This document describes a simplified SPICE model for simulating a DC motor. It includes 10 sections that describe: 1) the benefits of the model, 2) the model features, 3) how to set the model parameters, 4) an example motor specification, 5) simulating start up at normal load, 6) simulating start up at half load, 7) how the inductance parameter affects the model, 8) an application example, 9) how the inductance parameter affects the model, and 10) how the resistance parameter affects the model. The document provides information to help users set up and run simulations of a DC motor using the simplified SPICE model.

1. D.C. Motor
Simplified SPICE Model
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2. Contents
1. Benefit of the Model
2. Model Feature
3. Parameter Settings
4. D.C. Motor Specification (Example)
5. Motor Start Up Simulation at Normal Load
6. Motor Start Up Simulation at Half of Normal Load
7. Lj of the Motor Model (1/2)
8. Application Example
9. Winding Characteristic Parameters: Lm
10.Winding Characteristic Parameters: Rm
Simulation Index
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3. 1. Benefit of the Model
• The model enables circuit designer to use D.C. Motor as load in
their design which include: Back EMF, Torque(Nm) and Speed
(rpm) characteristics.
• The model can be easily adjusted to your own D.C. Motor
specifications by editing a few parameters that are provided in the
spec-sheet.
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4. 2. Model Feature
1 H1
TRQ
+
- 1
H
RM L1
{Rm} 100m
0
LM
2 E_EMF
{Lm}
N4 EMF
IN+ OUT+
OUT+ IN+ IN- OUT-
H2 Rdmy
OUT- IN- EVALUE
+
- GRB
2 H GVALUE
0 0
Equivalent circuit of the D.C. Motor model
• This D.C. Motor Simplified SPICE Model is for users who require the model
of D.C. Motor as a part of their system.
• Perform electrical (voltage and current) and mechanical (speed and torque)
characteristics at current load (Ampere) conditions.
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5. 3. Parameter Settings
D.C. Motor model and Parameters with Default Value Model Parameters:
If there is no measurement data, the default value will be used:
Rm: motor winding resistance []
Lm: motor winding inductance [H]
Data is given by D.C. motor spec-sheet:
V_norm: normal voltage [V]
mNm: normal load [mNm]
kRPM_norm: speed at normal load [kr/min]
I_norm: current at normal load [A]
Load Condition:
IL: load current [A]
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6. 4. D.C. Motor Specification (Example)
D.C. Motor Specification
Parameters are input
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7. 5. Motor Start Up Simulation at Normal Load (1/2)
Simulation Circuit and Setting
Input the Supply No
Load Voltage* and
Series Resistance
Simplified D.C. Motor with
RS-540SH Spec.
Current Sensing
Load Condition IL=I_norm
*No Load Voltage is adjusted until the D.C. motor voltage (VM) equals to the normal voltage (7.2V).
*Analysis directives:
.TRAN 0 400m 0 0.1m
.PROBE V(*) I(*) W(alias(*)) D(alias(*)) NOISE(alias(*))
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8. 5. Motor Start Up Simulation at Normal Load (2/2)
Torque Load= 19.6mNm
D.C. Motor Speed = 14.4krpm
D.C. Motor Voltage = 7.2V
D.C. Motor Current = 6.1A
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9. 6. Motor Start Up Simulation at Half of Normal Load (1/2)
Simulation Circuit and Setting
Input the Supply No
Load Voltage* and
Series Resistance
Simplified D.C. Motor with
RS-540SH Spec.
Current Sensing
Load Condition IL=I_norm
*No Load Voltage is adjusted until the D.C. motor voltage (VM) equals to the normal voltage (7.2V).
*Analysis directives:
.TRAN 0 400m 0 0.1m
.PROBE V(*) I(*) W(alias(*)) D(alias(*)) NOISE(alias(*))
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10. 6. Motor Start Up Simulation at Half of Normal Load (2/2)
Torque Load= 9.8mNm
D.C. Motor Speed = 18.4krpm
D.C. Motor Voltage = 8.725V
D.C. Motor Current = 3.05A
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11. 7. Lj of the Motor Model (1/2)
Simulation Circuit and Setting
Input the Lj value or input
“{Lj}” for parametric sweep
*Analysis directives:
.TRAN 0 400m 0 0.1m
.STEP PARAM Lj LIST 100m, 200m
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12. 7. Lj of the Motor Model (2/2)
The Motor Start-up Waveform is
changed by the Lj values.
Lj=200m
Torque Load= 19.6mNm
Lj=100m
D.C. Motor Speed = 14.4krpm
Lj=100m
Lj=200m
Lj=100m D.C. Motor Voltage = 7.2V
Lj=200m
Lj=200m
D.C. Motor Current = 6.1A
Lj=100m
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13. 8. Application Example (1/3)
Simulation Circuit and Setting
VDD VCC VCC
Simplified D.C. Motor with
RS-380PH Spec at No load.
Vdd Vcc
15V 15V
U2
SMPL_DC_MOTOR
Rm = 0.576
0 0 D2 + Lm = 165u
D4001 - V_norm = 7.2
mNm = 9.8
kRPM_norm = 14.2
I_norm = 2.9
U1
IL = 0.6
R1 NC VCC VDD
1u
A NC RG No load IL=0.6
120 D3
V1 = 0 V1 K VO DGT10J321_s
V2 = 1.8 U3
TD = 0 NC GND GT10J321
TR = 10n
TF = 10n
PW = 199.99u TLP350
PER = 400u
0 0 0 0
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14. 8. Application Example (2/3)
Measurement Simulation
14A 20V
1 2
12A 10V
10A 0V
Motor Voltage (10V/Div)
8A -10V
6A -20V
Motor Current (2A/Div)
4A -30V
2A -40V
0A -50V
>>
-2A -60V
-100ms 0s 100ms 300ms 500ms 700ms 900ms
1 I(U2:1) 2 V(U2:1,U2:2)
Time
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15. 8. Application Example (3/3)
Measurement Simulation
14A 50V 20V
1 2 3
12A 40V 10V
10A 30V 0V
IGBT: VGE IGBT: VGE (10V/Div)
8A 20V -10V
6A 10V -20V
4A 0V -30V
IGBT: VCE IGBT: VCE (10V/Div)
2A -10V -40V
0A -20V -50V
IGBT: IC >> IGBT: IC (2A/Div)
-2A -30V -60V
898.0ms 898.4ms 898.8ms 899.2ms 899.6ms
1 I(U3:C) 2 V(U3:C) 3 V(U3:G)
Time
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16. 9. Winding Characteristic Parameters: Lm
Winding Characteristic:2 Lm
1
14A 50V
3
20V
U2
SMPL_DC_MOTOR
12A 40V 10V
Rm = 0.576
+ Lm = 165u
- V_norm = 7.2 Motor Spec.
10A 30V 0V
mNm = 9.8
kRPM_norm = 14.2 Lm=100u
8A 20V -10V
I_norm = 2.9
Lm=165u
IL = 0.6 6A 10V -20V
Load Condition
4A 0V -30V
Lm=100u
The Motor Current Waveform is 2A -10V -40V
changed by the Lm values. Lm=165u
0A -20V -50V
>>
-2A -30V -60V
898.0ms 898.4ms 898.8ms 899.2ms 899.6ms
1 I(U3:C) 2 V(U3:C) 3 V(U3:G)
Time
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17. 10. Winding Characteristic Parameters: Rm
Winding Characteristic: Rm
U2
SMPL_DC_MOTOR
Rm = 0.576
+ Lm = 165u
- V_norm = 7.2 Motor Spec. Rm=0.1
mNm = 9.8
kRPM_norm = 14.2
I_norm = 2.9
IL = 0.6
Load Condition
Rm=0.576
The Motor Start-up is Current
changed by the Rm values.
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18. Simulation Index
Simulations Folder name
1. Motor Start Up Simulation at Normal Load................... Normal
2. Motor Start Up Simulation at Haft of Normal Load........ Half
3. Lj of the Motor Model.................................................... Lj
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