 1

from

VP = (Error Amp. Gain vFB )/d

•Error Amp. Gain = 100 (approximated)

•from the graph on the left, vFB = 25mV (15m - (-10m))•d = 1 – 0 = 1

VP ≈ ( 100 25mV )/1

≈ 2.5V


If the VP ( sawtooth signal amplitude ) does not informed by the datasheet, It can be approximated from the characteristics below.

LM2575: Feedback Voltage vs. Duty Cycle

Setting PWM Controller’s Parameters (Example)

vFB = 25mV

d = 1 (100%)

dMIN dMAX

vFBH

vFBL

1

If vFBH and vFBL are not provided, the default value, VP=2.5 could be used.

• Use the following formula to select the resistor values.

• Rlower can be between 1k and 5k.

Example

Given: VOUT = 5V

VREF = 1.23

Rlower = 1k

then: Rupper = 3.065k

C o m p

C 2

R 2 C 1

Type 2 Compensator

F B

R u p p e r

R lo we r

0

d

REF

PWM

1 / V p

-

+

U 3P W M _ C TR L

V P = 2 . 5V R E F = 1 . 2 3

Error Amp.

Vo

Setting Output Voltage: Rupper, Rlower


lower

upperREFOUT

R

RVV 1

2

Inductor Selection: L


Inductor Value• The output inductor value is selected to set the

converter to work in CCM (Continuous Current Mode) or DCM (Discontinuous Current Mode).

• Calculated by

Where

• LCCM is the inductor that make the converter to work in CCM.

• VI,max is input maximum voltage

• RL,min is load resistance at the minimum output current ( IOUT,min )

• fosc is switching frequency

L1 2

C

R lo a d

V o

E S R

max,

min,max,

2 Iosc

LOUTICCM

Vf

RVVL

3

Inductor Selection: L (Example)


Inductor Value

from

Given:

• VI,max = 40V, VOUT = 5V

• IOUT,min = 0.2A

• RL,min = (VOUT / IOUT,min ) = 25

• fosc = 52kHz

Then:

• LCCM 210(uH),

• L = 330(uH) is selected

L1 2

C

R lo a d

V o

E S R

max,

min,max,

2 Iosc

LOUTICCM

Vf

RVVL

3

Capacitor Selection: C, ESR


Capacitor Value• The minimum allowable output capacitor value should

be determined by

Where

• VI, max is the maximum input voltage.

• L (H) is the inductance calculated from previous step ( ).

• In addition, the output ripple voltage due to the capacitor ESR must be considered as the following equation.

L1 2

C

R lo a d

V o

E S R

F)H(

785,7max,

LV

VC

OUT

I

RIPPLEL

RIPPLEO

I

VESR

,

,

4

3

Capacitor Selection: C, ESR (Example)


Capacitor ValueFrom

and

Given:

• VI, max = 40 V

• VOUT = 5 V

• L (H) = 330

Then:• C 188 (F)

In addition:• ESR 100m

L1 2

C

R lo a d

V o

E S R

RIPPLEL

RIPPLEO

I

VESR

,

,

4

F)H(

785,7max,

LV

VC

OUT

I

• Loop gain for this configuration is

L1 2

R lo a d

C

0

C o m p

C 2

R 2 C 1

Type 2 Compensator

F B

R u p p e r3 . 0 6 6 k

R lo we r1 . 0 k

0

d

V in1 2 V d c

D

U 2B U C K _ S W

REF

PWM

1 / V p

-

+

U 3P W M _ C TR L

V P = 2 . 5V R E F = 1 . 2 3

V o

E S R

• The purpose of the compensator G(s) is to tailor the converter loop gain (frequency response) to make it stable when operated in closed-loop conditions.


PWMGsGsHsT )()()( GPWM

G(s)

H(s)

Stabilizing the Converter5

Stabilizing the Converter (Example)


Specification:VOUT = 5VVIN = 7 ~ 40VILOAD = 0.2 ~ 1A

PWM Controller:VREF = 1.23VVP = 2.5VfOSC = 52kHz

Rlower = 1k,Rupper = 3.1k,L = 330uH, C = 330uF (ESR = 100m)

Task: • to find out the element of the

Type 2 compensator ( R2, C1, and C2 )

L3 3 0 u H

1 2

C3 3 0 u F

R lo a d5

0

0

C O L1 k F

L O L

1 k H

C 2

R 2 C 1

F B

R u p p e r3 . 1 k

Type 2 Compensator

R lo we r1 . 0 k

0

d

V 31 V a c0 V d c

V in1 2 V d c

D

U 2B U C K _ S W

REF

PWM

1 / V p

-

+

U 3P W M _ C TR L

V P = 2 . 5V R E F = 1 . 2 3

V o

E S R1 0 0 m

G(s)

e.g. Given values from National Semiconductor Corp. IC: LM2575

5

1

34

2

L3 3 0 u H

1 2

C3 3 0 u F

R lo a d5

0

0

C O L1 k F

L O L

1 k H

R 20 . 7 5 6 k

F B

R u p p e r3 . 1 k

Type 2 Compensator

R lo we r1 k

0

d

V 31 V a c0 V d c

V in1 2 V d c

D

U 2B U C K _ S W

REF

PWM

1 / V p

-

+

U 3P W M _ C TR L

V P = 2 . 5V R E F = 1 . 2 3

V o

E S R1 0 0 m

C 21 f

C 11 k


Step2 Set C1=1kF, C2=1fF, and R2=calculated value (Rupper//Rlower) as the initial values.

Step1 Open the loop with LoL=1kH and CoL=1kF then inject an AC signal to generate Bode plot.

The element of the Type 2 compensator ( R2, C1, and C2 ), that stabilize the converter, can be extracted by using Type 2 Compensator Calculator (Excel sheet) and open-loop simulation with the Average Switch Models (ac models).

Stabilizing the Converter (Example)5

C1=1kF is AC shorted, and C2 1fF is AC opened (or Error-Amp without compensator).


Type 2 Compensator Calculator

Switching frequency, fosc : 52.00 kHzCross-over frequency, fc (<fosc/4) : 10.00 kHzRupper : 3.1 kOhmRlower : 1 kOhmR2 (Rupper//Rlower) : 0.756 kOhm (automatically calculated)

PWMVref : 1.230 VVp (Approximate) : 2.5 V


Step3 Select a crossover frequency (about 10kHz or fc < fosc/4 ), for this example, 10kHz is selected. Then complete the table.

Calculated value of the Rupper//Rlower

values from 2

values from 1

5

Parameter extracted from simulationSet: R2=R1, C1=1k, C2=1fGain (PWM) at foc ( - or + ) : -44.211Phase (PWM) at foc : 65.068


Frequency

100Hz 1.0KHz 10KHz 100KHzP(v(d))

0d

90d

180d

SEL>>

(10.000K,65.068)

DB(v(d))

-80

-40

0

40

80

(10.000K,-44.211)

Step4 Read the Gain and Phase value at the crossover frequency (10kHz) from the Bode plot, Then put the values to the table.


Tip: To bring cursor to the fc = 10kHz type “ sfxv(10k) ” in Search Command.

Cursor Search

Gain: T(s) = H(s)GPWM

Phase at fc

5

K-factor (Choose K and from the table)K 6q -199 ° (automatically calculated)

Phase margin : 46 (automatically calculated)

R2 : 122.780 kOhm (automatically calculated)C1 : 0.778 nF (automatically calculated)C2 : 21.600 pF (automatically calculated)



Step5 Select the phase margin at fc (> 45 ). Then change the K value (start from K=2) until it gives the satisfied phase margin, for this example K=6 is chosen for Phase margin = 46.

As the result; R2, C1, and C2 are calculated.

K Factor enable the circuit designer to choose a loop cross-over frequency and phase margin, and then determine the necessary component values to achieve these results. A very big K value (e.g. K > 100) acts like no compensator (C1 is shorted and C2 is opened).

5

Remark: If K-factor fail to gives the satisfied phase margin, Increase the output capacitor C then try Step1 to Step5 again.

R 21 2 2 . 7 8 0 k

Type 2 Compensator

C 22 1 . 6 p

C 10 . 7 7 8 n

L3 3 0 u H

1 2

C3 3 0 u F

R lo a d5

0

0

C O L1 k F

L O L

1 k H

F B

R u p p e r3 . 1 k

R lo we r1 k

0

d

V 31 V a c0 V d c

V in1 2 V d c

D

U 2B U C K _ S W

REF

PWM

1 / V p

-

+

U 3P W M _ C TR L

V P = 2 . 5V R E F = 1 . 2 3

V o

E S R1 0 0 m



The element of the Type 2 compensator ( R2, C1, and C2 ) extraction can be completed by Type 2 Compensator Calculator (Excel sheet) with the converter average models (ac models) and open-loop simulation.

The calculated values of the type 2 elements are, R2=122.780k, C1=0.778nF, and C2=21.6pF.

*Analysis directives: .AC DEC 100 0.1 10MEG

5

Frequency

100Hz 1.0KHz 10KHz 100KHzP(v(d))

0d

90d

180d

(9.778K,45.930)

DB(v(d))

-40

0

40

80

-100SEL>>

(9.778K,0.000)

• Phase margin = 45.930 at the cross-over frequency - fc = 9.778kHz.



Tip: To bring cursor to the cross-over point (gain = 0dB) type “ sfle(0) ” in Search Command.

Cursor Search

Gain: T(s) = H(s) G(s)GPWM

Phase at fc

5

Gain and Phase responses after stabilizing

Load Transient Response Simulation (Example)


R 21 2 2 . 7 8 0 k

C 22 1 . 6 p

Type 2 Compensator

C 10 . 7 7 8 n

L o a d

V o

I 1

TD = 1 0 mTF = 2 5 u

P W = 0 . 4 3 mP E R = 1

I 1 = 0I 2 = 0 . 8

TR = 2 0 u

R lo a d2 5

0

F B

R u p p e r3 . 1 k

R lo we r1 k

0

d

V in2 0 V d c

D

U 2B U C K _ S W

REF

PWM

1 / V p

-

+

U 3P W M _ C TR L

V P = 2 . 5V R E F = 1 . 2 3

L3 3 0 u H

1 2

C3 3 0 u F

E S R1 0 0 m

The converter, that have been stabilized, are connected with step-load to perform load transient response simulation.

5V/2.5 = 0.2A step to 0.2+0.8=1.0A load

*Analysis directives: .TRAN 0 20ms 0 1u

Simulation Measurement


Output Voltage Change

Load Current

• The simulation results are compared with the measurement data (National Semiconductor Corp. IC LM2575 datasheet).

Time

9.9ms 10.1ms 10.3ms 10.5ms 10.7ms 10.9ms1 V(vo) 2 I(load)

4.4V

4.5V

4.6V

4.7V

4.8V

4.9V

5.0V

5.1V

5.2V1

0A

0.5A

1.0A

1.5A

2.0A

2.5A

3.0A

3.5A

4.0A2

>>

Load Transient Response Simulation (Example)

A. Type 2 Compensation Calculation using Excel

Switching frequency, fosc : 52.00 kHz Given spec, datasheetCross-over frequency, fc (<fosc/4) : 10.00 kHz Input the chosen value ( about 10kHz or < fosc/4 )Rupper : 3.1 kOhm Given spec, datasheet, or calculated Rlower : 1 kOhm Given spec, datasheet, or value: 1k-10k OhmR2 (Rupper//Rlower) : 0.756 kOhm (automatically calculated)

PWMVref : 1.230 V Given spec, datasheetVp (Approximate) : 2.5 V Given spec, or calculated, (or leave default 2.5V)

Parameter extracted from simulationSet: R2=R2, C1=1k, C2=1fGain (PWM) at foc ( - or + ) : -44.211 dB Read from simulation resultPhase (PWM) at foc : 65.068 ° Read from simulation result

K-factor (Choos K and q from the table)K 6 Input the chosen value (start from k=2)q -199 ° (automatically calculated)

Phase margin : 46 (automatically calculated) Target value > 45

R2 : 122.780 kOhm (automatically calculated)C1 : 0.778 nF (automatically calculated)C2 : 21.60 pF (automatically calculated)



B. Feedback Loop Compensators

Type 1 Compensator

C 1

V O U T

F B

R u p p e r

R lo we r

0

d

REF

PWM

1 / V p

-

+

P W M _ C TR L

Type1 Compensator Type2 Compensator Type2a Compensator

Type2b Compensator Type3 Compensator

Type2b Compensator

C 1

V O U T

F B

R u p p e r

R lo we r

0

d

REF

PWM

1 / V p

-

+

P W M _ C TR L

R 2

Type2a Compensator

C 1

V O U T

F B

R u p p e r

R lo we r

0

d

REF

PWM

1 / V p

-

+

P W M _ C TR L

R 2

Type3 Compensator

C 1

F B

R u p p e r

R lo we r

0

d

REF

PWM

1 / V p

-

+

P W M _ C TR L

C 2

R 2

C 3

R 3

V O U T

Type2 Compensator

C 1

F B

R u p p e r

R lo we r

0

d

REF

PWM

1 / V p

-

+

P W M _ C TR L

C 2

R 2

V O U T


Simulations Folder name

1. Stabilizing the

Converter....................................................

2. Load Transient Response..................................................

ac

stepload

Libraries :1. ..\bucksw.lib2. ..\pwm_ctr.lib

Tool :• Type 2 Compensator Calculator (Excel sheet)

C. Simulation Index

Unipolar Stepping Motor Drive Circuit

Contents1. Concept of Simulation

2. Unipolar Stepping Motor Drive Circuit

3. Unipolar Stepping Motor

4. Switches

5. Signal Generator

6. Hysteresis-Based Current Controller

7. Unipolar Stepping Motor Drive Circuit (Example)7.1 One-Phase Sequence Drive, IPHASE=0.5A, IRIPPLE=0.1A

7.2 Two-Phase Sequence Drive, IPHASE=0.5A, IRIPPLE=0.1A

7.3 Half-Phase Sequence Drive, IPHASE=0.5A, IRIPPLE=0.1A

8. Drive Circuit Efficiency



Unipolar Stepping Motor Drive Circuit

B

Bc om

A

/B

A c om

/A

U ?U N I -P O L A R _ S TE P _ M O TRL = 2 . 5 MR = 4 . 2


Driver Unit:(e.g. Hysteresis-Based Controller)

Parameter:• I_SET• HYS

Switches(e.g. FET, Diode)

Parameter:• Ron

Stepping Motor

Parameter:• L• R

Control Unit (e.g. Microcontroller)

Sequence:• One-Phase• Two-Phase• Half-Step

U ?1 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

U ?2 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

U ?H A L F -S TE PP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

B

Bc om

A

/B

A c om

/A

U ?U N I -P O L A R _ S TE P _ M O TRL = 2 . 5 MR = 4 . 2

Models:

Block Diagram:

D I O D ED 1

0

+

-

+

-

S 1

SR O N = 1 0 m

V C C

C t rl_ A A

1.Concept of Simulation

U 2

A N D

+

-

REF

-+

FB.

U 1

H Y S _ I -C TR L

I _ S E T = {I _ S E T}V H Y S = {V H Y S }

C t r l_ AF A

2.Unipolar Stepping Motor Drive Circuit


Signal generator Hysteresis Based Current Controller

Switches Unipolar Stepping Motor Supply Voltage

B

Bc om

A

/B

A c om

/A

U 1U N I -P O L A R _ S TE P _ M O TRL = 2 . 5 MR = 4 . 2

U 8

A N D

U 9

A N D

R 11 k

0

F B

D I O D ED 1

D I O D ED 2

D I O D ED 3

D I O D ED 4

PARAMETERS:I _ S E T = 0 . 5

V H Y S = 0 . 1

B

0

PARAMETERS:R O N = 1 0 m

0

U 1 01 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

0

0

U 6

A N D

F A

+

-

REF

-+

FB.

U 2

H Y S _ I -C TR L


/ F A

/ F B

V C C

+

-

REF

-+

FB.

U 3

H Y S _ I -C TR L


+

-

REF

-+

FB.

U 4

H Y S _ I -C TR L


/ B

/ A

+

-

+

-

S 4

SR O N = {R O N }

A

+

-

REF

-+

FB.

U 5

H Y S _ I -C TR L


C L K

+

-

+

-

S 1

SR O N = {R O N }

+

-

+

-

S 2

SR O N = {R O N }

+

-

+

-

S 3

SR O N = {R O N }

V C C

V C C V C C

0

V C C

V c c1 2

V C C

V C C

U 7

A N D

3.Unipolar Stepping Motor


• The electrical equivalent circuit of each phase consists of an inductance of the phase winding series with resistance.

• The inductance is ideal (without saturation characteristics and the mutual inductance between phases)

• The motor back EMF is set as zero to simplified the model parameters extraction.

B

Bc om

A

/B

A c om

/A


Input the inductance and resistance values (parameter: L, R) of the stepping motor, that are usually provided by the manufacturer datasheet, to generally model the phase winding.

4.Switches


• A near-ideal DIODE can be modeled by using spice primitive model (D), which parameter: N=0.01 RS=0.

• A near-ideal MOSFET can be modeled by using PSpice VSWITCH that is voltage controlled switch.

D I O D ED 1

0

+

-

+

-

S 1

SR O N = 1 0 m

V C C

C t rl_ A A

The parameter RON represents Rds(on) characteristics of MOSFET, that are usually provide by the manufacturer datasheet. The value could be about 10m to 10 ohm.

5.Signal Generator

The signal generators are used as a microcontroller capable of generating step pulses and direction signals for the driver.

There are 3 useful stepping sequences to control unipolar stepping motor


One-Phase (Wave Drive)• Consumes the least power. • Assures the accuracy regardless of the winding imbalance.

Two-Phase (Hi-Torque)• Energizes 2 phases at the same time. • Offers an improved torque-speed result and greater holding

torque.

U ?1 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

U ?2 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

U ?H A L F -S TE PP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

Half-Step• Doubles the stepping resolution of the motor. • Reduces motor resonance which could cause a motor to stall at a resonant frequency. • Please note that this sequence is 8 steps.

Input PPS (Pulse Per Second) as a clock pulse speed(frequency).

5.1 One-Phase Sequence


Time

0s 40ms 80msV(/FB)

0V

5.0V

SEL>>

V(FB)0V

2.5V

5.0VV(/FA)

0V

2.5V

5.0VV(FA)

0V

2.5V

5.0VV(CLK)

0V

2.5V

5.0V

ON

ON

ON

ON

Clock

Phase A

Phase /A

Phase B

Phase /B

1 Sequence

Time

0s 40ms 80msV(/FB)

0V

5.0V

SEL>>

V(FB)0V

2.5V

5.0VV(/FA)

0V

2.5V

5.0VV(FA)

0V

2.5V

5.0VV(CLK)

0V

2.5V

5.0V

5.2 Two-Phase Sequence


ON

ON

ON

ON

1 Sequence

Clock

Phase A

Phase /A

Phase B

Phase /BON

Time

0s 80ms 160msV(/FB)

0V

5.0V

SEL>>

V(FB)0V

2.5V

5.0VV(/FA)

0V

2.5V

5.0VV(FA)

0V

2.5V

5.0VV(CLK)

0V

2.0V

4.0V

5.3 Half-Step Sequence


ON

ON

ON

1 Sequence

Clock

Phase A

Phase /A

Phase B

Phase /BON

6.Hysteresis-Based Current Controller


• Controlled by the signal from the microcontroller.

• Generate the switch (MOSFET) drive signal by comparing the measured phase current with their references.

Input the reference value at the I_SET (e.g. I_SET=0.5A) to set the regulated current level. The hysteresis current value is set at the VHYS (e.g. VHYS=0.1A).

U 2

A N D

+

-

REF

-+

FB.

U 1

H Y S _ I -C TR L

I _ S E T = 0 . 5V H Y S = 0 . 1

C t r l_ AF A

B

Bc om

A

/B

A c om

/A


U 8

A N D

U 9

A N D

R 11 k

0

F B

D I O D ED 1

D I O D ED 2

D I O D ED 3

D I O D ED 4


V H Y S = 0 . 1

B

0


0

U 1 01 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

0

0

U 6

A N D

F A

+

-

REF

-+

FB.

U 2

H Y S _ I -C TR L


/ F A

/ F B

V C C

+

-

REF

-+

FB.

U 3

H Y S _ I -C TR L


+

-

REF

-+

FB.

U 4

H Y S _ I -C TR L


/ B

/ A

+

-

+

-

S 4

SR O N = {R O N }

A

+

-

REF

-+

FB.

U 5

H Y S _ I -C TR L


C L K

+

-

+

-

S 1

SR O N = {R O N }

+

-

+

-

S 2

SR O N = {R O N }

+

-

+

-

S 3

SR O N = {R O N }

V C C

V C C V C C

0

V C C

V c c1 2

V C C

V C C

U 7

A N D

7.1 One-Phase Sequence Drive, IPHASE=0.5A, IRIPPLE=0.1A


*Analysis directives: .TRAN 0 40ms 0 10u

One-Phase Step Sequence Generator (100 pps)

Time

0s 10ms 20ms 30ms 40ms1 V(/FB) 2 -I(U1:/B)

0V

2.5V

5.0V1

0A

0.5A

1.0A2

SEL>>SEL>>

1 V(FB) 2 -I(U1:B)0V

2.5V

5.0V1

0A

0.5A

1.0A2

>>

1 V(/FA) 2 -I(U1:/A)0V

2.5V

5.0V1

0A

0.5A

1.0A2

>>

1 V(FA) 2 -I(U1:A)0V

2.5V

5.0V1

0A

0.5A

1.0A2

>>

V(CLK)0V

2.5V

5.0V



Clock

Phase A Current

I_SET=0.5A

I_HYS=0.1A

Phase /A Current

Phase B Current

Phase /B Current

B

Bc om

A

/B

A c om

/A


U 8

A N D

U 9

A N D

R 11 k

0

F B

D I O D ED 1

D I O D ED 2

D I O D ED 3

D I O D ED 4


V H Y S = 0 . 1

B

0


0

0

0

U 6

A N D

F A

+

-

REF

-+

FB.

U 2

H Y S _ I -C TR L


/ F A

/ F B

V C C

+

-

REF

-+

FB.

U 3

H Y S _ I -C TR L


+

-

REF

-+

FB.

U 4

H Y S _ I -C TR L


/ B

/ A

+

-

+

-

S 4

SR O N = {R O N }

A

+

-

REF

-+

FB.

U 5

H Y S _ I -C TR L


C L K

+

-

+

-

S 1

SR O N = {R O N }

+

-

+

-

S 2

SR O N = {R O N }

+

-

+

-

S 3

SR O N = {R O N }

V C C

V C C V C C

0

V C C

V c c1 2

V C C

V C C

U 7

A N D

U 1 02 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B



*Analysis directives: .TRAN 0 40ms 0 10u SKIPBP .OPTIONS ITL4= 40

Two-Phase Step Sequence Generator (100 pps)

Time

0s 10ms 20ms 30ms 40ms1 V(/FB) 2 -I(U1:/B)

0V

2.5V

5.0V1

0A

0.5A

1.0A2

SEL>>SEL>>

1 V(FB) 2 -I(U1:B)0V

2.5V

5.0V1

0A

0.5A

1.0A2

>>

1 V(/FA) 2 -I(U1:/A)0V

2.5V

5.0V1

0A

0.5A

1.0A2

>>

1 V(FA) 2 -I(U1:A)0V

2.5V

5.0V1

0A

0.5A

1.0A2

>>

V(CLK)0V

2.5V

5.0V



Clock

Phase A Current

I_SET=0.5A

I_HYS=0.1A

Phase /A Current

Phase B Current

Phase /B Current

B

Bc om

A

/B

A c om

/A


U 8

A N D

U 9

A N D

R 11 k

0

F B

D I O D ED 1

D I O D ED 2

D I O D ED 3

D I O D ED 4


V H Y S = 0 . 1

B

0


0

0

0

U 6

A N D

F A

+

-

REF

-+

FB.

U 2

H Y S _ I -C TR L


/ F A

/ F B

V C C

+

-

REF

-+

FB.

U 3

H Y S _ I -C TR L


+

-

REF

-+

FB.

U 4

H Y S _ I -C TR L


/ B

/ A

+

-

+

-

S 4

SR O N = {R O N }

A

+

-

REF

-+

FB.

U 5

H Y S _ I -C TR L


C L K

+

-

+

-

S 1

SR O N = {R O N }

+

-

+

-

S 2

SR O N = {R O N }

+

-

+

-

S 3

SR O N = {R O N }

V C C

V C C V C C

0

V C C

V c c1 2

V C C

V C C

U 7

A N D

U 1 0H A L F -S TE PP P S = 1 0 0

C L KF A

/ F A

F B

/ F B




Half-Phase Step Sequence Generator (100 pps)

Time

0s 10ms 20ms 30ms 40ms 50ms 60ms 70ms 80ms1 V(/FB) 2 -I(U1:/B)

0V

2.5V

5.0V1

0A

0.5A

1.0A2

SEL>>SEL>>

1 V(FB) 2 -I(U1:B)0V

2.5V

5.0V1

0A

0.5A

1.0A2

>>

1 V(/FA) 2 -I(U1:/A)0V

2.5V

5.0V1

0A

0.5A

1.0A2

>>

1 V(FA) 2 -I(U1:A)0V

2.5V

5.0V1

0A

0.5A

1.0A2

>>

V(CLK)0V

2.5V

5.0V



Clock

Phase A Current

I_SET=0.5A

I_HYS=0.1A

Phase /A Current

Phase B Current

Phase /B Current

B

Bc om

A

/B

A c om

/A


U 8

A N D

U 9

A N D

R 11 k

0

F B

D I O D ED 1

D I O D ED 2

D I O D ED 3

D I O D ED 4


V H Y S = 0 . 1

B

0


0

U 1 01 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

0

0

U 6

A N D

F A

+

-

REF

-+

FB.

U 2

H Y S _ I -C TR L


/ F A

/ F B

V C C

+

-

REF

-+

FB.

U 3

H Y S _ I -C TR L


+

-

REF

-+

FB.

U 4

H Y S _ I -C TR L


/ B

/ A

+

-

+

-

S 4

SR O N = {R O N }

A

+

-

REF

-+

FB.

U 5

H Y S _ I -C TR L


C L K

+

-

+

-

S 1

SR O N = {R O N }

+

-

+

-

S 2

SR O N = {R O N }

+

-

+

-

S 3

SR O N = {R O N }

V C C

V C C V C C

0

V C C

V c c1 2

V C C

V C C

U 7

A N D

W

W

8.Drive Circuit Efficiency (%)


*Analysis directives: .TRAN 0 40ms 0ms 10u SKIPBP .STEP PARAM RON LIST 10m, 100m, 1 .OPTIONS ITL4= 40

Half-Phase Step Sequence Generator (100 pps)

Time

10ms 15ms 20ms 25ms 30ms 35ms 40ms100* AVG(W(U1))/(-AVG(W(Vcc)))

94

96

98

100



at switches Ron = 10m, (99.6%)


at switches Ron = 1, (95.9%)

Note: Add trace 100*AVG(W(U1))/(-AVG(W(Vcc))) for the Efficiency.


Bipolar Stepping Motor Drive Circuit

A

/A

B/B

U ?B I -P O L A R _ S TE P _ M O TRL = 1 0 mR = 8 . 4

Bipolar Stepping Motor Drive Circuit

Contents1. Concept of Simulation

2. Unipolar Stepping Motor Drive Circuit

3. Unipolar Stepping Motor

4. Switches

5. Signal Generator

6. Hysteresis-Based Current Controller

7. Unipolar Stepping Motor Drive Circuit (Example)7.1 One-Phase Sequence Drive, IPHASE=0.5A, IRIPPLE=0.1A



8. Drive Circuit Efficiency



Driver Unit:(e.g. Hysteresis-Based Controller)

Parameter:• I_SET• HYS

Switches(e.g. FET, Diode)

Parameter:• Ron

Stepping Motor

Parameter:• L• R

Control Unit (e.g. Microcontroller)

Sequence:• One-Phase• Two-Phase• Half-Step

U ?1 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

U ?2 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

U ?H A L F -S TE PP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

Models:

Block Diagram:

D I O D ED 1

0

+

-

+

-

S 1

SR O N = 1 0 m

V C C

C t rl_ A A

1.Concept of Simulation

U 2

A N D

+

-

REF

-+

FB.

U 1

H Y S _ I -C TR L


C t r l_ AF A A

/A

B/B


Signal generator Hysteresis Based Current Controller V C C

0

V c c1 2

A

/A

B/B

U 1B I -P O L A R _ S TE P _ M O TRL = 1 0 mR = 8 . 4

OU

I

O L

U 2

G D R V

+

-

+

-

S 7S

V C C

0D I O D ED 7

/ B U

+

-

+

-

S 8

SD I O D ED 8

/ B L

0

OU

I

O L

U 3

G D R V

OU

I

O L

U 5

G D R V

B

+

-

REF

-+

FB.

U 1 1

H Y S _ I -C TR L


/ F B

+

-

REF

-+

FB.

U 7

H Y S _ I -C TR L


F A

+

-

+

-

S 5

S

V C C

0D I O D ED 5

B U

+

-

+

-

S 6

SD I O D ED 6

B L

0


+

-

+

-

S 1

S

V C C


V H Y S = 0 . 1

0

+

-

REF

-+

FB.

U 1 3

H Y S _ I -C TR L


D I O D ED 1

A U

+

-

+

-

S 2

SD I O D ED 2

A L

A

0

+

-

REF

-+

FB.

U 9

H Y S _ I -C TR L


+

-

+

-

S 3S

V C C

0D I O D ED 3

/ A U

+

-

+

-

S 4

SD I O D ED 4

/ A L

0

U 8

A N D

U 1 0

A N D

U 1 2

A N D

U 1 4

A N D

/ F A

R 11 k

F B

C L K

0

OU

I

O L

U 4

G D R V

/ A

/ B

U 1 51 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

2.Unipolar Stepping Motor Drive Circuit


Bipolar Stepping Motor Supply VoltageH-Bridge Switches (Driver)

3.Bipolar Stepping Motor


• The electrical equivalent circuit of each phase consists of an inductance of the phase winding series with resistance.

• The inductance is ideal (without saturation characteristics and the mutual inductance between phases)

• The motor back EMF is set as zero to simplified the model parameters extraction.

Input the inductance and resistance values (parameter: L, R) of the stepping motor, that are usually provided by the manufacturer datasheet, to generally model the phase winding.

A

/A

B/B


4.Switches


• A near-ideal DIODE can be modeled by using spice primitive model (D), which parameter: N=0.01 RS=0.

• A near-ideal MOSFET can be modeled by using PSpice VSWITCH that is voltage controlled switch.

• MOSFETs are used as a H-Bridge.

The parameter RON represents Rds(on) characteristics of MOSFET, that are usually provide by the manufacturer datasheet. The value could be about 10m to 10 ohm.

OU

I

O L

U 2

G D R V

OU

I

O L

U 3

G D R V

+

-

+

-

S 1

S0

V C C

D I O D ED 1

A U

+

-

+

-

S 2

S

R O N = 1 0 m

D I O D ED 2

A L

0

+

-

+

-

S 3S

V C C

0D I O D ED 3

/ A U

+

-

+

-

S 4

SD I O D ED 4

/ A L

0

C t rl_ A

C t r l_ / A

A

/ A

5.Signal Generator

The signal generators are used as a microcontroller capable of generating step pulses and direction signals for the driver.

There are 3 useful stepping sequences to control unipolar stepping motor


One-Phase (Wave Drive)• Consumes the least power. • Assures the accuracy regardless of the winding imbalance.

Two-Phase (Hi-Torque)• Energizes 2 phases at the same time. • Offers an improved torque-speed result and greater holding

torque.

U ?1 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

U ?2 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

U ?H A L F -S TE PP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

Half-Step• Doubles the stepping resolution of the motor. • Reduces motor resonance which could cause a motor to stall at a resonant frequency. • Please note that this sequence is 8 steps.

Input PPS (Pulse Per Second) as a clock pulse speed(frequency).

5.1 One-Phase Sequence


Time

0s 40ms 80msV(/FB)

0V

5.0V

SEL>>

V(FB)0V

2.5V

5.0VV(/FA)

0V

2.5V

5.0VV(FA)

0V

2.5V

5.0VV(CLK)

0V

2.5V

5.0V

ON

ON

ON

ON

Clock

Phase A

Phase /A

Phase B

Phase /B

1 Sequence

Time

0s 40ms 80msV(/FB)

0V

5.0V

SEL>>

V(FB)0V

2.5V

5.0VV(/FA)

0V

2.5V

5.0VV(FA)

0V

2.5V

5.0VV(CLK)

0V

2.5V

5.0V

5.2 Two-Phase Sequence


ON

ON

ON

ON

1 Sequence

Clock

Phase A

Phase /A

Phase B

Phase /BON

Time

0s 80ms 160msV(/FB)

0V

5.0V

SEL>>

V(FB)0V

2.5V

5.0VV(/FA)

0V

2.5V

5.0VV(FA)

0V

2.5V

5.0VV(CLK)

0V

2.0V

4.0V

5.3 Half-Step Sequence


ON

ON

ON

1 Sequence

Clock

Phase A

Phase /A

Phase B

Phase /BON

6.Hysteresis-Based Current Controller


• Controlled by the signal from the microcontroller.

• Generate the switch (MOSFET) drive signal by comparing the measured phase current with their references.

Input the reference value at the I_SET (e.g. I_SET=0.5A) to set the regulated current level. The hysteresis current value is set at the VHYS (e.g. VHYS=0.1A).

U 2

A N D

+

-

REF

-+

FB.

U 1

H Y S _ I -C TR L

I _ S E T = 0 . 5V H Y S = 0 . 1

C t r l_ AF A

V C C

0

V c c1 2

A

/A

B/B


OU

I

O L

U 2

G D R V

+

-

+

-

S 7S

V C C

0D I O D ED 7

/ B U

+

-

+

-

S 8

SD I O D ED 8

/ B L

0

OU

I

O L

U 3

G D R V

OU

I

O L

U 5

G D R V

B

+

-

REF

-+

FB.

U 1 1

H Y S _ I -C TR L


/ F B

+

-

REF

-+

FB.

U 7

H Y S _ I -C TR L


F A

+

-

+

-

S 5

S

V C C

0D I O D ED 5

B U

+

-

+

-

S 6

SD I O D ED 6

0

B L


+

-

+

-

S 1

S0

V C C


V H Y S = 0 . 1

+

-

REF

-+

FB.

U 1 3

H Y S _ I -C TR L


D I O D ED 1

A U

+

-

+

-

S 2

SD I O D ED 2

A L

0

A

+

-

REF

-+

FB.

U 9

H Y S _ I -C TR L


+

-

+

-

S 3S

V C C

0D I O D ED 3

/ A U

+

-

+

-

S 4

SD I O D ED 4

/ A L

0

U 8

A N D

U 1 0

A N D

U 1 2

A N D

U 1 4

A N D

/ F A

R 11 k

C L K

0

F B

OU

I

O L

U 4

G D R V

/ A

/ B

U 1 51 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B





Time

0s 20ms 40ms 60ms 80ms1 V(/FB) 2 I(U1:/B)

0V

2.5V

5.0V1

0A

500mA2

SEL>>SEL>>

1 V(FB) 2 I(U1:B)0V

2.5V

5.0V1

0A

500mA2

>>

1 V(/FA) 2 I(U1:/A)0V

2.5V

5.0V1

0A

500mA2

>>

1 V(FA) 2 I(U1:A)0V

2.5V

5.0V1

0A

500mA2

>>

V(CLK)0V

2.5V

5.0V



Clock

Phase A Current

I_SET=0.5A

I_HYS=0.1A

Phase /A Current

Phase B Current

Phase /B Current




V C C

0

V c c1 2

A

/A

B/B


OU

I

O L

U 2

G D R V

+

-

+

-

S 7S

V C C

0D I O D ED 7

/ B U

+

-

+

-

S 8

SD I O D ED 8

/ B L

0

OU

I

O L

U 3

G D R V

OU

I

O L

U 5

G D R V

B

+

-

REF

-+

FB.

U 1 1

H Y S _ I -C TR L


/ F B

+

-

REF

-+

FB.

U 7

H Y S _ I -C TR L


F A

+

-

+

-

S 5

S

V C C

0D I O D ED 5

B U

+

-

+

-

S 6

SD I O D ED 6

0

B L


+

-

+

-

S 1

S0

V C C


V H Y S = 0 . 1

+

-

REF

-+

FB.

U 1 3

H Y S _ I -C TR L


D I O D ED 1

A U

+

-

+

-

S 2

SD I O D ED 2

A L

0

A

+

-

REF

-+

FB.

U 9

H Y S _ I -C TR L


+

-

+

-

S 3S

V C C

0D I O D ED 3

/ A U

+

-

+

-

S 4

SD I O D ED 4

/ A L

0

U 8

A N D

U 1 0

A N D

U 1 2

A N D

U 1 4

A N D

/ F A

R 11 k

C L K

0

F B

OU

I

O L

U 4

G D R V

/ A

/ B

U 1 52 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B


Time


0V

2.5V

5.0V1

0A

500mA2

SEL>>SEL>>

1 V(FB) 2 I(U1:B)0V

2.5V

5.0V1

0A

500mA2

>>

1 V(/FA) 2 I(U1:/A)0V

2.5V

5.0V1

0A

500mA2

>>

1 V(FA) 2 I(U1:A)0V

2.5V

5.0V1

0A

500mA2

>>

V(CLK)0V

2.5V

5.0V



Clock

Phase A Current

I_SET=0.5A

I_HYS=0.1A

Phase /A Current

Phase B Current

Phase /B Current

V C C

0

V c c1 2

A

/A

B/B


OU

I

O L

U 2

G D R V

+

-

+

-

S 7S

V C C

0D I O D ED 7

U 1 5H A L F -S TE PP P S = 1 0 0

C L KF A

/ F A

F B

/ F B

/ B U

+

-

+

-

S 8

SD I O D ED 8

/ B L

0

OU

I

O L

U 3

G D R V

OU

I

O L

U 5

G D R V

B

+

-

REF

-+

FB.

U 1 1

H Y S _ I -C TR L


/ F B

+

-

REF

-+

FB.

U 7

H Y S _ I -C TR L


F A

+

-

+

-

S 5

S

V C C

0D I O D ED 5

B U

+

-

+

-

S 6

SD I O D ED 6

0

B L


+

-

+

-

S 1

S0

V C C


V H Y S = 0 . 1

+

-

REF

-+

FB.

U 1 3

H Y S _ I -C TR L


D I O D ED 1

A U

+

-

+

-

S 2

SD I O D ED 2

A L

0

A

+

-

REF

-+

FB.

U 9

H Y S _ I -C TR L


+

-

+

-

S 3S

V C C

0D I O D ED 3

/ A U

+

-

+

-

S 4

SD I O D ED 4

/ A L

0

U 8

A N D

U 1 0

A N D

U 1 2

A N D

U 1 4

A N D

/ F A

R 11 k

C L K

0

F B

OU

I

O L

U 4

G D R V

/ A

/ B





Time


0V

2.5V

5.0V1

0A

500mA2

SEL>>SEL>>

1 V(FB) 2 I(U1:B)0V

2.5V

5.0V1

0A

500mA2

>>

1 V(/FA) 2 I(U1:/A)0V

2.5V

5.0V1

0A

500mA2

>>

1 V(FA) 2 I(U1:A)0V

2.5V

5.0V1

0A

500mA2

>>

V(CLK)0V

2.5V

5.0V



Clock

Phase A Current

I_SET=0.5A

I_HYS=0.1A

Phase /A Current

Phase B Current

Phase /B Current

V C C

0

V c c1 2

A

/A

B/B


OU

I

O L

U 2

G D R V

+

-

+

-

S 7S

V C C

0D I O D ED 7

/ B U

+

-

+

-

S 8

SD I O D ED 8

/ B L

0

OU

I

O L

U 3

G D R V

OU

I

O L

U 5

G D R V

B

+

-

REF

-+

FB.

U 1 1

H Y S _ I -C TR L


/ F B

+

-

REF

-+

FB.

U 7

H Y S _ I -C TR L


F A

+

-

+

-

S 5

S

V C C

0D I O D ED 5

B U

+

-

+

-

S 6

SD I O D ED 6

0

B L


+

-

+

-

S 1

S0

V C C


V H Y S = 0 . 1

+

-

REF

-+

FB.

U 1 3

H Y S _ I -C TR L


D I O D ED 1

A U

+

-

+

-

S 2

SD I O D ED 2

A L

0

A

+

-

REF

-+

FB.

U 9

H Y S _ I -C TR L


+

-

+

-

S 3S

V C C

0D I O D ED 3

/ A U

+

-

+

-

S 4

SD I O D ED 4

/ A L

0

U 8

A N D

U 1 0

A N D

U 1 2

A N D

U 1 4

A N D

/ F A

R 11 k

C L K

0

F B

OU

I

O L

U 4

G D R V

/ A

/ B

U 1 52 -P H A S EP P S = 1 0 0

C L KF A

/ F A

F B

/ F B



*Analysis directives: .TRAN 0 80ms 0 10u SKIPBP .STEP PARAM RON LIST 10m, 100m, 1 .OPTIONS ITL4= 40


Time

10ms 20ms 30ms 40ms 50ms 60ms 70ms 80ms100*AVG(W(U1))/(-AVG(W(Vcc)))

85

90

95

100





at switches Ron = 1, (86%)

Note: Add trace 100*AVG(W(U1))/(-AVG(W(Vcc))) for the Efficiency.

Bee Technologies Group

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