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Braking resistor on AC servo motor driver is not working, and I cannot find power on it, but it is 100 Ohm (It burned up). Power on driver is 200W. Can I put 400W (100 Ohm) braking resistor on that motor?

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    \$\begingroup\$ The more the better... \$\endgroup\$ – po.pe Feb 7 '19 at 16:58
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    \$\begingroup\$ do you know why it burnt up? do you know if you just need to manage the regenerative energy due to decelerating the inertia or is there any aiding loads. what sort of duty is it experiencing. \$\endgroup\$ – JonRB Feb 7 '19 at 20:38
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To answer the question you asked, yes.

I just sized a few Brake resistors so while the information is fresh:

When a servo motor or any motor is slowing down a mechanical load, it is in essence transferring mechanical energy to heat and electrical energy. The added electrical energy can be seen by a rise in the DC bus voltage of the VFD (your servo motor driver). Often the amount of energy that would need to be stored in the DC bus raises the bus voltage to too high a level. To prevent damage most drives can be equipped with an external resistor to "bleed off" the excess energy.

Like any resistor the two most important parameters are resistance and rated power. The needed rated power is dependent on the application of course. A good start point is match the resistor rated power to the rated power of the motor your VFD is driving.

The resistance needed falls under a range and is dependent on your VFD and the DC voltage your VFD nominally runs at. The upper resistance limit is determined by your DC bus voltage. The resistance needs to be low enough that when the DC bus voltage is applied to it the needed power is consumed by the resistor.

Upper Resistance Limit:

\begin{equation} R_b = \frac{V^2}{P} \end{equation}

\begin{equation} Where\ R_b\ is\ the\ brake\ resistance,\ and\ P\ is\ the\ desired\ power. \end{equation}

The lower resistance limit is dependent on your VFD and should be in its datasheet. If the resistance is too low the current pushed through the brake resistor will burn out the VFD's brake driving circuit.

As long as the resistance is within this range you should be ok. Typically the VFD simply fires the Brake Resistor circuit when the DC bus voltage reaches some threshold, and turns it off when the DC voltage drops again.

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  • \$\begingroup\$ This doesn't actually answer the OP. While this does provide clear steps in determining the ohmage, this is known by the OP. The wattage and equally as importantly the pulse power is unknown. \$\endgroup\$ – JonRB Feb 7 '19 at 20:37
  • \$\begingroup\$ I know omhage is 100 ohm. Resistor in driver is 90W, and I found only 50W resistor (to low), and 400W resistor. I realy don't have time to order exact one, because waiting time for it is 30 days, and I have too much work. It's motor for one axis on CNC machine. \$\endgroup\$ – Davor Krpan Feb 8 '19 at 8:32
  • \$\begingroup\$ @DavorKrpan do you know what the power requirements of the load is? You could install the 400W and run for a few cycles, watch the temperature throughout and see if it reaches a safe steady state. Was the 100 watt resistor apart of the original install? \$\endgroup\$ – Clipboard_Waving_Enginerd Feb 8 '19 at 15:45
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It depends how fast you want to brake. Since the current in brake R converts deacceleration force F=ma where a is proportional to current.

You must consider the design of driver not it's heat dissipation limit and compare acceleration control speed ( maybe from V/f constant in a VFD)

\$ P=I^2Rce\$ for IGBT Rce=Vce(sat)/Ice then the acceleration current depends on motor impedance DCR+ωL vs brake resistance where the braking losses are shared between motor windings and brake dump load resistance and switch.

Simple choose R impedance and power rating according to Ohm's Law and Physics of stored mechanical rotating energy than the desired braking de-acceleration time from stored energy conversion at some RPM. If you expect a "speed brake", consider your load dump Power MUST increase to reduce the brake time for same stored kinetic energy. THen the duty factor of these operations of these rapid start-stops can be 10x rated motor power so the duty cycle average might be <10% of the heat rise time.

Servo with heavy inertial mass is the same as charging/ discharging a battery or supercap. Both store energy that result in excess heat if the I^2R exceeds the rating of current or thermal rise.

What is your stored energy and Watt Pd rating on the part and time interval during braking?

You may need a heatsink for the power R.

It appears from your previous questions that you did not tell us you chose a low inertia motor possibly with a heavy inertial mass. MSDA023A1A

If the power is applied with 200W for more than a second to motor, chances are you need a heat sink for the 400W resistor. This can be computed but you have no datasheets for every part with rotational mass.

Since you can control acceleration and braking by any method, there is is no need to increase W rating if you can wait longer. If you reduce R less than than servo DCR, the motor heats up more , it stops faster and the R heats up less than the motor @ 750W/Hp

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  • \$\begingroup\$ I don't have that informations, but resistor in driver is 90 W-100 Ohm. Can burning of that resistor cause malfunction of other components in driver, like IGBT? \$\endgroup\$ – Davor Krpan Feb 8 '19 at 8:24
  • \$\begingroup\$ What did you do to cause? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 8 '19 at 8:26

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