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Confirm Mosfet Power Dissipation Calculation

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LittleRain:
Can someone please confirm if I have calculated the power dissipation for a mosfet correctly?
I will be using an N-channel mosfet on high side for overvoltage/reverse polarity, with a boost converter driving the gate.
The part I used to test the formula is the AO3400.

This is the formula and values I used. Not sure if everything correct.


--- Code: ---I = Current = 5A
T = TJ(HOT) = 150°C
R = RDS(on) = 52 / 1000 (52mΩ)
S = TSPEC = 25°C
H = RDS(on)HOT = 1 + R * (T - S) [i]or R * 1.7[/i]
V = VIN = 3.3V
O = VOUT = V - (I *R)

P = PDResistive = ((I * I) * H) * ( 1 - (O / I)) = 0.166439W
--- End code ---

What I'm really unsure about is if I used the correct values for TJ(Hot), resulting in RDS(on)HOT also being incorrect.
Does everything check out?

Edit: Updated the equation.

T3sl4co1l:
An easy way to spot certain kinds of errors: attach a unit to everything, and follow the units through as you would any other algebraic factor.

Any time you have addition, the units must match.  If they do not, a conversion factor, or multiplier or ratio, is needed to get them to match.

Multiplication, division and exponentiation work as normal: you can have a unit raised to whatever power, including negative (flip sign if it's above/below the division line) or fractional (e.g. 0.5 i.e. sqrt).  All that matters is that it's done consistently, and the above condition holds.

Also, that whatever unit you get out, is somehow useful.

Which solves a whole heck of a lot of physics just as it is: how much weight does 1kg exert?  Well, 1kg is mass, it's not force.  We need a conversion factor to force.  F = ma tells us we need acceleration.  If it's [seemingly] unaccelerated sitting on the Earth, well a = 9.8 m/s^2, and we get kg m/s^2 == N, a unit of force, bingo.  And if we have several, well we can't just sum up the masses and forces willy-nilly, but we can take the sum of all one, or all the other, and convert to mass or weight as needed.

So here, you've got "R = RDS(on) = 52 / 1000 (52mA)", which I should assume the mA is just a typo for mOhm (mΩ if you like to be proper).  But lower down you have,
H = RDS(on)HOT = 1 + R * (T - S)
and presumably T and S are in °C (well, S is in C, but T isn't in anything at all!), so H is in units of ohm °C, which, isn't very useful, and definitely can't be multiplied by I^2 to get power.  (The 1 - (O/I) expression is a nice dimensionless ratio, no worries there.  Which, note there's a '1' in the above expression, which needs suitable units as well!)

So we can easily spot the problem: you've got temperature without a conversion factor to resistance.

And the "1 + R" makes it look like it was supposed to be a normalized change, which should actually be written: R' = R * (1 + normalized_change).  But most of the time they give normalized value, not change from nominal, so it's just R' = R * (value on the plot at temp T).  This is given in Figure 4.

Easier to not fiddle with a stack of equations and just pick a likely Rds(on) from the temp curve.  It's nonlinear anyway -- it increases faster at higher temperature -- so a simple (linear) arithmetic expression can't capture the details of it, and in general you'll need to solve for an initial value, then keep testing it against the curve, iteratively, until the result matches (that is, going between P and R, to solve for P and T together).  (On the upside, it's a fairly flat curve for this particular part -- lower voltage FETs have less bend in this curve -- and a linear fit wouldn't actually be bad at all.  But a fitted value, it is -- a magic number, a conversion factor, or calibration if you will, between T and R.  And so it has the combo units of Ω/°C.)

You also need to account for Rds(on)max, which is given in the data tables but only for certain points on the temp curve (usually 25C and Tjmax?).  Presumably the tempco is the same for nominal or max Rds(on)s, so you can use the same ratio (usually it's given as normalized resistance, not normalized change, so the curve is 1.0 at 25C and is simply the given multiple of Rds(on)nom elsewhere -- no "1+" needed).

Max is due to manufacturing variation between parts; you aren't likely to get one quite so bad, the average case really is quite typical -- and maybe it's worth rejecting them if they do perform so badly -- but you'll need to do that test to find out, so if you want a streamlined production test without full current and power and temperature, just design with max in mind.


(Another 'dimensional analysis' fact: most functions are dimensionless.  Or some special kind of ratio dimension, if you like.  e^x works with dimensionless x.  You can think of sin x as using units of radians, since its argument is angular.  They return dimensionless numbers -- pure ratios.  So you can have a sine wave y = sin x, in the abstract -- but a real wave that you measure, might be better expressed as: v(t) = A * sin(2π * (t * F + φ / 360)), with A == (V) (read: "in units of"), 2π == (rad/cycle), F == (cycles/s) and φ == (°), for t == (s) and v == (V).)

Tim

LittleRain:
Thanks you very much for the informative answer, it is much appreciated.
I was in the middle of writing a proper reply, but I keep falling asleep at my desk.
So I'm going to have to come back to this and respond tomorrow after a good nights sleep. :=\

Faringdon:
Your Vin is only 3.3V and you only have 5A....you  could use a 15v/20V SOT23 NFET and get much lower rdson.......rdson is never more than twice the 25degc value.

--- Quote --- come back to this and respond tomorrow after a good nights sleep
--- End quote ---
To sleep better.....reduce sugar intake (eg cakes, candy) and eat plenty of fibre, eg raw apples and carrots......never eat before bed..unless its high fibre food.
As an EE, sat at desk often, you(all of us)  will burn out if you dont follow the sleep advice.

To cool the SOT23....have it on the biggest pads possible...ie commensurate with your PCB assemblers maximum allowed pad size.....and sneak it up a bit more because they always ask for too small pad sizes......and the Standard ones you can download  are too small.....its amazing how much cooler a SOT23 will run if you increase the pad copper area by just 30%.

LittleRain:
I have updated the original post with the revised formula.
It seems extremely low, so I'm not sure if its correct.


--- Quote from: T3sl4co1l on May 10, 2022, 12:42:55 am ---So here, you've got "R = RDS(on) = 52 / 1000 (52mA)", which I should assume the mA is just a typo for mOhm (mΩ if you like to be proper).  But lower down you have,
H = RDS(on)HOT = 1 + R * (T - S)
and presumably T and S are in °C (well, S is in C, but T isn't in anything at all!), so H is in units of ohm °C, which, isn't very useful, and definitely can't be multiplied by I^2 to get power.  (The 1 - (O/I) expression is a nice dimensionless ratio, no worries there.  Which, note there's a '1' in the above expression, which needs suitable units as well!)

So we can easily spot the problem: you've got temperature without a conversion factor to resistance.

And the "1 + R" makes it look like it was supposed to be a normalized change, which should actually be written: R' = R * (1 + normalized_change).  But most of the time they give normalized value, not change from nominal, so it's just R' = R * (value on the plot at temp T).  This is given in Figure 4.

Easier to not fiddle with a stack of equations and just pick a likely Rds(on) from the temp curve.  It's nonlinear anyway -- it increases faster at higher temperature -- so a simple (linear) arithmetic expression can't capture the details of it, and in general you'll need to solve for an initial value, then keep testing it against the curve, iteratively, until the result matches (that is, going between P and R, to solve for P and T together).  (On the upside, it's a fairly flat curve for this particular part -- lower voltage FETs have less bend in this curve -- and a linear fit wouldn't actually be bad at all.  But a fitted value, it is -- a magic number, a conversion factor, or calibration if you will, between T and R.  And so it has the combo units of Ω/°C.)

--- End quote ---

You are correct that is a typo, I originally had it written as mohm, made some changes and accidentally put mA.

Ok that makes a lot of sense, I was following this technical document and the use of square brackets and round brackets was confusing me as to if RDS(ON)SPEC was included in the equation.
I suppose if I had known the rules you mentioned I would have realized it would create a useless value (omh/°C).
Re-reading the document it is very obvious that it was required, as almost all of the other equations include square brackets. I'm going to have to blame lack of sleep on that one.

This is the equation for H from the document.
RDS(ON)HOT = RDS(ON)SPEC [1 + 0.005 × (TJ(HOT) - TSPEC)]
For some reason I replaced 0.005 with R. I think it may have been that I didn't understand what the 0.005 was, and it being a value fairly close to R I ended up making a stupid assumption that it was an example for 5mΩ.

So I have changed the H equation to this.
H = R * (1 + 0.005 * (T - S))
The result checks out, as it is almost exactly the same as your advice to just multiple R by the value in figure 4 (0.0845 vs 0.0884).
Maybe I just got lucky because of the linearity of the part.

So now P equals 0.1664... Is that really all the power that is being dissipated, it seems extremely low to me.
If I just take the voltage drop (0.26V) and apply ohm's law using 5A I get 1.3W.
I originally was going to use TO-252 part, but if it is really this low I think I may actually use a SOT part.

If it is still incorrect, could you please show me the correct answer so I can check my result to it?


Oh and thanks again, there are a lot of very helpful tips here.

I hope to one day have math skills such as yours.
Math used to be my best subject in school, and I used to be able to do multiplication and division quickly in my head.
Then 7 or so years ago I ended up in a coma, and ever since then my math skills have been lacking.
Remembering 2 numbers at once is tough for me now.
Weirdly enough my memory is still great, writing programs I can remember where and what every function does and its name, its just if I have to remember more than 1 number at a time.
Can't blame that for being bad at algebra though, that's just because I haven't really done any in 15 years.
Only really started to again when I started working with FPGAs and battery operated circuits, when I needed to start worrying about power consumption.


--- Quote from: T3sl4co1l on May 10, 2022, 12:42:55 am ---Max is due to manufacturing variation between parts; you aren't likely to get one quite so bad, the average case really is quite typical -- and maybe it's worth rejecting them if they do perform so badly -- but you'll need to do that test to find out, so if you want a streamlined production test without full current and power and temperature, just design with max in mind.

--- End quote ---
Yes I have been designing with max in mind, I added all the IC's maximum current draw which ends up being around 5amps, as well as using the 2.5V RDS(on) at 150C.
The actual circuit will most likely require much less than half, as I made a generous estimate for the FPGA.


--- Quote from: Faringdon on May 10, 2022, 01:10:25 pm ---Your Vin is only 3.3V and you only have 5A....you  could use a 15v/20V SOT23 NFET and get much lower rdson.......rdson is never more than twice the 25degc value.

--- End quote ---
Just to confirm you mean 15/20V VDS?
I will take a look, but for 50 units for about $1 CAD excluding shipping its hard to beat.

I originally wasn't going to use this part, I had planned to use a TO-252, but now that I realize the power dissipation is a lot less than I thought it was I probably will use a SOT-23.


--- Quote from: Faringdon on May 10, 2022, 01:10:25 pm ---To cool the SOT23....have it on the biggest pads possible...ie commensurate with your PCB assemblers maximum allowed pad size.....and sneak it up a bit more because they always ask for too small pad sizes......and the Standard ones you can download  are too small.....its amazing how much cooler a SOT23 will run if you increase the pad copper area by just 30%.

--- End quote ---

Yes I am aware of that thanks.
A couple days ago I was looking at a SOT-23 that could dissipate up to 3 watts, but while reading the datasheet I found out it required a full square inch of copper to be able to handle 3 watts, gave me a chuckle.
I assume they were trying to abuse the parametric search, who knows.
Not sure why anyone wouldn't just upgrade to a larger package to save a ton of space at that point.
But I guess if you were already using it else where, and had a lot of extra board space, like on a UI board with buttons and switches that were far apart, it could make sense to keep part count down.
It would actually be nice if datasheets included something like that more often, but when it shows up in a parametric search you are under the impression it is for a standard package.


--- Quote from: Faringdon on May 10, 2022, 01:10:25 pm ---To sleep better.....reduce sugar intake (eg cakes, candy) and eat plenty of fibre, eg raw apples and carrots......never eat before bed..unless its high fibre food.

--- End quote ---
Yeah I don't eat much sugar at all, only decaf coffee as well. If I even eat a small piece of candy I'll be up until 5am.
I always eat a bowl of raisin bran before bed, the sugar in the raisins probably isn't the best.

Thanks though, it is good advice for many.

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