2
\$\begingroup\$

I recently came across an article explaining how to drive a high side MOSFET using a "separate isolated power supply whose ground and the ground of the MOSFET-based circuit are isolated". However after reading this article I still do not completely understand the driver circuit and had a number of questions:

  1. Why is it necessary to use an isolated power supply rather than a non isolated power supply that will simply boost the gate voltage to able 12V higher then the voltage of the "main supply" in this case 12V higher than +24V with reference to ground?
  2. the isolated power supply shown in the figure is represented by a battery, how can I instead replace this with a boost converter? (should the input to the boost converter be +24V and ground, with the output negative connected to the source of Q1 and positive terminal where the positive terminal of the battery was)
  3. what effect does having the negative terminal of the isolate power supply connected to the source of Q1 have?
  4. is there an isolated boost converter that you can recommend given the small current required to drive the gate.
  5. if applied to a H-bridge will i need 2 seperate isolated boost converters (one for the left high side drive and one for the right high side drive?)

The circuit suggested by the this article is below: enter image description here

\$\endgroup\$
  • \$\begingroup\$ This is not an answer but why a: Boost converter for high side Mosfet? Why not use a bootstrap circuit? \$\endgroup\$ – Weaverworm Jan 28 '16 at 9:57
  • \$\begingroup\$ The bootstrap will fail at d=100% .This may or may not be a problem. \$\endgroup\$ – Autistic Jan 28 '16 at 10:25
  • \$\begingroup\$ Not being able to achieve 100% futy cycle is the issue \$\endgroup\$ – user3095420 Jan 28 '16 at 11:58
2
\$\begingroup\$
  1. An isolated supply makes things really easy. You can easily get isolated DC-DC converters that will output 12V. By using an isolated DC-DC converter, you can treat the output just like the battery in your circuit. There is less chance for interaction between your isolated drive and the rest of your circuit. Starting at $3-4, it would probably be worth your time for a one-off or low volume solution.
  2. This is why you do not "simply" connect a boost converter to do the job. You could probably have the input 12V power supply be your source, and then have it run in a discontinuous mode. Voltage regulation of your gate driver supply is a significant concern, as you would probably need something like a zener diode to limit the maximum voltage. Past that, you'd also need a current return path for the boost charging loop, which may be an issue if your load decreases significantly or is disconnected. Even then, I'd probably start from a bootstrap topology and hack in the boost converter.
  3. Your gate driver, when connected to the source of Q1 with an isolated supply, is setting the gate to source voltage directly. It sets the gate-source voltage to either (Vsource + 12V) or (Vsource). Otherwise, you need to know what the source voltage is in order to drive the gate safely. If 36V is applied across your gate-source terminals, your MOSFET will probably be destroyed.
  4. Digikey/Mouser/etc... has a wide assortment of isolated DC-DC converters that generate a fixed output voltage. An "isolated" boost converter does not exist (because it would be called something else). You're probably on your own if you want to make your boost converter idea work.
  5. Yes.
\$\endgroup\$
1
\$\begingroup\$

If you use a boost converter to step the +24v supply up to +36v, then you don't necessarily need the isolated gnd. The circuit you drew uses an isolated gnd to allow the 12v battery to see gnd near the +24v supply rail, leaving the pos. battery terminal at ~+36v relative to your "main" (24v) ground. Using a boost converter can give you the same +36v without needing to "trick" the supply into providing a higher potential.

EDIT: Here's a (simplified) diagram of what I mean: simplified diagram

\$\endgroup\$
1
\$\begingroup\$
  1. Why not just have a boost circuit that gives 30-something-V and use that? One reason is that the 30V+ cannot be used to drive the gate of the MOSFET directly through a switch (a transistor), because the Vgs could be close to 30V+ which exceeds the ratings of most MOSFETs. But that can be overcome with some simple voltage regulation.

  2. With the configuration in the schematic, you need an isolated supply. The top two candidates are probably flyback and push-pull. The secondary winding of the transformer would provide the isolated supply and the ground of that would go to the source of the MOSFET.

  3. Don't know how to answer this question. It is part of the scheme.

  4. Just about any boost circuit can be easily used as flyback with primary side regulation just by replacing the inductor with a transformer. It is not difficult to get off-the-shelf power inductors with two equal isolated windings which can be used as the flyback transformer. This method is more flexible with the voltage levels as compare to the next.

Another method would be push-pull if the system has a convenient source voltage that can be matched to a convenient off-the-shelf push-pull transformer of certain winding ratio to give what is needed. (Check out TI SN6501 for example).

  1. For 2 seperate isolated converter, just parallel two transformers' primaries to the same converter circuit (with the proper component values). The regulation is not critical and the suggestions above are using primary side regulation anyway. And if your application already has a switching regulator, it may be easy to piggy back off of that.

By the way, what is given in the schematic may not (or may) be the optimal way for your application (which I don't know). For example, you mentioned H-bridge, which means you want to control the switching delay of the high and low side switches.

\$\endgroup\$

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.