enter image description hereI'd like to drive high side and low side MOSFET for buck converter, so I have used IR2184s,(the image bellow) to do the job,

The frequency of oscillation is 1Mhz, so DRVR is a square wave signal of 1MHz frequency and 5V peak. VBATT=23V and the other components are shown in the image, The duty cycle is 50% and the mosfet used to be driven id IRF540

My circuit doesn't work as expected, (ie: I do not get a squar wave at bothe HO and LO)

I have recalculated the value of C6 but this doesn't solve the problem, and I don't know why ?? enter image description here

  • 1
    \$\begingroup\$ show a sketch of the actual waveform. is duty cycle not 50%? are rise and fall times not ZERO_time? Do actual waveforms (from scope photos) have ringing? \$\endgroup\$ – analogsystemsrf Apr 13 at 10:52
  • 1
    \$\begingroup\$ I would use more than 100nF at Vcc of U1. Keep C33 (as close as possible to U1) and add 1uF closest to U1. Since you power it with a battery, it is likely there are long leads between the battery and U1 and other components fed by the battery. So, you might need more caps on other places. \$\endgroup\$ – Huisman Apr 13 at 11:17

The turn-on propagation delay is typically 680 ns and worst case 900ns. When driving at 1 MHz 50% duty-cycle,\$ t_{on}\$ = 500 ns. So, you already turn off the gate driver, before it could drive LO or HO high.
So, this IR2184 is too slow to be driven on 1 MHz.
Next, without connecting the low-side mosfet, the bootstrap cap will not be charged, and therefore the high side driver will not work.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.