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Hello,
Mounting the mosfets in the same heatsink will double the current handling of one mosfet, due the positive temperature coefficient of channel resistance.
Best regards
djailli
the mosfet are IRF3205, upper part act as the high side voltage and lower side mosfet for low side. drived using a digital input [0,1]. Each mosfet can out a power of 200W,
what is the maximum power i can transfer ? or i can drive the load?
(the load is inductor of the brushless motor)
Each power transistor (including power mosfet) has a SOA (Safe Operating Area). Even it's rated as 200W (20V/10A), it does not mean you will be safe to use it for 20V/10A application. The power handling capability of the power mosfet is dependent on the switching frequency and duration of ON/OFF. The SOA of power mosfet should be provided in the datasheet. You can find the power handling (that suit your application) of the mosfter based on the SOA.
u should be able to simulate the power consumption of your device by using Orcad (it looks like this is the software you use). i see this one phase of a BLDC motor, it looks like you have an exotic topology. i'd like to see the whole thing, with explination of how it works, some time
to analyze the power dissipation realize that MOSFETs have two types of power loss. switching loss & conduction loss. usually only one term is calculated (the dominant) term and the other term is considered insignificant.
i recall from memory (i.e. u should double check this)
1) switching loss = (Eon + Eoff)*Fsw ; where Eon/off is the energy to turn ON or OFF the device, usually stated in the datasheet.
2) conduction loss = (Ids)^2 * Rdson
now u must determine for your application which is the dominant term. for motor drives usually it is conduction, but a simple calculation will tell you.
the power loss in your motor is primarily from conduction loss. (I_line_rms)^2 * R_armature ; per phase.
yes your circuit will carry double the current because you are paralleling mosfets. but remember that with engineering you never get a free lunch. you will see a tradeoff with efficiency as your MOSFETs will ring/oscillate in that configuration. orcad is not likely to pick this up because it depends greatly on parasitic capacitance & the quality of your gate drive.
also, do not expect your mosfets to share current 50% each. it can drift as much as 75% in one and 25% in other.
How did you know, when the mosfet burnt, the current from battery was low? 1ms of overcurrent might have burnt your mosfet.
I'm not sure the circuit shown in your first post is complete or part of the full circuit. If it's complete,
1) Why do you need the upper switch (mosfet)? In fact, lower switch is enough to drive the inductive load.
2) There should be a freewheeling diode across the inductive load. Otherwise, voltage spike when the switches are turned off might damage the mosfter.
3) Is the driving circuit for upper switch designed properly (i.e. to drive the mosfet as a switch, not amplifier)?
4) Did you set a dead time to prevent 'shoot-through' problem?
By the way, did you use PWM to control the mosfets? Or, other methods?
As requested by other repliers, pls upload the full schematic if the one shown in your first post is not the full schematic. Thanks.
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