2
\$\begingroup\$

I've been designing a test bench for RC propulsion systems and for that I thought it would be nice to be able to switch the power to the ESC on/off. For this I designed a simple circuit which I thought would do the trick. Vaa is used for both driving the gate and the ESC. However, as it turns out, the ESC GND is internally connected to the GND of its servo connection. First this means that the ESC never turns off. Second, this is quite dangerous when the MOSFET source isn't pulled to GND (open switch)! In that case if someone was to open throttle, the current would most likely kill the flight controller instantly. Another downside is that it is only possible to use/simulate a 3S or 4S battery pack (maybe 2s but not optimal) since Vgs of the MOSFET can't go above 20V (5s is 21V when fully loaded).

First switch for drone ESC

I think I've come up with a solution to both these problems by switching in the high side of the ESC and using an isolated DC/DC converter to drive the gate. The converter converts a 5V signal to a 12V signal which is high enough for the Rds(on) to be low enough to support a current draw up to 40 Amps with appropriate cooling (6 W/mK thermal pad and 2.7 K/W heatsink). However, I'm not sure this setup will work as I've never worked with such a converter before.

My question is basically if there is an obvious mistake I've made and what this mistake might be. Also if you have any tips as to how this circuit might be improved you are more than welcome to give them!

Second switch for drone ESC

PS: This is my first ever post on this website. Feel free to comment on the contents of my question and how I might improve future questions!

\$\endgroup\$
  • \$\begingroup\$ It seems that your second circuit should work, except I don't understand why you are connecting -Vout and 1.5k resistor to ESC+, instead of ESC-? \$\endgroup\$ – Guill May 19 '17 at 23:28
  • \$\begingroup\$ @Guil Is this a trick question? Suppose the MOSFET conducts then (almost) the full source voltage of Vaa is found over the ESC. If I was to connect -Vout and the 1.5k resistor to ESC- the MOSFET will never conduct since +Vout is then referred to GND and Vgs would most likely be negative. However if I connect -Vout to ESC+ then +Vout is referred to ESC+. Suppose the MOSFET conducts then Vgs will always equal (+Vout + Vaa) - (-Vout + Vaa) = 12V. Therefore the MOSFET will always conduct (that is, if CTC+ is high). Again I might be totally wrong. If so, convince me! I am more digitally inclined :) \$\endgroup\$ – Robbe Elsas May 20 '17 at 16:32
  • \$\begingroup\$ I'm not an RC guy, so I don't know what an ESC is, but there is not a thing missing from your approach from a standpoint of switch design - it will turn on when CTC current is present, and off when removed. If the ESC is a load, it will work just as you intended, with about 8-10 W lost in the FET at 40 amps once it warms up. One caveat: if there is any chance of ESC+ exceeding Vaa, The FET will conduct backwards. \$\endgroup\$ – John Birckhead May 31 '17 at 22:26
  • \$\begingroup\$ This is a fundamentally bad idea. Electronic switches always have loss, and ESCs already put two in series with the motor, which have to be carefully chosen to minimize this. You now want to add a third, and of course circuit topology means you need to do it on the more difficult high side where P channel is weak and N channel means complexity. Instead, look to improving the control signal to the existing FETs. It usually comes from a microcontroller and typically is only enabled by an arming sequence. Replacement firmware is available for common implementations. \$\endgroup\$ – Chris Stratton Nov 6 '17 at 13:47
  • \$\begingroup\$ @ChrisStratton I don't see why this is a terrible idea, the MOSFET would only dissipate 6.4 W at 40 A while causing a 140 mV drop. Sure, there are significantly better MOSFETs out there, but that isn't too bad. What would you use, some huge relay? \$\endgroup\$ – jms Jan 23 '18 at 18:25
0
\$\begingroup\$

The circuit looks OK, provided that converter -VIN is connected to ground.

For switching 40A I would put two MOSFETs in parallel, then you only need a small heatsink to dissipate the 4W of heat produced by both FETs.

\$\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.