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I have understood all the parameters except Rds and conduction loss. Please explain them to me.

I am using a lithium-ion battery pack of 36 V and its maximum voltage is 42 V to drive a motor.

Motor specifications:
Minimum current: 500 mA
Maximum current: 15 A
Ipulse: 30 A

The MOSFET I am planning to use is the Vishay SQD50N05-11L

MOSFET specifications:

MOSFET specifications part 1

MOSFET specifications part 2

So based on the above specification, I have calculated the power dissipation:

Power dissipation: I2 * Rds = (15 A)2 * 11 mohm = 2.47 watts

And the maximum power the transistor can dissipate without a heat sink:

(Tmax - Tambient) / R(jA) = (175 °C - 25 °C) / 60 °C/W = 2.5 watt

I am confused that how to check and relate the power dissipation with Rds and select the MOSFET based on it. Is the MOSFET that I have chosen okay, based on the above specifications or should I go for a MOSFET with a low Rds?

And please explain how to select the MOSFETs based on Rds.

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    \$\begingroup\$ What does "Ipulse" mean? \$\endgroup\$ – Bruce Abbott Apr 1 at 2:20
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    \$\begingroup\$ You seem to have given an equation that relates power loss and needing a heatsink to Rds. You seem to already understand the implications. What more do you need? Probably just cost more for lower Rds which will save power and weight. Probably doesn't matter if you're just making something for yourself a few extra $$ won't matter. \$\endgroup\$ – Carl Gilbert Apr 1 at 2:54
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    \$\begingroup\$ Please, pretty please, allow for more margin for the heat loss. You have less than 1% and I am sure you won't be always operate it in 25C or less. The simplest way is to just add a second transistor in parallel. You will get half the heat distributed over 2 transistors. \$\endgroup\$ – fraxinus Apr 1 at 13:17
  • \$\begingroup\$ Or use some nice big solder pads on your PCB and glue on a heatsink. \$\endgroup\$ – Michael Apr 1 at 19:23
  • \$\begingroup\$ An extra FET in parallel might be cheaper, and more reliable. Or use a FET with lower Rdson (eg. 5 milliohms). \$\endgroup\$ – Bruce Abbott Apr 1 at 19:36
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Your understanding of RDson and conduction loss is fine. Just know RDson varies with Vgs and increases with temperature, usually by 150% to 200% at its hottest, so you must account for this if you plan on running lots of current or hot weather. You can find the RDSon vs. temperature graph in the datasheet.

But know there are switching losses if you are switching your transistor at high frequency (or have too long a transition time). Motor PWM is one of those cases where you have to pay attention to switching losses. It is more complex than conduction loss and is supposed to equal conduction loss in an optimal design.

You select one that can handle your current at your switching frequency where conduction loss = switching loss. Trying to select based on conduction loss without switching loss is like trying to pick a car based on top speed and nothing else: if you only care about top speed don't care about cost, fuel efficiency, handling, features, appearance, or anything else. You would just pick as fast as possible, because faster is better and you don't care about compromises made anywhere else, but that's not practical, because you would end up with a drag racer goes fast, but it can't even turn a corner.

Sizing near maximum using only conduction loss when operating at high frequency will burn out you MOSFET due to the heat from switching losses.

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  • \$\begingroup\$ Thank you So much DKNguyen for your reply. The Problem is i can understand the calculation of Rds and total Power i dont know how to compare the result and Select the MOSFET. and in some of the MOSFET data sheets like IPD088N06N3 G which has less Rds resistance but.The Rthja is given in Kelvin . like 62 K/W . for how will we calculate and select the MOSFET. kindly explain me \$\endgroup\$ – Muthu Apr 1 at 12:57
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    \$\begingroup\$ @Muthu You select one that can handle your current at your sswitching frequency where conduction loss = switching loss. Trying to select based on conduction loss without switching loss is like trying to pick a car based on top speed and nothing else: if you only care about top speed don't care about cost, fuel efficiency, handling, features, appearance, or anything else you would just pick as fast as possible because faster is better and you don't care about compromises made anywhere else but that's not practical because you would end up with a drag racer goes fast but can't even turn a corner \$\endgroup\$ – DKNguyen Apr 1 at 13:13
  • \$\begingroup\$ What is "c current"? \$\endgroup\$ – Peter Mortensen Apr 1 at 15:49
  • \$\begingroup\$ @PeterMortensen Just current. Corrected \$\endgroup\$ – DKNguyen Apr 1 at 15:57
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The MOSFET with the minimal Rdson is not always the most suitable. Those with a very low Rdson have a large Qg gate charge. It means that they need a powerful gate driver if you want to do PWM. A higher gate charge makes turn on/off slower, thus higher switching losses. The good MOSFET is the one with the lowest product: Rdson * Qg

The total loss of a MOSFET is the conduction loss + switching loss. If your application needs only to turn on/off a motor, then the MOSFET you choose will suit your demand.

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Ideally, a transistor should have ZERO resistance when turned on, but we don't live in an ideal world, so we're stuck with just getting the lowest resistance we can.
I don't understand why would you even ask such a question, seeing that you understand that higher RdsON means more losses. OF COURSE you will try to get the lowest RdsON available.
There are very few special cases where a higher RdsON would be preferred.
Since you are able to calculate the power loss (the amount of heat generated) based on the RdsON, you are able to see if it will be too much for your circuit.
In the case above, the transistor is pretty good for its purpose, all you need is a small heatsink.
Sometimes, the limiting factor in choosing a MOSFET is also its price, its capacitance and/or speed. Higher capacitance will require a more powerful MOSFET driver and will increase losses at higher switching frequencies, so in that case we don't go for the lower RdsON if it means higher losses overall.

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  • \$\begingroup\$ Thank you So much Edin for your reply. The Problem is i can understand the calculation of Rds and total Power i dont know how to compare the result and Select the MOSFET. and in some of the MOSFET data sheets like IPD088N06N3 G which has less Rds resistance but.The Rthja is given in Kelvin . like 62 K/W . for how will we calculate and select the MOSFET. kindly explain me \$\endgroup\$ – Muthu Apr 1 at 12:47
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    \$\begingroup\$ @Muthu Thermal resistance and drain-source resistance are entirely different things. \$\endgroup\$ – Hearth Apr 5 at 21:41
  • \$\begingroup\$ @Muthu: Hearth is right. And I missed your question here, sorry! The thermal resistance is what is used to describe the transfer of heat from the component to its environment (air) or its heat-sink. It has nothing to do with the electrical resistance like RdsON, but it shows how easily will the component transfer its heat to whatever it is in physical contact with. The lower its thermal resistance, the more easily it loses its heat. The K/W value tells you how many degrees will the component temperature rise for each watt of heat being dissipated in that component. \$\endgroup\$ – Edin Fifić Apr 5 at 22:00

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