Revert to the theta junction to case and do your thermal math from there.

I'm sorry, I'm still new at this, what do you mean by that?

 

Basic thermodynamics: For every energy wasted in a transistor there is a heat caused. For every temperature difference there is a flow of heat to someplace else. Heat flow = temperature difference divided by thermal resistance. Theta JC is the thermal resistance from the tiny transistor to the outside of the case which contains it.

If power = 2 watts and the theta Junction to Air is 83 Centigrades per watt, that tiny transistor inside the case will heat up to 166 C hotter than the air around it. (That's a bad day for a transistor.)

If the theta JC is 3C/W that transistor will heat up by 6 centigrade at the case, assuming the case is connected to an infinite escape route, like you have it submerged in a bucket of water. Water isn't good for transistors, but you better add something for the case to attach to or the heat flow will stop and the tiny transistor will melt.

So...2 watts has to get through 3 C/W to get to the outside of the case and you attach the case to a heat sink labeled 5 C/W and the result is:

2 watts times (3C + 5 C) = 30 degrees C hotter than the surrounding air.

 

2 watts times (3C + 5 C) = 30 degrees C hotter than the surrounding air.

Are those some new variety of watts?

 

Are those some new variety of watts?

Those are centigrade degrees per watt as described in paragraphs 2, 3, and 4 and defined as thermal resistance in many text books. Sorry if I confused you by only defining the units 3 out of 4 times.

 

Is it possible for you to measure the voltage drop from drain to source under load? I bet if you put a scope on it you would see a lot of 'linear action' going on. It is probably apples to oranges, but I just connected a SMP50N06-25 MOSFET' drain lead to its gate lead and its source to ground and applied a constant current of .5 amps with 4v dropped across the transistor. It was dissipating 2 Watts internally and reached 120C within 5 mins. This device is rated to supply a 125W load. Obviously the watts cannot be dissipated internally.

marc

 

This device is rated to supply a 125W load. Obviously the watts cannot be dissipated internally.

Exactly the correct conclusion.

ps, you aren't allowed to tag on to the end of somebody else's conversation. The moderators will get you your own spot and you will get better answers.

 

Exactly the correct conclusion.

ps, you aren't allowed to tag on to the end of somebody else's conversation. The moderators will get you your own spot and you will get better answers.

Thanks for the info. Could you explain...I do not exactly understand what I did.

By the way do you know about this website?
http://www.petervis.com/electronics...nk/Heatsink_for_TO-220_Voltage_Regulator.html

marc

 

Basic thermodynamics: For every energy wasted in a transistor there is a heat caused. For every temperature difference there is a flow of heat to someplace else. Heat flow = temperature difference divided by thermal resistance. Theta JC is the thermal resistance from the tiny transistor to the outside of the case which contains it.

If power = 2 watts and the theta Junction to Air is 83 Centigrades per watt, that tiny transistor inside the case will heat up to 166 C hotter than the air around it. (That's a bad day for a transistor.)

If the theta JC is 3C/W that transistor will heat up by 6 centigrade at the case, assuming the case is connected to an infinite escape route, like you have it submerged in a bucket of water. Water isn't good for transistors, but you better add something for the case to attach to or the heat flow will stop and the tiny transistor will melt.

So...2 watts has to get through 3 C/W to get to the outside of the case and you attach the case to a heat sink labeled 5 C/W and the result is:

2 watts times (3C + 5 C) = 30 degrees C hotter than the surrounding air.

Thanks a lot for taking the time to write this, very helpful.

Meanwhile, I've found a couple of FQP33N10 and built the double MOSFET circuit with 150 ohm gate resistors and 10K pull down gate resistors.
It sort of works, the issue however is that the MOSFET temps vary from 60C to 110C and I didn't let it work for very long.
I put small heat sinks on the MOSFETs and the diode this time. Tomorrow I want to get some more high amp MOSFETs and one large heat sink, hope it'll work...

 

@Flug540: Look at post #47. Good link about heat and heat sinks. When you can do the math, you don't have to hope. You will KNOW.
@marcf: Flug540 owns this conversation (rules of the website). You are allowed to answer him, but you are supposed to ask your own questions in a new thread. Otherwise we get too many cooks in the kitchen and it all goes crazy.

 

We gotta teach you how to read a data sheet.
Take a look at the FQP33N10 and you will see Rds on is .053 ohms. This is the resistance of the FET when it is turned on "hard". Meaning gate to source voltage is 10 volts.
So now you can calculate the power in the FET just like you did the resistor. I^2 X R. I think you measured 22 amps and 2 volts across the resistor. So your fan probably draw 25 amps if you remove the resistor. So 25 squared time .053/2 (since you have 2 in parallel they act like parallel resistors). 16 watts is being dissipated in the FET. If you look a little further down the data sheet you will see a couple of lines called thermal characteristics. The one labeled junction to ambient is how hot it will get without a heat sink. In this case 62.5C per watt. 16 X 62.5 - well it won't work. The ratings can be misleading because they assume a temperature of 25C, so you have to go thru all this derating stuff.
The other thing you need to be careful with in your application is the gate to source voltage. Most FETs will turn on good with 10 volts from gate to source, but you only have 5 volts, so you need what is commonly called a logic level FET where it will turn on with 5 volts or less.
So try the same thing on the one I posted earlier.

 

We gotta teach you how to read a data sheet.
Take a look at the FQP33N10 and you will see Rds on is .053 ohms. This is the resistance of the FET when it is turned on "hard". Meaning gate to source voltage is 10 volts.
So now you can calculate the power in the FET just like you did the resistor. I^2 X R. I think you measured 22 amps and 2 volts across the resistor. So your fan probably draw 25 amps if you remove the resistor. So 25 squared time .053/2 (since you have 2 in parallel they act like parallel resistors). 16 watts is being dissipated in the FET. If you look a little further down the data sheet you will see a couple of lines called thermal characteristics. The one labeled junction to ambient is how hot it will get without a heat sink. In this case 62.5C per watt. 16 X 62.5 - well it won't work. The ratings can be misleading because they assume a temperature of 25C, so you have to go thru all this derating stuff.
The other thing you need to be careful with in your application is the gate to source voltage. Most FETs will turn on good with 10 volts from gate to source, but you only have 5 volts, so you need what is commonly called a logic level FET where it will turn on with 5 volts or less.
So try the same thing on the one I posted earlier.

Wow, thank a lot!

So, let's say I have one PSMN1R1-30PL:
Rdson = 0.0012 ohm (for ambient slightly higher then 25C)
P = 25^2 * 0.0012 = 0.75
θJA = 60
OpTemp = 60 * 0.75 = 45C
Is that correct?

One thing that confuses me is that it says "Suitable for logic level gate drive sources", but later in the sheet they use Vgs = 10V, why is that?

 

@Flug540: Look at post #47. Good link about heat and heat sinks. When you can do the math, you don't have to hope. You will KNOW.
@marcf: Flug540 owns this conversation (rules of the website). You are allowed to answer him, but you are supposed to ask your own questions in a new thread. Otherwise we get too many cooks in the kitchen and it all goes crazy.

Thanks, I will do this from now on...
Marc

 

Now your cookin!
Don't forget to add the 25C for the room temperature. So it will still be hot, but will work.
The FETs perform better with gate voltages up to about 10 or 12 volts so they use the value that makes them look best.
Here is a handy tidbit.
To figure out what a small heat sink will do you can use 50 / square root of the surface area in cm. So if you attached your FET above to a 2X2 cm piece of copper it would have a thermal resistance of about 18C per watt. Then the FET would be cool.

 

Now your cookin!
Don't forget to add the 25C for the room temperature. So it will still be hot, but will work.
The FETs perform better with gate voltages up to about 10 or 12 volts so they use the value that makes them look best.
Here is a handy tidbit.
To figure out what a small heat sink will do you can use 50 / square root of the surface area in cm. So if you attached your FET above to a 2X2 cm piece of copper it would have a thermal resistance of about 18C per watt. Then the FET would be cool.

Awesome, thanks a lot!
Would you recommend using two or even three FETs in parallel, or just add a heat sink?
In my environment I need to account for ambient temps as high as 45C, although in average it'll be probably around 25C.

 

We gotta teach you how to read a data sheet.
Take a look at the FQP33N10 and you will see Rds on is .053 ohms. This is the resistance of the FET when it is turned on "hard". Meaning gate to source voltage is 10 volts.
So now you can calculate the power in the FET just like you did the resistor. I^2 X R. I think you measured 22 amps and 2 volts across the resistor. So your fan probably draw 25 amps if you remove the resistor. So 25 squared time .053/2 (since you have 2 in parallel they act like parallel resistors). 16 watts is being dissipated in the FET. If you look a little further down the data sheet you will see a couple of lines called thermal characteristics. The one labeled junction to ambient is how hot it will get without a heat sink. In this case 62.5C per watt. 16 X 62.5 - well it won't work. The ratings can be misleading because they assume a temperature of 25C, so you have to go thru all this derating stuff.
The other thing you need to be careful with in your application is the gate to source voltage. Most FETs will turn on good with 10 volts from gate to source, but you only have 5 volts, so you need what is commonly called a logic level FET where it will turn on with 5 volts or less.
So try the same thing on the one I posted earlier.

Another thing I just noticed... In the case of two parallel FETs shouldn't we divide the current as well?
So it would be 12.5^2*0.026=4W, still too much for FQP33N10 though.

 

Another thing I just noticed... In the case of two parallel FETs shouldn't we divide the current as well?
So it would be 12.5^2*0.026=4W, still too much for FQP33N10 though.

Also, looking at the FQP33N10 data sheet indicates current will begin to go into linear operation at around 11 Amps with a Vgs to 5v. I would expect the device to start really going linear at 12A and tend to warm up. Also be aware that the gate of this device has a capacitance of around 0.0015 uf. Looking at the arduino web site, they indicate the default frequency for pwm is 32Kz (not sure of rise time). A input resistance of 1k would cause some (not much) rounding of the output at 32khz. Again causing some linear operation. Avoid any series resistance and use low value (10k ) gate to drain pulldown resistors. Any liner operation in the input around the threshold voltage (4v) would also cause rather dramatic changes in Rds (increasing it), causing further heating.

The goal is to cause the input voltage to go from Min_V to Max_V and Max_V to Min_V, instantaneously, and use a device with a Rds of zero.

If you could do this, you would have a low cost superconducting device and be a multibillionaire.

I would suggest finding a device that would be capable of an Ids of 20 amps with a Vgs of 5volts, check out its thermal resistance specs and use a heatsink with an appropriate thermal resistance.

Look at the specs for a SMP50N06-25 device

Just because a device may operate at 125C, does not necessarily mean you should let it. 125C is 70F above boiling and I am sure can cause serious pain. There is a rule of thumb for hot things. 140F is the point where people go from feeling something is hot to feeling pain and 2nd degree burns. I always try to set my hot water heater thermo to 140F just for this reason.

marc

 

Usually heat sinks are cheaper than more FETs.

Another thing I just noticed... In the case of two parallel FETs shouldn't we divide the current as well?

That would be like... well cheating.
While it is true each would have 1/2 the current each would still have it's original resistance. I always found it easier to just cut the resistance in half, but you can do it either way.

 

Also, looking at the FQP33N10 data sheet indicates current will begin to go into linear operation at around 11 Amps with a Vgs to 5v. I would expect the device to start really going linear at 12A and tend to warm up. Also be aware that the gate of this device has a capacitance of around 0.0015 uf. Looking at the arduino web site, they indicate the default frequency for pwm is 32Kz (not sure of rise time). A input resistance of 1k would cause some (not much) rounding of the output at 32khz. Again causing some linear operation. Avoid any series resistance and use low value (10k ) gate to drain pulldown resistors. Any liner operation in the input around the threshold voltage (4v) would also cause rather dramatic changes in Rds (increasing it), causing further heating.

Thanks for the analysis, marc. It's now clear that FQP33N10 is the wrong part to use for my device, but out of curiosity, which graph were you looking at (trying to understand what you mean exactly by going linear)?
Also, do you mean I should not use any resistor between Arduino PWM output and the gate?

Look at the specs for a SMP50N06-25 device

I looked at the specs of this FET, but accordingly to the power and temp estimations suggested by ronv, it will not survive the task

 

Usually heat sinks are cheaper than more FETs.


That would be like... well cheating.
While it is true each would have 1/2 the current each would still have it's original resistance. I always found it easier to just cut the resistance in half, but you can do it either way.

But it isn't the same though is it, since the current is squared?

 

I think it comes out the same. Using parallel resistors: 25^2 X .028 = 17.5 watts. 1/2 in each FET = 8.75 watts per FET.
Using 1/2 current 12.5^2 X .056 = 8.75 watts in each FET.
You can halve 1 but not both.