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Old 31st October 2008, 05:30 AM   #1
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Default hide side mosfet driving

Dear all,

Attached is a circuit to drive a single high side driver, the circuit works fine when the load is not connected to the parallel capacitor (470uF/400V). but when the capacitor is connected in parallel to the load the IR2110 is burnt,

can any one advise on this.

Many thanks
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Old 31st October 2008, 06:11 AM   #2
Eva is offline Eva  Spain
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In fact, your circuit as it is drawn is specifically designed to blow the IR2110
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Old 31st October 2008, 06:35 AM   #3
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Dear Eva,

Can you please tell me the mistakes that i have done. sorry the diode was a drawing error. now i have corrected and post the schematic again

Many Thanks
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Old 31st October 2008, 06:59 AM   #4
mflorin is offline mflorin  Romania
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you must have a low impedance path for charging the bootstrap capacitor.

If your circuit is a buck converter, then you forgot to put a diode or a mosfet for preventing brutally inductor current interruption.

My suggestion is to add a low side mosfet. IR2110 is designed to work in a half bridge configuration.
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Old 31st October 2008, 07:41 AM   #5
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I have added another diode to prevent the inductor current interuption, yet when the capacitor 470uF/400V is added the IR2110 is burnt.
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Old 31st October 2008, 07:54 AM   #6
mflorin is offline mflorin  Romania
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the diode in the mosfet's gate is connected in a wrong way. Usually mosfet turn-on (charging of gate-source capacitance) is done through a resistor (20Ohm in your case). Fot turning off, sometimes a faster discharge current is requred, so the gate resistor is bypassed by a diode.

Also, you can remove the diode after the inductor
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Old 31st October 2008, 08:01 AM   #7
mflorin is offline mflorin  Romania
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but bootstrap capacitor is still not properly charged....
Search for IR app note AN978
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Old 31st October 2008, 08:55 AM   #8
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Actually my main application is as attached when there is a step up of load and the output voltage drops so the microcontroller detects the voltage drop and gives pulses to the mosfet and high voltage is injected to recover the voltage drop...

As from simulations i have succesfully got the result for this method ... but now the problems is switchiong the high side mosftet... any ideas or solutions are very much appreciated.
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Old 31st October 2008, 11:09 AM   #9
mflorin is offline mflorin  Romania
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you didn't understand how IR2110 works...
In a half bridge configuration, when the low side switch is in ON state, the bootstrap capacitor finds a low impedance path (through the mosfet) and charge to the driver supply voltage.
When low side switch turns OFF, the mid point between the two mosfets became floating, bootstrap capacitor being charged at vcc "above" floating point voltage.

Now, in your circuit, there's no way to charge the bootstrap capacitor because you do not have a low side switch (path). In this situation, you may consider an auxiliary circuit capable to generate the required floating voltages.

IR app note AN978 suggests a kind of a voltage doubler (figure 16)
http://www.irf.com/technical-info/appnotes/an-978.pdf
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Old 31st October 2008, 11:33 AM   #10
Eva is offline Eva  Spain
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Everything is wrong. The IR2110 blows because there is no discharge path for the inuductor other than through the internal diodes of the own IC.

On the other hand, the isolated converter can regulate its output without additional switching regulators.

Furthermore, it seems that you are trying to switch non-isolated 300V from rectified mains to an isolated +/-25V output, which is completely crazy. Please give up ultil you learn enough electronics to know what you are actually doing
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