D RI RI V E 9 1 3 - 0 3 v e r B 2 6/ 0 9/ 2 00 7 Swit Sw itze zerl rlan and d
ETEL S.A.
Tel. Te l.+4 +41 1 (0 (0)) 32 86 862 2 01 00
http ht tp:/ ://w /www ww.e .ete tel. l.ch ch
B r a k e r e si si st o r A motor coupled with a load has a certain amount amoun t of energy. This Thi s energy is mainly kinetic when the load is moving or rotating. When the system brakes, the energy must be either stored or dissipated. It may be gravitational potential energy in addition to kinetic energy if the load movement is not horizontal (in case of a linear motor), or could be stored in a spring or in any outer system. In this case, the energy must be either stored or dissipated when the system is braking, and sometimes also when the system is at constant speed in descent direction. The DSB2P, DSC2P, DSCDP and DSC2V position controllers as well as the DSO-PWR power supply (used for the DSB2P rack format) and the DSO-PWS power supply (used for the DSC2P and DSCDP rack format) contain capacitors that are capable of storing a certain amount of energy. If the energy is too big, then a brake resistance is needed.
E T In this application note, we consider three phase motors. O NHo How w big is the energy energy st ored in my sys ystt em? LIn a standard direct drive application, the energy balance can be written like this: A E = ( E + E ) − ( E + E ) Equati on 1 C I Where: E = Total energy of motor/load minus the system losses [J] E = Kinetic energy of motor/load [J] N E = Gravitational potential energy of motor/load [J] E = Energy lost in the motor copper (Ohm losses) [J] H E = Energy lost by friction [J] C1. Torque motor case, standard configuration ( E term equal to zero): E T E = 1 ⋅ ( J + J )⋅ − 3 ⋅ I ⋅ ⎛ ⎜ R ⎞⎟ ⋅ t − t ⋅ ⋅T Equat ion 1a M
K
P
Co
F
1 4 24 3
1 4 24 3
System energy
System losses
M K K
P
Co Co
F F
P
2
M
2
M
L
ω M
144 4 244 4 3
Kinetic energy
Where:
2 M
d
M
⎝
2
d
⎠
144 244 3
Copper losses
ω M
2
F
14 24 3
Friction losses
2
J M = Rotor inertia [kgm ] 2 J L = Load inertia [kgm ] ω M
I M R M t d d T F F
= = = = =
Motor speed before deceleration [rad/s] Motor current during deceleration [A RMS /phase] RMS Motor resistance [Ω] terminal to terminal Time to decelerate [s] Friction torque [Nm]
Note: A rotary axis may have in addition: • Gravitational potential energy (in case of non-direct drive, if the load is non horizontal). • Spring stored energy. • …
2. Linear motor case with gravitational potential energy: E M
t ⋅ v R ⎞ 1 2 = ⋅ (m M + m L ) ⋅ v M + (m M + m L ) ⋅ g ⋅ (hinitial − h final ) − 3 ⋅ I M 2 ⋅ ⎛ ⎜ M ⎟ ⋅ t d − d M ⋅ F F 2 2 ⎝ 2 ⎠ Gravitational potential energy 144 4 244 4 3
144 4 4 4 244 4 4 4 3
144 244 3
Kinetic energy
Where:
Copper losses
Equ . 1b
14 24 3
Friction losses
m M = Motor mass [kg] moving part of motor only m L = Load mass [kg] v M = Motor speed before deceleration [m/s] = Gravitational acceleration [m/s2] g hinitial = Initial load altitude [m] h final = Final load altitude [m] I M = Motor current during deceleration [A RMS /phase] RMS R M = Motor resistance [Ω] terminal to terminal = Time to decelerate [s] t d d = Friction force [N] F F F
Note: For a constant speed speed syst syst em (like a long stroke conveyor for example), all the terms of the equation 1b have the same meaning except for: v M = Motor speed (constant) during the travel [m/s] travel at constant speed [A RMS /phase] I M = Motor current during travel RMS = Time to travel [s] t d d
I s a regenerat ive resist resist ance neede needed? d? If the condition here after (equation 2) is 2) is t r u e , then a brake resistance is needed . For a n axes system plugged on the same power supply: (All negative E M terms are set to zero in order to have the worst case) n
∑ E
Mj
1
>
2 2 ) ⋅ C ⋅ (U MAX − U Nom
2 144 4 244 4 3
j =1
Equati on 2
Maximal energy storable in the capacitors
Where:
E M C U MAX U Nom
= = = =
Total energy of motor/load minus the system losses [J] Total capacitance seen from the BUS [F] (see table 1) Maximal allowed BUS voltage [V] (see table 1) Nominal BUS voltage [V] (see table 1)
How t o determ ine th e resis resistt anc ance e value value For an n axes system plugged on the same power supply: 2
R MAX =
U MAX n
∑U
Bj
j =1
⋅ I Mj ⋅
Equati on 3 3
Where:
R MAX U MAX U B I M
= Brake resistance maximal value [ Ω] = Maximal allowed BUS voltage [V] (see table 1) = Motor back EMF less motor losses [V] (see here after) = Deceleration current in motor [A RMS /phase] RMS
1. Torque motor case: U B
R ⎞ = K u ⋅ ω M − I M ⋅ ⎛ ⎜ M ⎟ ⋅ ⎝ 2 ⎠ Back EMF
3
Equat ion 3a
1 4 2 4 3
14 4 244 3
Phase voltage
Where:
K u = Back EMF constant [V/(rad/s)] terminal to terminal
2. Linear motor case: U B
R ⎞ = K u ⋅ v M − I M ⋅ ⎛ ⎜ M ⎟ ⋅ ⎝ 2 ⎠ Back EMF
3
Equat ion 3b
123
14 4 244 3
Phase voltage
Where:
K u = Back EMF constant [V/(m/s)] terminal to terminal
How Ho w t o determ ine t he resis resistt anc ance e diss dissipated pow er For an n axes system plugged on the same power supply: (All negative E M are set to zero) n
∑ E
Mj
P AV =
j =1
1 2 2 − ⎛ − U HYS ) ⎞⎟ ⎜ ⋅ C ⋅ (U MAX ⎝ 2 ⎠
t CYCLE
Equati on 4
Where: P AV = Average power to be dissipated by the brake resistance [W] U HYS = Hysteresis point of power supply [V] (see table 1) t CYCLE CYCLE = Longest (time between two consecutive decelerations) of the n axis system [s] When the time between two consecutive decelerations becomes very large, the average power is not a meaningful number. In this case, the peak power is the main concerned: PPK =
2 V MAX
R REGEN
Where: PPK R REGEN
Equat ion 5
= Peak power dissipated by the regenerative resistance [W] = Brake resistance value [Ω]
ETEL elect elect ron ics specifi specifi cati ons ( t able 1)
Housed version DSB2P Rack version DSB2P Power supply (DSO-PWR) Housed version DSC2P / DSCDP Rack version DSC2P / DSCDP Power supply (DSO-PWS) DSC2V
C
Umax
R internal internal
Unom
Uhys
2100μF
385VDC
3k3Ω, 50W
120-340VDC
365VDC
300μF
385VDC
-
24-340VDC
365VDC
2250μF
385VDC
3k3Ω, 50W
120-340VDC
365VDC
660μF
450VDC
40k Ω, 12W
120-400VDC
420VDC
165μF
450VDC
100k Ω, 6W
24-400VDC
420VDC
990μF
450VDC
22k Ω, 15W
120-400VDC
420VDC
450μF
750VDC
47k Ω, 12W
400-600VDC
650VDC
Note: In the case of a rack case including one power supply (DSO-PWR) and n DSB2P, C is calculated as follows: C [ μ F ] = 2250 + n ⋅ 300 Equati on 6 Note: In the case of a rack case including one power supply (DSO-PWS) and n DSC2P or DSCDP, C is calculated as follows: C [ μ F ] = 990 + n ⋅ 165 Equati on 7 Note: In the case of a rack case including one power supply (DSO-PWS) and n DSC2P or DSCDP, R internal internal is calculated as follows: R int[Ω] =
1
⎛ 1 + n ⎞⎟ ⎜ ⎝ 22k 100k ⎠⎟
Equat ion 8
