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I've constructed the following circuit: High-side Driver

From this blog: Tahmid's Blog

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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  • \$\begingroup\$ The IRF540N is only rated for 100V max. \$\endgroup\$
    – Tut
    May 26, 2015 at 11:34
  • \$\begingroup\$ That 300V supply could be wrong. \$\endgroup\$
    – user16324
    May 26, 2015 at 11:34
  • \$\begingroup\$ @BrianDrummond I didn't use the 300V supply. I used 5V. \$\endgroup\$ May 26, 2015 at 11:41
  • \$\begingroup\$ @Tut I used 5V. \$\endgroup\$ May 26, 2015 at 11:42
  • \$\begingroup\$ It's a bootstrap driver. It needs to be continually switched on and off to work correctly. It won't work at 100% duty cycle as C1 and C2 lose the charge that keeps the gate above the source. \$\endgroup\$
    – Jon
    May 26, 2015 at 12:12

2 Answers 2

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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.

enter image description here

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  • \$\begingroup\$ This was a long time ago and I have since abandoned the project, but when I find time, I will resume it with your comments taken into account. I will let you know what happens :) \$\endgroup\$ Nov 5, 2015 at 16:45
  • \$\begingroup\$ @Undertherainbow Good luck :o). I did not even notice that the question is so old! \$\endgroup\$ Nov 5, 2015 at 18:20
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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.

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