You didn't show the VR, how did you connected it?

this is my VR circuit


R1: protection
C1: stabilize

Did you still have used the heatsink or not?

That is about the Wattage,
W = V*I = 24V*3A = 72W
2.5A ... 60W
2A ... 48W
You better messure the draw current of motor when the duty cycle is 100%.

I have heatsink! but i don't use now because this FET's degree is about 30cº (less then 30cº)

Post #32 FET too hot.
3A motor? Is the motor really pulling 3A? The current very much is related to load and RPM.
What is the Gate voltage. It needs to be at least 5V.

Assuming Gate voltage = 5V, Motor current 3A, MOSFET = 0.1 ohms internal resistance, 3A X 0.1V = 0.3V across the transistor, = 0.9 watts. The part will get a little hot. Now if the Gate voltage is only 4V the heat will be up 1.4X.

Gate resistor is a little large. So the part is switching slow which will increase the heat some.

yes i think my main power source might be a problem.

and motor 24v/3A/3800RPM
but it dosen't matter if motor current 2.5A (2.5A is powerful enough).

i checked gate voltage, and it was 4.9v

maybe my VR is damaged.

What’s the PWM source? Microcontrollers have a max current on output pin of 20-40mA. Be careful going smaller, 100 Ohms is pushing it. I’m sure it’s a power supply issue or poorly specd NMOS. What is too hot? @3A assuming it’s as the spec you’re dissipating 1/3 watts at the MOSFET. Its going to get hot. Brownout are probably power supply issues. Could be your source or your cruddy 7805 Did you decouple the regulator?

my source is

TCCR1A=0x82;
TCCR1B=0x1A;
ICR1=49;
//16000000 / 8 / 49 = 40816 Hz
// 25us

ORCA1=49; //change value to duty cycle

and this is my regulator curcuit



C1 is 3216 size and C2 is 1608 size (i have no PCB i did this on my breadboard)

As mentioned in post 37, the gate resistor could be smaller, so less voltage drop, and higher voltage at the gate. In addition, a pull down resistor should be connected from gate to ground to ensure the mosfet turns off with no signal from MCU (gpio will also float at boot-up or reset).

I'm thinking a method to protects the MCU and that is to use a diode as 1n4001 or maybe 1n4148 to replace the R1, and in papallel with a 2.2K for Vgs.

Edit : 2.2K is designed to discharge the Cgs and provides Vg a low level when the MCU output low, so the values need to try, specially when the input frequency up to 50 Khz.

okay i'll try add 2.2k resistor to gate (pull down)

 

7805 can't provides output current over 1A, and also the input voltage is too high, the better input voltage is Vin<=12V, when the input voltage over 12V then you can in series with a power resistor or a Cement resistor to get the heat from 12V and current, and then the temperature of 7805 will be not that high.

Did you measure the draw current of motor?

Did you measure how much current that the 7805 could provides, if over the current then you have to change to another Power regulator.

 

7805 can't provides output current over 1A, and also the input voltage is too high, the better input voltage is Vin<=12V, when the input voltage over 12V then you can in series with a power resistor or a Cement resistor to get the heat from 12V and current, and then the temperature of 7805 will be not that high.

Did you measure the draw current of motor?

Did you measure how much current that the 7805 could provides, if over the current then you have to change to another Power regulator.

i need regulator output current under 1A.

so i choose lm7805 instead of lm2576

i only use MCU, 2Digit-Segment,

now i need run FET(IRLR024N)

IRLR024N

Rds(on) Vgs 10v -> 0.065 R
Vgs 5v -> 0.080 R (this is a value which use)

"if i have a little extra current to turn on the FET, then lm7805 gets no heat."

this is what i thinking.

motor is connected to external power

 

this is my VR circuit

R1: protection
C1: stabilize

You can take R1 to the right side, so the input of ADC has RC integrator that it can be slow down the voltage from potentiometer.

Edit : You can modify the values of R1 or C1 to match the rotation speed that you want.

 

You can take R1 to the right side, so the input of ADC has RC integrator that it can be slow down the voltage from potentiometer.

Edit : You can modify the values of R1 or C1 to match the rotation speed that you want.

yes, thank you!

 

When the VR is 10K then the resistor values changing will be small, so the C1 charging time and voltage rise time will be fast when you rotate the potentiometer.

If you want to slow down the C1 charging time and the voltage rise time, then you can try to increase the C1 values as 1~4.7μF.

 

When the VR is 10K then the resistor values changing will be small, so the C1 charging time and voltage rise time will be fast when you rotate the potentiometer.

If you want to slow down the C1 charging time and the voltage rise time, then you can try to increase the C1 values as 1~4.7μF.

thank you i'll try!
i have VR 10K, 50K
i'll test it rotate 10K,50K