VFD - DC Bus Connections with Dynamic Braking

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Buccaneer

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Trying to trace the path of current on a VFD . The DC bus has +3 & - , that feeds into (2) braking units. The braking units feed a Power Ohm Resistor that has (2) 13,600 watt strips. Each Braking Resistor feeds (1) 13,600 watt section of the Power Ohm. Inside of the Braking Resistor there is a Level Detector that senses across DC + & DC - from VFD. The VFD was energized and within 10 seconds , smoke came from the top. After taking the covers of, we found that the Pre Charge Resistor ( on Vfd )was open. Upon further investigation we found that the Gate Board had also a burnt spot on it. My question is if the DC + and DC - were crossed; without the VFD Contactor pulling in and the motor not running, how did the power make it pass the Transistor ( or through the braking unit ) with reversed polarity and no current to be sensed. This was the first start up on new equipment. Everyone that has seen the drawing of the braking unit diagram said it looks impossible. The braking resistor is a . Any help in understanding this would be appreciated.
 

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What you are not seeing in this diagram is that the actual braking is being done by running the motor as a DC generator through the VFD's switching elements and all that the braking resistor unit has to do is bleed off the excess DC voltage developed on the internal bus.
More efficient (and expensive) VFD's can convert that DC to backfed power to the AC source instead.

Without the motor connected and running your VFD is still supplying line-sourced power to the DC bus and apparently the high voltage of reverse polarity damaged some of the components on the braking board, allowing the VFD to supply uncontrolled constant power to the resistor.
I will leave the more detailed circuit analysis to someone else. (Like you, maybe?)


Tapatalk...
 
what he said. when the motor starts making juice instead of using it up, it accumulates on the dc bus capacitor.

the brake unit looks at the dc bus voltage and turns on when the bus voltage is higher than desired, and turns back off when the dc bus voltage drops to an acceptable level. it burns the excess energy off as heat.
 
What I did not mention, and probably should have, is that the normal maximum voltage on the DC bus is controlled, but is also limited by the peak AC line voltage and the rectifier circuit.
The only time the voltage can possibly rise above that maximum is when the motor is being run as a generator (braking).
That is why a simple voltage sensor is all that the braking module needs
to have. No need for a control input.

PS: when the VFD is not actively operating the motor, I would expect the DC bus to be powered anyway (through the precharge circuit ?) so that the VFD is ready to start the motor with no delay and also has the needed control power source.

Tapatalk...
 
What I can not figure out is that no damage was done to the braking unit. And being that it was powered up for less than 10 seconds under a no load condition, how the level detector allowed the voltage to pass through the transistor with out damage. To the best of my knowledge , the level detector is not polarity sensitive and if there was not any counter EMF to sense, why would it have closed. Maybe I am not seasoned enough on the overall function of the braking unit. In addition, the voltage on the DC bus +3 & - are a slow rise from the rectifier. NOTE: The VFD panel did not even show the power indicator light.
 
Most if not all power transistors (IGBT, MOSFET, etc.) have an 'reverse diode' associated with them. This reverse diode may be intentional, and optimized for use, or it may be parasitic, meaning that it is there as a consequence of the way the device is manufactured. This reverse diode may or may not be shown in a schematic for the switching element.

If the braking unit was reverse connected to the DC bus, then it would have acted as a short circuit even if the level sensor was not turning on the transistor. Because of the diode across the braking resistor, you would essentially have a two diode drop short, with very little resistance.

-Jon
 
Buccaneer said:
What I can not figure out is that no damage was done to the braking unit.
Click to expand...
1. The braking unit may have acted as a near short circuit, with the current being limited in the other (supply) part of the VFD. No power dissipation, and limited current = no burnout.
2. If the current did actually end up flowing through the braking resistor, that is a very high power unit compared to whatever was limiting the current supplied by the main VFD. Current within the resistor rating but above the rating of the supply circuit also = no burnout.
FWIW, the transistor is diagrammed as an insulated gate device rather than a conventional two-junction transistor.
 
Thanks for the input, we are going to experiment with another VFD . Don't if the Diodes are BIAS, and still puzzled. But I have learned not to argue with Electricity, She's always right.
 
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