I want to turn on and off both the mosfets at same time and I used IR2110 gate drivers for high and low MOSFETS. Gate driver IR2110 problems. Ask Question. Up vote 0 down vote favorite. I want to turn on and off both the mosfets at same time and I used IR2110 gate drivers for high and low MOSFETS. I am doing my project on 9 level multilevel inverter. For this, I want a gate driver circuit for MOSFET and at the same time circuit diagram of 9 level multilevel inverter. The IR2110 IC is one of the high speed and high voltage gate driver ICs for IGBT and power MOSFET. The IC is having independent low and high side output channel. By using a single IC, a half bridge circuit can be operated in which one MOSFET is in high side configuration and another one is in the low side configuration.
Active3 years, 10 months ago
I've constructed the following circuit:
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I put an LED in series with a 330 ohm resistor at the OUT terminal.I have the 5 V being supplied by an Arduino and the 12 V from a lead acid battery. The arduino is not programmed, I'm just using it for its output, and I am manually moving HIN from ground to +5V to test the circuit.
When I connect the lead-acid battery, the IR2110 gets really hot. I've measured the voltage between 3 and 5 and it is about 1.5 V, when HIN is at ground. This doesn't change when I move it to +5.The LED turns on brightly then dims when I connect the 12V and HIN is at ground. It turns off if HIN is moved to +5V. I am using the IRF540N for the MOSFET.
I have 5V at the drain of the MOSFET.
Please help. I don't know what could be wrong.
Oh and all the grounds are common.
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UndertherainbowUndertherainbow
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IMHO, the blogger is not a very good power electronics engineer. His blog seems to be well read but most posts are of poor quality.
In Fig. 9 we see the IR2110 being used as a single high-side driver. The circuit is simple enough and follows the same functionality described above. One thing to remember is that, since there is no low-side switch, there must a load connected from OUT to ground. Otherwise the bootstrap capacitors can not charge.
That is not true. To charge the bootstrap cap, there needs to be a low-impedance path, such as a turned-on power switch would be.
A series combination of a LED with a resistor, on the other hand, has resistance equal to the resistor value.
The IR2110 and IR2113 are bootstrap gate driver circuits.
The charge reservoir for the upper gate gets charged when the lower switch it turned on.
This schematic from the IR2110 datasheet does not even show the current path that would enable your circuit to work. I assume your OUT terminal is connected to ground via a resistor in series with a LED.
There must be some stray current path inside the chip that is used when the chip is not configured as intended.
I suggest to put the switch to the low side and reconfigure the IR2110 to the LO output.
Btw. I have burned a dozens of these gate drivers in the past. They are notorious for being easy to burn out due to static electricity.
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SunnyBoyNYSunnyBoyNY
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First of all, all the 1K is wrong. It damages IRF540 (it did in my case). And other one is VS pin, this should be connected to GND for your case. Ref AN-978 page 23, also related figure 23. Also the pages last para mentions 'if the output voltage of the buck converter is between 10 and 20 V' , as an example using buck converter. That means if the voltage is low range like 10~12v DC, UV locks the current state of the high side, probably and mostly leaving it ON according to AN-978. The boot capacitor 0.47uF is enough for 110 nC gate charge instead of 22uF.
Biddut MitraBiddut Mitra
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Active1 year, 6 months ago
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I want to turn on and off both the mosfets at same time and I used IR2110 gate drivers for high and low MOSFETS.
When I try to simulate the circuit, after few seconds it shows error saying time step too small. What is the problem with circuit or simulations?
In the simulation graph,
Green represents the inductor currentBlue represents gate voltage of M1Pink represents gate voltage of M2
lomesh pudipeddi
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Not sure about the LTSpice error, but I do see one problem. In order for the high side IR2110 to work, the bootstrap capacitance (C1 and C3) must be charged so that 12 volts appears across it. This is done by pulling the VS pin to ground, which happens every time you switch the low side on a normally-configured half bridge. In your circuit, this capacitor will not ever be charged. C1 and C3 must be refreshed periodically by pulling VS to ground; the charge on the cap provides the current and voltage necessary to get the high side gate drive voltage to a voltage greater than your 12 V supply. If the charge on this cap is less than 8 volts or so, the high side switch will shut off. You could put in the low side FET instead of D2 and pulse it at the start and then again periodically.
John BirckheadJohn Birckhead
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