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I would like to use the example circuit, from TL5001's guide, for a wider voltage range. I need it to work with almost 50V input.

Here's the circuit:

enter image description here

Now, the problem with this is that Q3 pulls the MOSFET's gate to GND. While the source voltage is around 40-50V, and this would destroy the gate. How can I limit this voltage?

As a secondary question: TL5001 max input voltage is 40V. Will a simple shunt regulator, with a zener diode, be enough to supply it?

I know this isn't the ideal IC to do this job, and I know that there are easier ways to drive a MOSFET gate nowadays, but please, I'd like to avoid answers suggesting the use os MOSFET driver ICs or other controller chips. There are reasons why I need to work with this particular IC.

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    \$\begingroup\$ You should probably lay out the complete set of reasoning. Just saying that you need to work with this IC or that particular circuit topology is not the way to approach an engineering problem. You should search out the best solution with ALL of the cards on the table. You can never expect to get the best design advice if you try to hide some nonsense ideology. \$\endgroup\$
    – Michael Karas
    Commented Feb 27, 2014 at 5:38
  • \$\begingroup\$ @MichaelKaras I live in a country where imports are restricted to TWO packages a year per person. That means I can't just go to DigiKey and order any IC I want. Even if i could, shipping from the US to here is, minimum, $100. Do you have anything else to say? \$\endgroup\$
    – hjf
    Commented Feb 27, 2014 at 13:10
  • \$\begingroup\$ @hjf: There are step-down designs which use only an NE555, a few transistors, the switching FET, and a Schottky as semiconductors. \$\endgroup\$
    – Laszlo Valko
    Commented Feb 28, 2014 at 0:14

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Well, your choices are limited. You could use a TVS diode from gate to source, and a resistor in series with the gate to clamp Vgs. Of course that will slow down the switching and you'll have higher switching losses. Same goes with the supply voltage- The TL5001 only draws a couple of mA, so a shunt regulator could work at the cost of more power dissipation.

John

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