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I am trying to create a simple buck converter on a breadboard using this topology:

L6385E gate driver buck converter

I am familiar with the theory of this circuit and believe I have connected everything correctly. The gate driver I am using is the L6385E.

http://www.st.com/content/ccc/resource/technical/document/datasheet/b2/ab/e2/82/c8/bc/42/fe/CD00169717.pdf/files/CD00169717.pdf/jcr:content/translations/en.CD00169717.pdf

Some of the parameters are:

Vcc = 16V

H.V. = 2-24V

L = 33uH

Cout = 3300uF

Cboot = 1uF

R = 10K

The FETS are IRF100B202. I will link the datasheet down below.

I apply a PWM of 62kHz to the Hin, and connect Lin to GND. With Vcc at 16V, Vboot becomes about 13V due to some internal losses. With H.V. around 2V this is fine and the high-side MOSFET is driven fine with a Vgs above 10V. Except, when I increase H.V., I would expect Vboot to rise the same amount, so the high-side FET can still be driven. But the problem is that Vboot does in fact not rise at all when I increase H.V.. This causes the high-side gate driver to stop working, because it can no longer supply Vgs = 10V to the FET.

I have tried a variety of different components and the problem still persists. Am I missing something crucial in the workings of this circuit, or am I maybe using some wrong components? I would appreciate some feedback!

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  • \$\begingroup\$ infineon.com/dgdl/… \$\endgroup\$ – Thomas Gerrits Jan 31 '17 at 15:08
  • \$\begingroup\$ connect Lin to GND So the bottom NMOS is never on ?!? Then the Drain-Source diode of that NMOS will act as the flyback diode. Is that what you want ? Do you get a proper squarewave ( 0 Volt to HV Volt) at out (pin 6) ? \$\endgroup\$ – Bimpelrekkie Jan 31 '17 at 15:24
  • \$\begingroup\$ Yes, the bottom NMOS is basically only there to function as a diode, i have also replaced it by a normal diode, but the problem persists. Pin 6 does not go down to 0V, it stays at H.V. due to this. \$\endgroup\$ – Thomas Gerrits Jan 31 '17 at 15:32
  • \$\begingroup\$ Then the boost function will not work, pin 6 has to be pulled to ground so that Cboot can be charged (to almost Vcc). You have a sort of chicken-egg problem because to make pin 6 low, the inductor has to be charged so that it can discharge through the diode (in bottom NMOS) but since Cboost is not charged, top NMOS does not turn on so the inductor is never charged enough. \$\endgroup\$ – Bimpelrekkie Jan 31 '17 at 15:35
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    \$\begingroup\$ You need enough of a load on the output of the buck to guarantee that your low-side diode conducts long enough to charge the bootstrap cap. A non-sync bootstrapped buck has a minimum load requirement. \$\endgroup\$ – John D Jan 31 '17 at 15:36

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