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Silicon rectifier failure |
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| Feb28-13, 01:51 AM | #1 |
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Silicon rectifier failure
hi there,
I'm currently facing a problem with the silicon rectifier, http://www.alldatasheet.com/datashee...CTRON/6A8.html with the current setup which is at the attachment. The source is a AC current of 230V 50Hz , while the cap is at 0.3μF. The setup is to provide a half-wave rectification to the universal motor. The problem is that sometimes the rectifier breakdown, but when does it breakdown. sorry if it might be a simple question, i'm not good in electrical circuits |
| Feb28-13, 09:05 AM | #2 |
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How much current does the motor draw?
Do you have an oscilloscope? Can you measure the peak to peak ripple when the switch is closed? If you do have an oscilloscope be very careful when making this measurement. You need to use a 1:1 isolation transformer between the oscilloscope and the outlet. When does the diode blow? When you open the switch, when the motor has a heavy load or a light load? When you open the switch, how long does it take for the motor to coast to a stop? |
| Feb28-13, 11:26 AM | #3 |
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Which rectifier are you using - there is a range of voltage specs for that rectifier.
The univ motor is a DC motor and you are only giving it 1/2 of a waveform - my guess is the inductive Kick back from the motor when the diode goes into reveres bias. Best way to check is add a diode across the motor, normally reverse biased (it will look like it is reverse polarity of the existing diode. ) Also - note per your diagram the "top" of the motor is negative polarity - nothing wrong with that just not conventional drawing method. |
| Feb28-13, 06:34 PM | #4 |
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Silicon rectifier failure |
| Feb28-13, 07:12 PM | #5 |
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I was trying to determine by my questions whether the rectifier was being destroyed by over voltage or by over current.
The peak to peak ripple voltage would give an indication of how much current the diode is conducting. It only conducts during the time the slope of the ripple voltage is positive. That time could easily be 1/10 or less of the whole cycle time. That means if the motor is drawing 1 amp, during a whole 50 Hz cycle, during the time when the diode is conducting, it could easily be conducting 10 amps or more. Since the diode blows upon opening or closing of the switch, it probably is not blowing because of over current. Nevertheless, it never hurts to overrate the rectifiers. Most likely the diode is blowing because of inductive spikes when the switch is opened and I agree with Windadct's analysis. The capacitor isn't nearly large enough to absorb the inductive kick so a reverse biased rectifier across the motor is the best solution. Why do need to use a rectifier with a universal motor? |
| Mar1-13, 12:09 AM | #6 |
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regarding the question of the usage of a rectifier with a U.motor, its because the necessity to reduce the motor power output. |
| Mar1-13, 06:09 AM | #7 |
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Connect the diode like this.
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| Mar1-13, 06:34 AM | #8 |
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Recognitions:
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I would add a small (68pF) high voltage (say, 1.5kV) capacitor across each diode. That's cheap insurance.
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| Mar1-13, 07:35 AM | #9 |
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How big is that universal motor?
Be aware a vacuum cleaner motor might draw more current than your 6 amp rectifier can handle,,, starting current may be easily 5X running. Vacuum cleaner motors are the better part of 1 horsepower. |
| Mar1-13, 02:21 PM | #10 |
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This may be a reverse recovery issue.
If the switch closes during the peak of the AC waveform, but in the reverse biased direction, the diode will conduct, in the reverse direction, for some period of time before it snaps off abruptly. Large power diodes can have substantial reverse recover times. If significant current builds up in the motors inductance before the diode snaps off, you can get huge surge in reverse voltage across the diode. Why do you want to use the motor on only half of the AC waveform? |
| Mar1-13, 03:14 PM | #11 |
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Silver, there are triac controllers that will allow you to control the power to the motor and even power it down without using the switch. This might be a better solution albeit more expensive.
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