Power MOSFET for Motor Driver
- Quantum
- Topic Author
14 years 11 months ago #10084889
by Quantum
What kind of MOSFET transistors are used on the AX1500?
Power MOSFET for Motor Driver was created by Quantum
What kind of MOSFET transistors are used on the AX1500?
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- cosma
14 years 11 months ago #10094355
by cosma
Replied by cosma on topic Re:Power MOSFET for Motor Driver
<P style="MARGIN: 0px">We use the IRF1405. This is a MOSFET with 5mOhm RDSon - one of the most efficient there is.
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- Quantum
- Topic Author
14 years 11 months ago #10141626
by Quantum
If I used the IRF1405 to control a 12V motor at 10A, how can I calculate the thermal resistance rating of the heatsink that I'd need, and how can I tell whether or not a heatsink is necessary?
Replied by Quantum on topic Re:Power MOSFET for Motor Driver
If I used the IRF1405 to control a 12V motor at 10A, how can I calculate the thermal resistance rating of the heatsink that I'd need, and how can I tell whether or not a heatsink is necessary?
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14 years 11 months ago #10174697
by tonysantoni
Robotony
Replied by tonysantoni on topic Re:Power MOSFET for Motor Driver
<P style="MARGIN: 0px">Assuming you have one MOS controlling a 12V 10 A motor connected to ground, then the power dissipated in the MOS is:
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">P = R * (PWM * Im ^2)
<P style="MARGIN: 0px">R = resistance drain source plus the resistance of the pcb power tracks. RDS=8 mohm worst case. A good guess is a total 16 mohm.
<P style="MARGIN: 0px">PWM is the duty cycle on the motor
<P style="MARGIN: 0px">Im is the peak current through the motor.
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">In case of continuous current PWM = 1 then
<P style="MARGIN: 0px">P = 0.016 * 100 = 1.6 Watt
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">The junction to air thermal resistance is 62 deg/watt therefore the junction delta temperature DT would be:
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">DT = 1.6 * 62 = 100 deg.
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">The junction temerature assuming an ambient temperature of Tamb will be:
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Tj = Tamb + DT
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Assuming the MOS is mounted on a pcb, a good rule is to limit the board temperature to 100 deg, since most other components like electrolitic capacitors and IC's need to be kept below 100 deg. Since the junction temperature and the case temperqture of the MOS a nearly same, this means to limit Tj to 100 Deg. The MOS is the main heating element for the board.
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Assuming Tamb = 30 deg then:
<P style="MARGIN: 0px">Tj = 130 deg which is above the limit.
<P style="MARGIN: 0px">A heat sink is needed.
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">The heatsink at PWM = 1 can be sized by the formula:
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Ths = Tamb + Rth-ba * (Poc+R*Im^2)
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Ths = temperature of the heat sink.
<P style="MARGIN: 0px">Rth-ba is the thermal resistance boad to air, mostly the therrmal resistance of the heat sink.
<P style="MARGIN: 0px">Poc = power dissipated in all other components but MOS.
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Estimating R (Rds plus power tracks plus eventual power connector), Poc power dissipated by all board except the MOS, Ths (example 90 deg), Tamp (example 30 deg) then Rth-ba can be calculated.
<P style="MARGIN: 0px">Tony
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">P = R * (PWM * Im ^2)
<P style="MARGIN: 0px">R = resistance drain source plus the resistance of the pcb power tracks. RDS=8 mohm worst case. A good guess is a total 16 mohm.
<P style="MARGIN: 0px">PWM is the duty cycle on the motor
<P style="MARGIN: 0px">Im is the peak current through the motor.
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">In case of continuous current PWM = 1 then
<P style="MARGIN: 0px">P = 0.016 * 100 = 1.6 Watt
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">The junction to air thermal resistance is 62 deg/watt therefore the junction delta temperature DT would be:
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">DT = 1.6 * 62 = 100 deg.
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">The junction temerature assuming an ambient temperature of Tamb will be:
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Tj = Tamb + DT
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Assuming the MOS is mounted on a pcb, a good rule is to limit the board temperature to 100 deg, since most other components like electrolitic capacitors and IC's need to be kept below 100 deg. Since the junction temperature and the case temperqture of the MOS a nearly same, this means to limit Tj to 100 Deg. The MOS is the main heating element for the board.
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Assuming Tamb = 30 deg then:
<P style="MARGIN: 0px">Tj = 130 deg which is above the limit.
<P style="MARGIN: 0px">A heat sink is needed.
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">The heatsink at PWM = 1 can be sized by the formula:
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Ths = Tamb + Rth-ba * (Poc+R*Im^2)
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Ths = temperature of the heat sink.
<P style="MARGIN: 0px">Rth-ba is the thermal resistance boad to air, mostly the therrmal resistance of the heat sink.
<P style="MARGIN: 0px">Poc = power dissipated in all other components but MOS.
<P style="MARGIN: 0px">
<P style="MARGIN: 0px">Estimating R (Rds plus power tracks plus eventual power connector), Poc power dissipated by all board except the MOS, Ths (example 90 deg), Tamp (example 30 deg) then Rth-ba can be calculated.
<P style="MARGIN: 0px">Tony
Robotony
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