1
\$\begingroup\$

I would like to produce a buck/boost converter that can convert 24V to 48V with a 40A max. output current. It would be used to control a DC motor, specifically the MY1016Z3 DC motor.
I cannot use a 48V source directly because it is for a competition which requires that the battery voltage is 24V.
I have done some testing with the Texas Instruments power stage designer, and according to it, at 20Khz PWM, I would need at least a ~45 μH 100A inductor which I can get.
I would use a 4 switch buck boost converter so that I can decrease losses and also the output voltage.

I just want to check that such a converter is possible with a reasonably high efficiency.

\$\endgroup\$
1
  • \$\begingroup\$ You should also check the other conditions of the competition thoroughly. The limitation of the battery voltage is there most likely for a reason. It may be that behind that limit is an overall limitation of voltages in the system. Most likely because of safety considerations. \$\endgroup\$
    – Ariser
    Apr 22, 2022 at 10:25

2 Answers 2

2
\$\begingroup\$

Looking at the motor you want to use I would not advise on doing what you have planned for a few reasons :

  1. Conversion efficiency can be as high as 90-95%, but 5% loss at the power you talking that is a lot of heat.
  2. Motor nominal voltage is 24V, running it at 48V to get more power/rpm out of it will most likely result in over-heating of its winding. You are better off changing the gearing to adapt its nominal speed/torque to your need.
  3. Reliability, if you are planning to build a vehicle, such a boost regulator will be the source of many possible issues down the road.
  4. Weight, again if we are talking vehicle, the added weight of your boost contraption alongside all the required cooling, will weigh down your vehicle (maybe more that what you gain from overdriving your motor)

If the efficiency is your ultimate goal, I would not advise trying to boost at such power level the battery voltage. Have you looked into rewinding the motor windings or adapting the gearing. I would do that first before looking into your proposed solution. Also if we are talking vehicle (you need to save weight !!)

\$\endgroup\$
3
  • \$\begingroup\$ Unfortunately the competition doesn't allow modification to the motor. I do not believe heating will be an issue as I only plan to use the converter for a short time such as a nitro in a gas car. It would only be operating at its maximum power for 5 seconds maximum. The events are also very short so the boost regulator would not need to last for extended periods of time. I also estimate that the controller would only weigh around 750g which won't be very significant. I am more worried whether the converter would work at all with such a high power. \$\endgroup\$ Mar 12, 2020 at 9:21
  • \$\begingroup\$ For some reason I felt the vibe of endurance racing in your initial post, thus my answer was wrapped around the efficiency/weight performance of the system. With the new insights in mind and with the over-heating of the motor out of the equation, I would say that it is entirely feasible. The converter would most likely work but you will want to evaluate the influence of the surge current when you turn things on. I would precharge a capacitor bank that goes from parallel ( 24 volts) to series (48 volts) in order to help the converter at startup or something. That's a very courageous endeavour ! \$\endgroup\$
    – benguru
    Mar 12, 2020 at 9:42
  • \$\begingroup\$ Thanks a lot for the advice! \$\endgroup\$ Mar 12, 2020 at 9:45
2
\$\begingroup\$

It is possible. You need a huge battery though. Below you may find a simple boost design. Pay attention to the input current. There are MOSFETs/IGBTs and diodes that can handle this current available on the market. However, heat dissipation may be an issue.

enter image description here

Traces

\$\endgroup\$
4
  • \$\begingroup\$ Thank you for this, I am planning that the converter will only work as a quick boost lasting for about 10 seconds at the maximum. I am planning to replace the diode with a synchronous mosfet and use 2 hy4008 mosfets in parallel for each switch. this would give a rdson of 1.5mohm which at the input rms current of around 80a, would give me 4.8w per mosfet and since 2 sets of 2 mosfets are being used at any given time, 19.2w of mosfet losses at full power which I think would be fine with decent heatsinking. However I am unsure whether this mosfet is ideal due to the high gate charge. \$\endgroup\$ Mar 12, 2020 at 9:38
  • 3
    \$\begingroup\$ I would NOT rely only on such a quick simulation to conclude that it's possible! In that simulation a lot has been left out and considered ideal. There's for example no series resistance in the wiring, the power source, the inductor and the diode. In the simulator it all looks fine and works. Then you actually build it and all those "forgotten"resistances and losses will bite you. Look at existing designs for currents of 40 A, realize what needs to be done to avoid having too high losses. It is a common trap that beginners fall into! \$\endgroup\$ Mar 12, 2020 at 10:14
  • 1
    \$\begingroup\$ That is absolutely true. This model is not a base for an actual design. This is purely to indicate the magnitude of current and power that we're dealing with. It is however an indication that a concept (i.e 24/48V with 40A load) is feasible. \$\endgroup\$
    – Krzych
    Mar 12, 2020 at 10:35
  • \$\begingroup\$ @Krzych that is so nice of you to go above and beyond and do a quick simulation ! LTspice for the win ! One thing that have always bothered me when dealing with voltage regulators of any kind, is that when driving an inductive load (a motor for instance) you always have sort of a short circuit at startup. And it may if not considered complitely prevent your supply to even start. But I am no expert it is just something I had noticed. Now I take great care designing all my systems with the battery voltage as voltage source for all motors, so that I avoid the issues of high power conversion. \$\endgroup\$
    – benguru
    Mar 13, 2020 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.