I have designed a synchronous buck converter to step down a 17.4V voltage source to 12V at the output. I have attached the schematic as shown. Problem: The Buck is not operating as expected at all. Here are the measurements I was able to get from testing it:

  • PWM frequency: 62.5kHz
  • Vg of M1 = 8.5V
  • Vds = 0.03V
  • VL (Voltage drop across inductor) = 17.37V
  • Vload {Voltage at the load) = 0V

First off, I cannot understand why I am only getting 8.5V to the gate of M1. I am using a half bridge gate driver with bootstrap topology, but yet the voltage I am getting out of the HIGH pin of the driver is just the 8.5V. And not to mention the fact that I am even getting a significant voltage drop across the inductor. I can't understand why that is so at all. I am thinking that the buck may be operating in DCM which may explain it, but I'm not so sure. Could anyone help me make sense of what could be wrong here?

P.S.(I have attached a copy of the datasheet of the gate driver showing how I have hooked it up. See page 2) DRV_HI pin feeds into the gate of M1, and DRV_LO pin feeds into M2. C4 and C3 are both 0.1uF ceramic capacitors. Vcc is 11V, and Vin from the microcontroller is roughly 5V

I apologize, here is the URL for the sheet: http://www.onsemi.com/pub_link/Collateral/NCP5111-D.PDF


simulate this circuit – Schematic created using CircuitLab

  • 2
    \$\begingroup\$ If your question is about the gate driver, you should probably include the gate driver in your schematic. (Also, whatever datasheet you meant to attach, it didn't come through) \$\endgroup\$
    – The Photon
    Commented Oct 20, 2015 at 5:18
  • \$\begingroup\$ You might also want to share your oscilloscope traces of the signal at the gate of M1. \$\endgroup\$
    – The Photon
    Commented Oct 20, 2015 at 5:20
  • 1
    \$\begingroup\$ You should provide details on L1, also. Does it have a high series resistance? It looks like you are planning to supply 2A to your 6 Ohm load, so you want something with pretty low DCR. \$\endgroup\$
    – user57037
    Commented Oct 20, 2015 at 5:41
  • 1
    \$\begingroup\$ Driving M1 with 8.5 V is no good, M1 will then behave as a source follower so the node at the left side of the coil will never rise above about 4 V. Can the gate driver supply more than 8.5 V when the gate of M1 is not connected? What is the current capability of the coil, can it handle 4 A (double the current through R1) without saturating? When saturated a coil behaves like a short! \$\endgroup\$ Commented Oct 20, 2015 at 7:41
  • \$\begingroup\$ related question by the same O.P.: electronics.stackexchange.com/questions/195526/… \$\endgroup\$ Commented Oct 20, 2015 at 8:08

1 Answer 1


(Voltage drop across inductor) = 17.37V Vload {Voltage at the load) = 0V

Inductors are capable of being destroyed and having 17.37 volts across it totally indicates that it has failed open circuit. This means you will see no voltage on the output and it will likely affect how your bootstrapping operates.


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