How to calculate accurate Mosfet power dissipation.

Thread Starter

laco22

Joined Oct 31, 2017
18
Hello.

I am studying a commercial product of LED DIMMER controlled based on PWM.

The product can dim the LED with the following specifications.

Maximum Spec : 12VDC, 50A, 600W

And as the core device, SM6020NS Mosfet is used. (http://www.sinopowersemi.com/temp/SM6020NSFP_datasheet.pdf)


I have a question here.

Q1. The device can dissipate up to 83W of power at a case temperature of 25 degrees,
But, The device's Rds(on) is 4.5m Ohm.
And, Mosfet power consumption formula I think is as below.


I = 50A
R = 0.0045 Ohm

W = V * I

V = I * R

V = 0.225V

W = 11.25W


Why does the manufacturer of the product not use a product that can only consume 20W, but use a device that can consume 83W?

Q2. How can you help me with Thermal Resistance-Junction to Case or Thermal Resistance-Junction to Case?

Q3. The picture below is a commercial product that I am currently studying.
I would like to make a product like this myself.
Can you give me a link to find a starting point for me to study?

WIN_20210523_17_50_28_Pro.jpg
 

Ian0

Joined Aug 7, 2020
2,192
Why does the manufacturer of the product not use a product that can only consume 20W, but use a device that can consume 83W?
So that it runs cooler and lasts longer.
can you help me with Thermal Resistance-Junction to Case or Thermal Resistance-Junction to Case?
Add up all the thermal resistances between junction and ambient and multiply by the power dissipation. That gives you the difference in temperature between junction and ambient.
Using a sensible estimate of ambient temperature allows you to estimate the junction temperature.

The power dissipation is the sum of the conduction losses (which you have correctly calculated above) and the switching losses, which approximate to 0.5 * V * I * switching_time * frequency.

You also should use the value of Rds(on) for the temperature of the MOSFET. It might be twice as much at 70°C
 

MrAl

Joined Jun 17, 2014
8,241
Hello.

I am studying a commercial product of LED DIMMER controlled based on PWM.

The product can dim the LED with the following specifications.

Maximum Spec : 12VDC, 50A, 600W

And as the core device, SM6020NS Mosfet is used. (http://www.sinopowersemi.com/temp/SM6020NSFP_datasheet.pdf)


I have a question here.

Q1. The device can dissipate up to 83W of power at a case temperature of 25 degrees,
But, The device's Rds(on) is 4.5m Ohm.
And, Mosfet power consumption formula I think is as below.


I = 50A
R = 0.0045 Ohm

W = V * I

V = I * R

V = 0.225V

W = 11.25W


Why does the manufacturer of the product not use a product that can only consume 20W, but use a device that can consume 83W?

Q2. How can you help me with Thermal Resistance-Junction to Case or Thermal Resistance-Junction to Case?

Q3. The picture below is a commercial product that I am currently studying.
I would like to make a product like this myself.
Can you give me a link to find a starting point for me to study?

View attachment 239364
If you see an over qualified MOSFET in a product count your blessings because this is not always typical. For a long time i have wanted to see better MOSFETs in simple type switching regulator IC's with built in power MOSFETs. As of late we have been seeing this finally after many years of seeing poor built-in power devices that limited efficiencies to some 80 percent when 90 percent is possible and would make a very attractive product. It could be that global energy concerns changed the trend over the years.

The MOSFET used in this device seems to be fairly decent though also keep in mind that 5mOhms is quite small and is going to be comparable to wire resistances and possibly even some connection resistances so there will be other losses too.

Another loss that it looks like Ian has mentioned is the switching loss. When the transistor (mosfet, bipolar, igbt, etc.) turns on and turns off there is a transient period where we will see both high levels of current through the transistor and high levels of voltage across the transistor meaning that there will be brief periods of time when there is much higher power dissipation. Luckily, it only lasts for a fraction of the entire switching period. This is because as the transistor turns on the voltage across it starts to decrease as the current starts to increase and so somewhere in the middle we might see half the full voltage across the device at the same time we see half the current through the device. This of course means a large amount of power is absorbed by the transistor for that short period of time.
Because this happens for only part of the cycle (once turning on and once turning off) the average power is usually much lower, as long as the transient switching interval is short compared to the entire cycle.
An approximation often used is that the voltage rise (or fall) and the current fall (or rise) is linear in that it starts at zero and ends at the maximum or it starts at the maximum and ends at zero, either way as a straight line. This creates a power profile that is quadratic. As a rough guide, if the transient time is 1/10 of the entire switching time and the max voltage is 12 volts and max current is 50 amps then the extra power dissipation will be about 20 watts because of the on and off transient periods where the device is in an intermediate state partially on and partially off.
I think it would be unusual to see a transient period longer than that so that could be taken as a maximum. There are however also losses associated with the gate drive which could add another 2 watts.

The main point however is probably that the manufacturer wanted to make a product that was not going to eat up too much power because that makes the product look bad. Back when i worked in the power product industry the trend was to try to get 90 percent efficiency and that was quite a long time ago so now i would think that would be more of a standard and it may even be written down somewhere. You may have noticed that the efficiency of regular small 10-ish watt wall warts has risen too in an effort to reduce global energy waste and of course we all know the laws that came into play over the use of very inefficient incandescent light bulbs.

We can go over the derivation of the switching power calculation if you wish. It's not all that difficult. It is just an averaging of the power during the transient switch period using the approximation above. If you want better than that then we would have to dig into the more exact switching waveforms.
 
Last edited:

andrewmm

Joined Feb 25, 2011
1,466
So this sort of circuit,
if you want to make it as a commercial product, be prepared for a few thousand dollars of qualification / approvals, unless your in the far east ,,, when it seem fine just to stick a pass lable on the product !

also be aware, mains can kill.

So warnings over,

The general way to "solve" this sort of question , is by trial and error.
The reason is the parts are cheap, easy to get hold of, easy to modify the circuit, and to test, and the simulation to come up with a real good result is very problematic.


For instance, you can trade off the resistance and thermal resistance of the part, against the cost of the part and the amount of heat sinking you have,

A bigger,, lower resistance part with lower thermal resistance might be cheaper than the basic part calculated due to volume, and the smaller heat sink needed.

So play ,

Just remember , not to measure temperature with your finger
a) The thing can be at mains volts,
b) a finger stuck to the hot part and burning is real painful for a long time .

yes I have the T shirt...
 

Thread Starter

laco22

Joined Oct 31, 2017
18
So that it runs cooler and lasts longer.

Add up all the thermal resistances between junction and ambient and multiply by the power dissipation. That gives you the difference in temperature between junction and ambient.
Using a sensible estimate of ambient temperature allows you to estimate the junction temperature.

The power dissipation is the sum of the conduction losses (which you have correctly calculated above) and the switching losses, which approximate to 0.5 * V * I * switching_time * frequency.

You also should use the value of Rds(on) for the temperature of the MOSFET. It might be twice as much at 70°C
I remember you.

Thank you very much for the answer.

Thank you for helping me understand correctly.
I will study in more detail based on what you said.

Thank you.
 

Thread Starter

laco22

Joined Oct 31, 2017
18
If you see an over qualified MOSFET in a product count your blessings because this is not always typical. For a long time i have wanted to see better MOSFETs in simple type switching regulator IC's with built in power MOSFETs. As of late we have been seeing this finally after many years of seeing poor built-in power devices that limited efficiencies to some 80 percent when 90 percent is possible and would make a very attractive product. It could be that global energy concerns changed the trend over the years.

The MOSFET used in this device seems to be fairly decent though also keep in mind that 5mOhms is quite small and is going to be comparable to wire resistances and possibly even some connection resistances so there will be other losses too.

Another loss that it looks like Ian has mentioned is the switching loss. When the transistor (mosfet, bipolar, igbt, etc.) turns on and turns off there is a transient period where we will see both high levels of current through the transistor and high levels of voltage across the transistor meaning that there will be brief periods of time when there is much higher power dissipation. Luckily, it only lasts for a fraction of the entire switching period. This is because as the transistor turns on the voltage across it starts to decrease as the current starts to increase and so somewhere in the middle we might see half the full voltage across the device at the same time we see half the current through the device. This of course means a large amount of power is absorbed by the transistor for that short period of time.
Because this happens for only part of the cycle (once turning on and once turning off) the average power is usually much lower, as long as the transient switching interval is short compared to the entire cycle.
An approximation often used is that the voltage rise (or fall) and the current fall (or rise) is linear in that it starts at zero and ends at the maximum or it starts at the maximum and ends at zero, either way as a straight line. This creates a power profile that is quadratic. As a rough guide, if the transient time is 1/10 of the entire switching time and the max voltage is 12 volts and max current is 50 amps then the extra power dissipation will be about 20 watts because of the on and off transient periods where the device is in an intermediate state partially on and partially off.
I think it would be unusual to see a transient period longer than that so that could be taken as a maximum. There are however also losses associated with the gate drive which could add another 2 watts.

The main point however is probably that the manufacturer wanted to make a product that was not going to eat up too much power because that makes the product look bad. Back when i worked in the power product industry the trend was to try to get 90 percent efficiency and that was quite a long time ago so now i would think that would be more of a standard and it may even be written down somewhere. You may have noticed that the efficiency of regular small 10-ish watt wall warts has risen too in an effort to reduce global energy waste and of course we all know the laws that came into play over the use of very inefficient incandescent light bulbs.

We can go over the derivation of the switching power calculation if you wish. It's not all that difficult. It is just an averaging of the power during the transient switch period using the approximation above. If you want better than that then we would have to dig into the more exact switching waveforms.
Thank you for investing a lot of time for me and responding.

I'm not an English speaking person, so I don't understand everything right now, but I'll understand and learn correctly.

Thank you very much for the answer.

If there is anything I don't know, I'll ask you additional questions in that post.
 

Thread Starter

laco22

Joined Oct 31, 2017
18
So this sort of circuit,
if you want to make it as a commercial product, be prepared for a few thousand dollars of qualification / approvals, unless your in the far east ,,, when it seem fine just to stick a pass lable on the product !

also be aware, mains can kill.

So warnings over,

The general way to "solve" this sort of question , is by trial and error.
The reason is the parts are cheap, easy to get hold of, easy to modify the circuit, and to test, and the simulation to come up with a real good result is very problematic.


For instance, you can trade off the resistance and thermal resistance of the part, against the cost of the part and the amount of heat sinking you have,

A bigger,, lower resistance part with lower thermal resistance might be cheaper than the basic part calculated due to volume, and the smaller heat sink needed.

So play ,

Just remember , not to measure temperature with your finger
a) The thing can be at mains volts,
b) a finger stuck to the hot part and burning is real painful for a long time .

yes I have the T shirt...
Thank you for answer.

I will keep in mind your list of cautions.

I thought similarly to you.

"Can I make it with only the Internet and my thoughts without trying it myself?"

After all, it is the fastest way to go through trial and error yourself in any field.

Also, you tell me that these are relatively easy to part and modify, so I'd like to try them out right away.

Thank you for saying me about this.

I will not touch it with my hand, I will measure it with the measuring equipment.

Thank you.
 
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