Editing Overvoltage and undervoltage faults
Your changes will be displayed to readers once an authorized user accepts them. (help) |
Warning: You are not logged in. Your IP address will be publicly visible if you make any edits. If you log in or create an account, your edits will be attributed to your username, along with other benefits.
The edit can be undone.
Please check the comparison below to verify that this is what you want to do, and then save the changes below to finish undoing the edit.
Latest revision | Your text | ||
Line 17: | Line 17: | ||
[[File:Regenres.jpg|thumb|250 Watt 82 ohm regenerative resistor suitable for Argon drive and sufficient for large machines]]The needed regenerative resistor value can be calculated by equation: | [[File:Regenres.jpg|thumb|250 Watt 82 ohm regenerative resistor suitable for Argon drive and sufficient for large machines]]The needed regenerative resistor value can be calculated by equation: | ||
− | <math>R_{regen}=\frac{U_{DCBusVoltage}}{I_{PeakMotorCurrent}}</math> | + | <math>R_{regen}=\frac{U_{DCBusVoltage}}{I_{PeakMotorCurrent}}</math> |
I.e. if supply voltage is 48VDC and peak current is 10A, then a resistor of 4.8 ohms may be needed to consume all current that is returning from the motor. However, in most practical cases the regenerative current is less than the motor peak current, which allows using higher resistance thus reducing risk of overloading the MOSFET switch operating the resistor. It is recommended to experiment with higher resistor values first, and gradually move to lower resistances if problem persists. | I.e. if supply voltage is 48VDC and peak current is 10A, then a resistor of 4.8 ohms may be needed to consume all current that is returning from the motor. However, in most practical cases the regenerative current is less than the motor peak current, which allows using higher resistance thus reducing risk of overloading the MOSFET switch operating the resistor. It is recommended to experiment with higher resistor values first, and gradually move to lower resistances if problem persists. | ||
Line 53: | Line 53: | ||
For short current surges/spikes, a capacitor added to HV DC bus might provide a solution for filtering out the spikes. Capacitor can be sized by equation: | For short current surges/spikes, a capacitor added to HV DC bus might provide a solution for filtering out the spikes. Capacitor can be sized by equation: | ||
− | <math>C_{filter}=t_{duration}\frac{I_{PeakMotorCurrent}}{U_{MaxVoltageChange}}</math> | + | <math>C_{filter}=t_{duration}\frac{I_{PeakMotorCurrent}}{U_{MaxVoltageChange}}</math> |
I.e. if current is 20A, surge duration 0.005 seconds and maximum allowed voltage change (increase or drop) during that surge/spike is 10 VDC, then capacitance becomes <math>C_{filter}=0.005s\frac{20A}{10V}=0.01F=10000\mu F</math>. | I.e. if current is 20A, surge duration 0.005 seconds and maximum allowed voltage change (increase or drop) during that surge/spike is 10 VDC, then capacitance becomes <math>C_{filter}=0.005s\frac{20A}{10V}=0.01F=10000\mu F</math>. | ||
Line 60: | Line 60: | ||
If the duration is unknown, and we're dealing with fast reversing torque setpoint on a relatively large motor, the capacitor size may be calculated based on the stored energy inside motor inductance: | If the duration is unknown, and we're dealing with fast reversing torque setpoint on a relatively large motor, the capacitor size may be calculated based on the stored energy inside motor inductance: | ||
− | <math>C_{filter}=scaler*L_{MotorInductance}*(\frac{I_{PeakMotorCurrent}}{U_{MaxVoltageChange}})^2</math> | + | <math>C_{filter}=scaler*L_{MotorInductance}*(\frac{I_{PeakMotorCurrent}}{U_{MaxVoltageChange}})^2</math> |
I.e. if current is 20A, motor inductance 5 mH (0.005 H) and maximum allowed voltage change in capacitor, and we choose scaler 0.5, then capacitance becomes <math>C_{filter}=0.5*0.005H*(\frac{20A}{10V})^2=0.01F=10000\mu F</math>. | I.e. if current is 20A, motor inductance 5 mH (0.005 H) and maximum allowed voltage change in capacitor, and we choose scaler 0.5, then capacitance becomes <math>C_{filter}=0.5*0.005H*(\frac{20A}{10V})^2=0.01F=10000\mu F</math>. | ||
Line 70: | Line 70: | ||
= See also = | = See also = | ||
* [[Configuring drive voltage limits FUV and FOV]] | * [[Configuring drive voltage limits FUV and FOV]] | ||
− | |||
− | |||
− | |||
− | |||
− |