Hi all,
I'm watching to drive 64 led in parallel with a mosfet and I see the AO3401...I do some calculations and 64 leds in parallel consume 20mAx64=1.28 A so.....this P-channel mosfet has a low Rds(on),in the worst case 120 mOhm, so the power dissipation on this mosfet is Rds(on)x (Id)² = 196.6 mW only....is it possibile? in this case with only TO-263 package I can drive 1.28 A probably with low heat. Is possible? Thank you in advise
I ma not good at maths like this, but i drove single 10W LED diode with one MOSFET in TO-220 package. It required heatsink about 10 x 10 x 3cm and never become hot.
BTW. what color of leds are you driving?
So i guess that this is possible. But be carefull, TO-263 package is quite dependent to good solede in order to properly cooling.
Helium328PU:
BTW. what color of leds are you driving?
green leds
the power dissipation of ao3401 is 1,4 W at 25°C with 1 inch² on FR-4 board so with this calculations there will be no problem but in my opinion it seems that 1.28 A is a big current for this small package...but I never try this on practice
so the power dissipation on this mosfet is Rds(on)x (Id)² = 196.6 mW only....is it possibile?
Yes but that is only the static dissipation. That is the dissipation when the FET is on. During the time when the FET turns on there will be more as the FET resistance goes from very high to very low.
This becomes an issue when you use PWM to control the brightness because you are switching it on an off several hundred times a second. This is much harder to calculate and you need to know things like the rise time of your driving signal under the capacitive load that is the gate.
Grumpy_Mike:
During the time when the FET turns on there will be more as the FET resistance goes from very high to very low.
...
This is much harder to calculate and you need to know things like the rise time of your driving signal under the capacitive load that is the gate.
Forgot to mention that 
As Grumpy_Mike told really nicely. Its harder to calculate. You have to count percentage when FET is on and off. Not easy realy, as these components arent linear.
If you dont have good practise with calculating this, better to use common sense and start prototyping with DIP MOSFETS with huge heatsing. This is a way how I am doing it when douing PWM controls for 10W leds
Once circuit is on, watch temperature on heatsing and use smaller if cool enough 
Then you can move to smaller packages.
Grumpy_Mike:
This is much harder to calculate and you need to know things like the rise time of your driving signal under the capacitive load that is the gate.
and how can I calculate the dynamic dissipation?
That sounds reasonable to me. I wouldn't even bother with the calculations for the increase when using PWM; anything below 1Mhz and your total power dissipation should be well under 500mW. I doubt you'll even need a heatsink.
ralphd:
That sounds reasonable to me. I wouldn't even bother with the calculations for the increase when using PWM; anything below 1Mhz and your total power dissipation should be well under 500mW. I doubt you'll even need a heatsink.
Then you'll be surprised. MOSFET gates are highly capacitive and unless you drive them
hard they might take several microseconds to turn of or off, which ain't looking good for
your 1MHz figure.
Drive them hard and it might be a nice snappy 50ns switch on or switch off, but that
means large gate currents - then you have to worry about dissipation in the gate
resistance...
Basically you have to do the maths for the particular MOSFET and drive circuit and load.
Stick to 1kHz PWM and you'll be fine of course.
The OP said the AO3401 has an on-resistance of 120mOhm, but its a worst-case of
60mOhm in fact. Since its a small MOSFET its effective gate capacitance is quite
low at 1.4nF, larger devices can be much more.
MarkT:
ralphd:
That sounds reasonable to me. I wouldn't even bother with the calculations for the increase when using PWM; anything below 1Mhz and your total power dissipation should be well under 500mW. I doubt you'll even need a heatsink.Then you'll be surprised. MOSFET gates are highly capacitive and unless you drive them
hard they might take several microseconds to turn of or off, which ain't looking good for
your 1MHz figure.
No, I wouldn't be surprised because I'm not suggesting 999Khz, although for the pedantic it is "below 1Mhz". Plus it would take some work to get 1Mhz PWM output from an Ardunio board. With the 8bit counter for PWM and a 16Mhz clock, the PWM frequency would be 62.5Khz.
And since you're being pedantic, I'll point out that there's lots of MOSFETs available that have fast (10's to a few hundred ns) switching time with modest drive requirements (i.e. ~5V). Just look on any PC motherboard from the last 10 years.
The first datasheet I pulled up for fast swtiching MOSFETs was this one:
10ns rise time, 10ns fall time with 4.5V gate drive.
First of all check the datasheet to find out whether the MOSFET supports that much of current. To make sure that the MOSFET dissipates minimum energy, make sure that you operate it in the saturation region i.e apply a gate voltage that will turn it on completely. Any gate voltage below that will increase the MOSFET resistance and hence power dissipation.
Thanks all for answers..
so from datasheet the mosfet provide a max of 4.3 A so for current is ok...the static dissipation from the calculation below is correct and only ~200mW the problem is the formula to calculate the dynamic dissipation...I don't want random to it...anyone know a formula to calculate it...maybe a simplified formula thanks
this is the datasheet : AO3401 pdf, AO3401 Description, AO3401 Datasheet, AO3401 view ::: ALLDATASHEET :::
ralphd:
The first datasheet I pulled up for fast swtiching MOSFETs was this one:
http://www.vishay.com/docs/73216/si7108dn.pdf
10ns rise time, 10ns fall time with 4.5V gate drive.
Those figures are only the rise and fall times of the FET when fed with a 'perfect' signal. What you have to take into account is the real rise and fall time you are going to get when you apply a driving voltage to the gate with real impedance.
geppou:
anyone know a formula to calculate it...maybe a simplified formula thanks
Sorry there is no simple formula, as I said you need to know the real rise time you will be applying not what the data sheet says.
what is your "constant current generator" in that first diagram.
Can you have a mosfet and then leds and then a "constant current generator", all between 5V and 0V ??
michinyon - you are right. I was too busy answering the asked question to pay much attention to the schematic.
No you can not power LEDs in parallel like this. For a start the LEDs will not share current you need some form of current regulation for each one. So that would mean 64 constant current regulators if you want to go down that route. But with 20mA LEDs there is no point, resistors will do instead.
You are also better using a P-channel FET if you want to top switch, or an N-channel FET for bottom switching.
I put the costant current generatore but Yes you are rigth because is 20 ma current generator x64 leds in practice I use 4 TLC5940 that sink 20 mA for each led..sorry for the incorrect diagram.
but the pick of power dissipation in this case that I switch from 0V to 5V is possible 1.28 A for 5V so 6.4W?
this value is out of the value declared on datasheet but the dissipation depend of the frequency of switch because if I switch rapidly the mosfet must support more pick of current than a low frequency
in practice I use 4 TLC5940 that sink 20 mA for each led.
If you are using a TLC5940 then you don't need the FET? Or am I missing something?
I only put a part of the complete circuit in practice I have 4 TLC5940 where I connect 64 RGB leds(all 3 colors of a led in one pin of TLC) and 3 of these mosfet for drive the 3 color once at time(so via software I drive each mosfet one turn on and the other turn off ) and the worst case for power dissipation for mosfet is in the case that I turn on all 64 green leds or 64 blue leds.
Look at the schematic I used for this project:-
http://www.thebox.myzen.co.uk/Hardware/Hexome.html
geppou:
but the pick of power dissipation in this case that I switch from 0V to 5V is possible 1.28 A for 5V so 6.4W?
Power dissipation is the voltage drop * current flow. When there is a 5V drop across the MOSFET (i.e. when it is turned off), the current flow is near 0, and therefore the power dissipation is virtually 0.
When on, the voltage drop will be near zero (Rds-on * current), so power dissipation will be very low.
When switching between on and off, the voltage drop will be in between. So for a few ns (or tens of ns for a slow MOSFET) the voltage drop will be 2.5V, the current will be ~ 1A, so the power dissipation will be 2.5W for that brief instant.
In a slow MOSFET the turn on and turn off would add up to ~200ns. If your switching rate is 10kHz, the MOSFET is switching for 2 million ns of every second, or 0.2% of the time. Even assuming 5W switching power dissipation 0.2% of the time, that works out to only 10mW average.