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Mineral Resources
Solution for overcurrent problem in rope saw machine applications
Mineral Resources View volume: 26

1. The frequency inverter model:

YD3045 T4 N

2. Problems Description:

a. Report an overcurrent(OC) during the operation.

b. Report an overcurrent(OC) at the moment of startup.

3. System Illustration:

http://www.chengjinxi.cn/en/Data/upload/20250329/3a8d88b95d6c0d0a63417246ad39be08.png

4. The basic principle of the system

a. YD3045T4N+AC Motor(45kW/90A) spindle drive, the disc drives the steel rope (red line) ① to cut the rock.

b. 101 frequency inverter + the small motor perfroms translational driving and controls the tension through the PID controller②. When the steel rope is too loose, the vehicle will go back along the track (to left) to maintain the tension.

c. At the moment of startup, the frequency command is 10Hz and the startup current is about 90-130A, The value fluctuates according to the rope's status.

d. During automatic operation, the frequency command is 35Hz, the operating current fluctuates between 80~120A, ( PID control the tension of steel rope and drive the vehicle in translation).

5. On-site problem analysis 1:

a. The fault-record U3 shows that the OC was triggered 4 times.

b. The last fault record U2 shows that the frequency command is 10Hz, the operating frequency is 6.43Hz, the output current is 156.8A. ( triggered OC at the moment of startup)

c. When it triggered overcurrent (OC) alarm during startup, the steel rope was suddenly pulled, which lead a violent vibration of the vehicle .

d. Found the frequency inverter was set to “free stop”. Under this mode, the frequency inverter didn’t give any output. When the steel rope and disc out of control, it may block the rock hole.

This will cause the steel rope stuck and the motor stall during the next startup, resulting in a large starting current and trigger the over current (OC) alarm.

e. Regarding the parameter set up, change the stop mode from “free running “ to “deceleration stop”. Set deceleration time as 10 second.

6. On-site problem analysis 2:

a. During the automatic operation at 35Hz, the current occasionally rises from 80A to 140A and trigge an over-current alarm.

b. It’s analyzed that this might be too fast speed of the vehicle to cause during the PID controlling process.(During PID tension control, the vehicle moves backwards too fast) ( It reflects the insufficient “overload capacity” of the frequency inverter.)

c. The parameters of the frequency inverter were adjusted as below( the system was without PG vector control ):

①. Accelerate/stall prevention: from 150% down to 100%.

②. Torque limitation: each four-quadrants from 200% down to 120%.

d. For the PID controller, decrease the P gain to reduce the backwards speed and instantaneous stiffness.

7. After the parameter adjustment, the frequency inverter operates in normal. The current has decreased from 80 - 120A to 50 - 80A.The start current can be controlled within 100A.

After two days cutting test, it has been confirmed that the inverter can run normally and meet the customer's requirement. (it was concern that a reduction in torque would affect usability.)


8. Technical data:

a. The model of heavy-duty frequency inverter: 150% for 60 seconds and 180% for 1 second. Which means:
The model of heavy-duty frequency inverter: 150% for 60 seconds and 180% for 1 second. Which means:

96A for continuous operation.

144A for 60 seconds.

172A for 1 second.

Summary: Frequency inverters, especially the heavy - duty, must meet the requirement of twice the rated current. it is reasonable for the over - current (OC) alarm to be triggered at 96 * 2 = 192A.

b. Calculation of torque limitation:

  • Rated current of motor: 96.0A (E2 - 01); set up by the customer.
  • unload current of motor: 33.6A (E2 - 03); result of self - learning.
  • Rated current of IGBT: (200A/1.414)*0.95 = 134A; the factor 0.95 for ripple, and the customer set the carrier wave at 3Khz.

1. Maximum torque of the frequency inverter = √(1342 - 33.62) = 129 N·m

2. Rated torque of the motor = √(962 - 33.62) = 90 N·m

129 / 90 = 143%, which means the maximum torque limit of L7 - 01/2/3/4 can be set to 143%.

Considering the margin, it was set to 120%.

c. Acceleration stall prevention limit
The default value is 150%, which means 45kW motor, when the current reaches the rated current multiplied by 1.5 (i.e., 96 * 1.5 = 144A), the frequency will limit the acceleration.
As a safety measure, we set it to 100%. during the acceleration process, when the current reaches the rated current of 96A, the frequency will limit the acceleration.

d . The parameters of the frequency inverter have been changed as follows:

No. Parameters Original Setttings Later settings Instructions
1 C2-02 20 10 Shutdown / deceleration time
2 B1-03 1 0 Change from free stop to deceleration stop
3 L3-02 150% 100% Acceleration prevention level
4 L7-01 200% 120% Forward torque limit
5 L7-02 200% 120% Reverse torque limit
6 L7-03 200% 120% Forward rotation regeneration limit
7 L7-04 200% 120% Reverse regeneration limit
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