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I have a NE555 producing about 250 kHz that runs at 5 V. I want to control the MOSFET driver IR2104 with it. Here's my schematic:

Schematic

I tried reducing/increasing the frequency (from 1 kHz to 250 kHz) and tried capacitor values like 22 pF, 470 pF, 1 nF, 10 nF, 100 nF and even an electrolytic 4.7 μF capacitor.

I don't quite understand the value required for the capacitor (any calculation will be very helpful). I also tried removing the BC547 acting as a NOT gate and connecting pin 3 of the IR2104 to VCC directly as suggested by some random forums and YouTube videos.

Anyway, the ground of the NE555 is connected to the IR2104 and others as well.

The problem is that the circuit doesn't work at all. It seems all cold and dead, except when I swap the capacitor; then I see a blink of light. Am I doing something wrong or is the IR2104 dead?

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    \$\begingroup\$ Decoupling capacitor is missing. Show oscillogram of IN, VB, HO and LO. \$\endgroup\$
    – winny
    Commented Jan 13, 2023 at 8:36
  • \$\begingroup\$ That's the thing I ignored. Shouldn't it work without the decoupling capacitor? \$\endgroup\$
    – 15 Volts
    Commented Jan 13, 2023 at 9:05
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    \$\begingroup\$ In addition to the decoupling caps as winny stated, always use gate stopper resistors (from driver output to the gate) to limit the charge and discharge currents for the sake of the chip's output drivers' health. For the value of the bootstrap cap check my answer here, but generally, for frequencies above 50 kHz, 20 to 100 times the Ciss of the MOSFET is enough. IRFZ44N's Ciss is 1.5n so 100n should be alright. \$\endgroup\$
    – Rohat Kılıç
    Commented Jan 13, 2023 at 9:16
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    \$\begingroup\$ The way the circuit is wired can be very critical. Add the capacitor and eliminate a potential problem. \$\endgroup\$
    – Kartman
    Commented Jan 13, 2023 at 9:18
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    \$\begingroup\$ Probably not. Why risk it? \$\endgroup\$
    – winny
    Commented Jan 13, 2023 at 12:28

2 Answers 2

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The problem is that the circuit doesn't work at all.

Why does your circuit put the IR2104 into shutdown every time your input tries to go high? This doesn't seem at all right! You should leave the \$\overline{SD}\$ pin high to enable the chip: -

enter image description here

Then, if you were correctly activating the \$\overline{SD}\$ pin (which you aren't unfortunately), you'd need a base resistor adding like this: -

enter image description here

Without that base resistor, the \$IN\$ input can't be raised any higher than about 0.7 volts and, that won't likely activate the IR2104.

Then do all the other stuff mentioned in comments under the question.

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remove the transistor and resistor network from the input side because it will shut down the ic every half cycle and the circuit wont work

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    \$\begingroup\$ This is what Andy already stated above. \$\endgroup\$
    – winny
    Commented Jan 3 at 11:41

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