SANEMB
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« on: November 29, 2012, 12:03:30 12:03 » |
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In L6562AD PFC ckt , 200uF/450v Electrolytic cap at the PFC o/p takes longer time to discharge when we switch OFF the mains in No load / no lamp condition. .
Switching on the mains while Cap is discharging gives high voltage spike and also hissing noise which may damage our HID ckt.
Ckt charges cap at 450v .
i need strng suggestion on this.
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PaulC
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« Reply #1 on: November 30, 2012, 01:08:27 01:08 » |
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Maybe Normally closed relay that when mains is on goes open, so when mains disconnected it will close allowing a high wattage resistor to discharge faster..
this relay - resistor will be only on one phase, so does not matter if single or three phase ..
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pickit2
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« Reply #2 on: November 30, 2012, 01:27:01 01:27 » |
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Demo board EVL6562A-400W output cap don't have bleed resistor, that are in the datasheet.
if this is the case, add two 1M (2M) resistors to discharge the cap.
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SANEMB
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« Reply #3 on: November 30, 2012, 03:34:02 03:34 » |
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I will try with 1M resistor and let you know.
bt i have tried with few resistor,bt still the cap is nt discharging fastly.
if their is another way to disacharge the cap,please share it.
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metal
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« Reply #4 on: November 30, 2012, 04:20:05 04:20 » |
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The cap in the datasheet is 47uF, did you try it to see if spikes occur?
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solutions
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« Reply #5 on: November 30, 2012, 05:32:45 05:32 » |
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t=rc
Smaller cap, or smaller R. NO magic solution that won't completely swamp your cost,reliability, failure mode, and board space.
If 1M doesn't do it, take it to 100k...it's only 2mA of load and 1/10th the time constant. It also does not affect your power factor. Go lower if you have to have the discharge time..it's only power.
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alexxx
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« Reply #6 on: November 30, 2012, 08:07:48 20:07 » |
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Concerning capacitors I would stay close to the values given by datasheet. I would focus more on the resistor value.
Of course that means that if your Watts are high, you will have larger capacitance so slower RC. If you drop too much with the resistance, you will pay a penalty in board space (resistor size), as well as temperature. You must experiment a bit to find the optimum solution.
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pickit2
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« Reply #7 on: November 30, 2012, 10:16:56 22:16 » |
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Running the L6562AD without a load in the real world, would not be done, even if your load is removed before switch off, a 1/2meg bleed resistor would make supply safe in a known time frame.
The design load of the power supply should be part of the discharge, most units built with this ic, are not switched on/off in rapid succession, typical Tv power, ect.. If your design needs to have this function, then you need to switch a load that will discharge the cap in the time you need.
a 6" nail comes to mind a 6" nail comes to mind a 6" nail comes to mind
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miserable
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« Reply #8 on: February 23, 2013, 06:52:26 18:52 » |
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Tekno1
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« Reply #9 on: May 30, 2013, 07:35:11 07:35 » |
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I am not familiar with your application but you may also consider soft-start function to limit the inrush current during the start-up (turn on) of your circuit. Soft-start function is normally implemented with power control IC (or PFC ) so it is a part of IC that yo may be using.
If soft-start is not an option than another consideration you may have is employing an active bleeder circuit. When circuit is ON than bleeder resistance current path will be open circuited by a series switch and when circuit turns off than the switch will be closed enabling the resistance current path. Since active switch circuit will be only series with bleeding resistor (not on main circuit current path) power consumption will not be an issue. Of course this active switch circuit will cost some money and design time.
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