I'm using a 1A constant current buck LED driver (the TPS92518 from TI), in order to implement "shunt FET dimming", where the load (1A LEDs) is short-circuited using a MOSFET in order shut-off a LED string very fast (but without ripple, hence the need for the output capacitor).

However, since an output capacitor will be problematic when shorting the LED string (because it will discharge through the Q2 switch when it's closed, and when Q2 will open it will sink current for awhile before any current goes to the LEDs again), I want to add a second switch, Q1, in order to disconnect the capacitor when the LEDs are shorted (Q1 OFF when Q2 is ON and vice-versa).

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Ideally, Q2 needs to close a few nanoseconds after Q1 opens so that the capacitor doesn't begin to discharge through Q2.

However, I'm not sure what kind of switching topology to use. Q2 seems relatively easy, only a NMOS would suffice. However, for Q1 I'm lost. If I only use an NMOS for Q1 I won't be able to switch it because the body diode will conduct afaik.

I was wondering if anybody has a good suggestion for a fast circuit to implement Q1 (and make Q2 slower a bit). Basically, Q1 needs to be a few nanoseconds faster than Q2. Q2 needs to be able to switch for approx. 1us at worse.

There is in my circuit a voltage that is much higher than the voltage at the buck output, so this could be used to switch Q1, but since Q1 et Q2 must be triggered by an FPGA, there needs to be some kind of transistor arrangement. However I'm not too sure what to do to optimize the speed. Perhaps somebody has some clever transistor tricks?


EDIT: For those asking why shunt LED driving would even be a thing, see answer here.

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    \$\begingroup\$ You only need one switch for LEDs. A series inductor and parallel capacitor can remove ripple and flicker, or go to a very fast switching speed. You seem to be over complicating something rather simple to solve without added MOSFETs. \$\endgroup\$ – user105652 Aug 14 at 2:24
  • \$\begingroup\$ Think more about your RLC filter damping ratio for a step change in current with overshoot and ripple voltage. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 14 at 2:56
  • \$\begingroup\$ I'm thinking similar - Seems overly complicated. Why do you need to shut off a string of LED's so fast that shorting their power supply is better than just cutting their power off??? Seems to me, unless I'm missing some 2nd order effect, that the result would be virtually indistinguishable. Q2 would have to be a big expensive device in a shunt implementation because it'll be taking the full brunt of a 35V 1A power supply whereas a series-cutoff implementation of Q2, you could use a MOSFET that costs pennies. \$\endgroup\$ – Kyle B Aug 14 at 3:55
  • \$\begingroup\$ I prefer not delving too much into the details of my application as I fear explaining more than getting solutions, but LEDs aren't lasers so trying to speed up their optical switching speed is important here. I edited my answer to include the link to an answer explaining why shunt LED driving is useful for high switching speeds. The LED driver I linked is indeed recommended for this application, so I thought it would be a good solution, if I can lower the output ripple in some way. \$\endgroup\$ – Yannick Aug 14 at 4:25
  • \$\begingroup\$ As for overshoots/undershoots, the buck is current regulated, and recommended for this application. The control loop is optimized to be fast, however the manufacturer only using an inductor (and no cap) somewhat makes the ripple difficult to manage. So I hope the inductor DC resistance should be enough to damp it. \$\endgroup\$ – Yannick Aug 14 at 4:36

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