Not all pins on the Mega and Mega 2560 support change interrupts, * TX is digital pin 11 (connect to RX of other device) * RX is digital pin 10 (connect to TX of other device) Receives from software serial, sends to hardware serial. Receives from the hardware serial, sends to software serial. With the diode providing braking the motor should stop quickly.OVERVIEW.

After switching direction You should not apply PWM until the motor is stationary, else a very high current will flow through the FET. If you switch direction while the motor is spinning the diode will become forward biased and 'brake' the motor. If throttle is reduced quickly then the motor will continue to generate voltage until it stops, but as that voltage is the same polarity it won't hurt the rest of the circuit. When the motor is a generator, the diode is initially reverse biased. At this frequency the flyback diode needs to be a fast switching type, preferably a high current Schottky type to keep voltage drop down. To improve the switching speed you could add a push-pull driver using an NPN and PNP transistor, like this:-īoth transistors are connected in 'Emitter Follower' configuration, so they amplify the current by about 100 times (Hfe of the transistors) and reduce switching time by about the same amount.įor efficiency the PWM frequency should be high enough (>3kHz) that motor inductance smooths out the current, but low enough to not cause too much switching loss. With this driver circuit the Gate voltage should rise reasonably quickly (assuming the optocoupler is being driven hard enough) but will fall quite slowly due to the large amount of charge stored on the Gate. Your circuit shows the Gate being driven by an optocoupler switching +12V to a (1k?) resistor. As PWM frequency is increased there are more transitions and the FET will get hotter, so the rise and fall times need to be shorter. Each transition causes the FET to heat up a little. If you are applying PWM then you must ensure that the Gate voltage rises and falls quickly. This also applies when switching it on and off. If the FET is only partly turned on then it will drop more voltage and heat up. A 14.4V drill motor can easily draw over 30A at startup, so the FET needs to turn on as hard as possible to maintain a low Drain-Source resistance. To be fully turned on it needs 5V or more (see datasheet fig 1. Threshold voltage is just the point at which the FET starts to turn on. It has a threshold voltage of min 1V and max

Is there a better circuit that will not have these issues? If my circuit is fine, please also leave a comment. The problem is that this does not work when the motor is quickly reduced from full speed to low speed. This then allows the energy in the circuit to be returned to ground. Therefore in the code we have it set so that every time the speed becomes zero, the relays change position. This is because the position of the relay switches when driving the motor determines the polarity the generator will be, and this position always means the diode is reverse biased when the motor is a generator. The other issue is that we have blown a few of the MOSFETs, and I'm not sure if its because the diode is in the wrong place. Is the left half (the gate turn on circuit) of my schematic correct? Our guess was that this is because the Gate to Source voltage needs to be at least 1V higher than the voltage across the motor, and not just 1V higher than ground. We removed the 1K resistor so that the gate gets the full 12V and it then worked. This potential divider meant that the gate to ground voltage was a 10th of the 12V, i.e 1.2V, which according to the datasheet should be sufficient to create a conducting path between the source and drain, however it didn't. In my very first attempt at this circuit RG was 100 ohm and I had another 900 ohm resistor between the optocoupler and the 12V source. It has a sufficient heat sink attached to it, as shown in the picture, along with the motor and gear box bellow. It has a threshold voltage of min 1V and max 2V. The relay module I have has built in optocouplers and pull down resistors for the direction switching. Pin 1 and pin 2 of the relay are connected to the same digital-out pin on the arduino. Ignore the yellow pen, this was a suggestion by someone else. My first attempt at a circuit is shown bellow: We are using relays instead of a H bridge is because of the high current.

I need a circuit that can control the speed and direction of the motors, using an Arduino with PWM. I have a 14.4V DC drill motor that can draw 10 amps when under full load.

Bidirrection motor code arduino relay