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Overvoltage and undervoltage faults

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Drive faulting due to voltage fluctuations in HV DC bus are commonly experienced with servo systems. These faults occur when drive measures a HV DC bus supply voltage that is not within the range defined by Under voltage fault thresholdFUV and Over voltage fault thresholdFOV parameters. The deviation of voltage may be impossible to notice with multimeter as length of these voltage surges can be in millisecond range.

Overvoltage faults[edit | edit source]

Servo drive attached to a motor can act two ways: energy supply and energy consumer. The energy consumer behavior occurs during decelerations and during fast torque reversals, and this causes current flow from motor to drive power supply capacitors. If the generated energy is not absorbed anywhere, the voltage of HV DC bus capacitors will rise above overvoltage threshold (Over voltage fault thresholdFOV) and trigger an software cleanable overvoltage fault. Overvoltage faults that are caused by returned energy from motor, can be dealt with a regenerative resistor and with optional extra capacitance in HV DC bus.

Scenarios where returned energy is causing the rise of HV DC bus voltage:

  • Deceleration of motor speed when there is significant amount of energy stored in mechanical motion (rotating inertia or moving mass). This typically occurs with spindles and linear axes.
  • Sudden reversal of torque setpoint. This can generate voltage spike even when motor is standing still. This typically occurs in high bandwidth torque control applications (such as Force feedback system (FFB)). These spikes are very short and an added capacitor to HV DC bus and/or low resistance regenerative resistor can provide a solution.

Sizing regenerative resistor[edit | edit source]

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:

$ R_{regen}=\frac{U_{DCBusVoltage}}{I_{PeakMotorCurrent}} $ click to enlarge equations

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.


Resistor wattage[edit | edit source]

Resistors generally have two properties: resistance and power dissipation rating. Resistance we have determined in the previous chapter which leaves power to be determined. Determining proper power rating is not very straightforward as estimating the amount of returned energy is difficult without practical experiments (i.e. try one resistor and monitor it's temperature). Only rough starting guide is provided here.

Small machines, such as desktop machining tools and lightweight mechanical axes
0-50W (0W = no resistor may be needed)
Medium sized machines, such as cast iron machine tool
50-200W
Large and heavy machines (>kilowatt of motor power)
>200W
Force feedback systems
10-50W, in these systems the energy is mainly from motor inductance and not inertia

Most suitable type of resistors are the ones that can handle high peak power without damage. These are common wire wound power resistors that can handle typically ~10 times more power in short periods than their actual rating.

Undervoltage faults[edit | edit source]

Undervoltages come from drop of power supply voltage during surges. The easiest solution is to set undervoltage paramater [FUV] to a lower value and use power supply that doesn't shut down or drop to near zero under power surges. For very short current surges (millisecond range), an added capacitor to HV DC bus might provide a solution.

Using additional capacitor in HV DC bus[edit | edit source]

Variables used in equations[edit | edit source]

  • $ t_{duration} $ = time duration in seconds which capacitor should be able to help at maximum current surge
  • $ L_{MotorInductance} $ = motor coil inductance in Henrys (use value of Coil inductanceML/1000)
  • $ I_{PeakMotorCurrent} $ = motor peak current in Amps (use value of Peak current limitMMC/1000)
  • $ U_{MaxVoltageChange} $ = maximum voltage change in HV DC bus during this current surge/peak. I.e. if drive is set to fault at 56V and supply voltage is 48VDC, then 56-48V=8V should be used.
  • $ scaler $ = a user chosen value between 0.1 to 1.0. 1.0 is for the worst case where we assume instantaneous torque reversal (rare) and lower values can be used with slower change of torque direction.

Surge duration based method[edit | edit source]

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:

$ C_{filter}=t_{duration}\frac{I_{PeakMotorCurrent}}{U_{MaxVoltageChange}} $ click to enlarge equations

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 $ C_{filter}=0.005s\frac{20A}{10V}=0.01F=10000\mu F $.

Motor inductance based method[edit | edit source]

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:

$ C_{filter}=scaler*L_{MotorInductance}*(\frac{I_{PeakMotorCurrent}}{U_{MaxVoltageChange}})^2 $ click to enlarge equations

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 $ C_{filter}=0.5*0.005H*(\frac{20A}{10V})^2=0.01F=10000\mu F $.

Keeping it practical[edit | edit source]

Targeting low value in max voltage change will increase capacitor size significantly, and may become unpractical. I.e. if supply voltage is 48V and max voltage is 56V, the max voltage change would be only 8V. Reducing supply voltage by few volts, say 44V, the allowed voltage change becomes 12V which yields much smaller required capacitance (in the above inductance based method example this change would make 2.2 times difference).

See also[edit | edit source]


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