Thanks for the tip Matter45

the longer the mosfet is on, the more potential difference it developes between source and drain

I assume you mean "the longer the mosfet is OFF..."?

Point Two.

I use a high side gate driver so that should be covered.

Point three. having the N channel mosfet on the negative side of the circuit

I have tried this but I couldn't get it to work because I then had two different ground potentials and therefore I was not able to sense the right voltage for the feedback loop. But I'd really appreciate it if you'd help me with it.
Here is the datasheet for the mosfet: http://www.google.no/url?sa=t&rct=j...e5Mi1MgWJ8R-2VGohynpDPg&bvm=bv.46340616,d.Yms

 

wow. so many factors here to consider. why is it that on your schematic you have ground declarations on both the pmw and the input if there is a potential difference? This is an issue!

The longer the mosfet is ON, the more potential difference it gains. do you not know how inductors work?

when the mosfet is off, its potential difference between source and drain is 300v. but the reason it doesnt blow up in this state is because almost 0 current is going through it. say 5ua. watts in heat = V * I. in this case, 300v * 0.000005 = 0.0015w of heat. nothing to be concerned about.

but when the mosfet is on, the component in your circuit with the most impedance is your inductor. the more inductance it has, the longer it takes for current to flow, holding back that potential difference. but as current increases, the potential difference across the inductor decreases. so where is the potential difference going to go? lets look at your schematic... oh yeah, the mosfet!

here is a noobie equation to help you find out how fast the potential difference dies off an inductor as current goes through it. if you have input voltage of 1v, and you have 1H inductor in series with 1Ohm resistor, then the voltage across the inductor will get to half the input voltage in about 0.69 seconds.

In a circuit, if you have 1H + 1R + 1V = then it takes 0.69316 seconds for the inductor to decrease potential difference to half of the input voltage. (this is not a standard formulas, rather to help people understand, use as a guide only)

if the input voltage was 10v, it will still take 0.69 seconds to get to half input voltage (5v)

decreasing H decreases time based on %. if you half H (50%), then you half 0.69316 seconds. if 0.1h then time to half input V = 0.069316 seconds instead of 0.69316.

if you add R, then you must divide 0.69316 by R for total time to get to half input voltage between H and R

lets have a look at the on resistance of the mosfet. on the datasheet it says 38mohms.

lets do some math.

H = 10uh
R = 0.038

equation:
calculate time to half input voltage between inductor and resistor with variable R = 1R / 0.038R = 26.32 times more time. SO:
Time after R = 26.32 x 0.69 = 18.16 seconds for 1H and 0.038 ohm resistor to reach half input voltage.

but your inductor isnt 1H, its 10uh!

More math:

1H / 0.00001H = 100000 times less!!!!

18.16 seconds / 100000 = 0.0001816 seconds to reach half input voltage.

So @ 300v, after the mosfet is turned on for 0.0001816 seconds, it will reach 150v! 150v * I = heat!

1 amp means 150w of heat. 3 amps mean 450w of heat. your mosfet is rated absolute max 400W.

So the basic conclusion of all this, is that if you want to safely draw more amps at high volts, you need a bigger inductor. not a bigger mosfet.

And you dont want a potential difference on the mosfet while its on, otherwise it is inefficient.

 

And after a short search it appears the higher inductor value the lower current it is able to handle. So how to get a good inductor for high currents?

Wrong, the more inductance, the more amps that can flow through the inductor and the slower the potential difference changes.

 

Thanks for the explanation Matter45! This is something I have really not thought of. So briefly summarized I would want a large inductor to make the potential difference over the mosfet smaller? I will try this as soon as possible.
Do you recommend an value around 1H inductor?

 

If you did find a 1H inductor, it would be expensive and quite large if it had to support large currents. But the general idea when building these circuits is that the bigger the inductance, the better it is. it all comes down to input voltage, current, how big your inductor can be and how much u can afford. Building your own is more cost effective. Using the math above and an oscilloscope, you can roughly find out how much inductance an inductor has, or close to it.

The Math above is a Universal Constant, like E = mc2

if you have 1H and 1R in a circuit, half input voltage will always be about 0.69 seconds. Another law of the universe.

more info here:
http://www.kean.edu/~asetoode/home/tech1504/acdc/rltc.htm

 

So @ 300v, after the mosfet is turned on for 0.0001816 seconds, it will reach 150v! 150v * I = heat!

1 amp means 150w of heat. 3 amps mean 450w of heat. your mosfet is rated absolute max 400W.

The math appears to be very incorrect.
What will reach 150V? The simplified circuit consists of Voltage Source, MOSFET (switch), Inductor and the load.
The voltage across the MOSFET, when it is on, is Rds(on) X Id, and Pd is Id^2 X Rds(on), without considering the switching losses.
The 50% 150V, is not the MOSFET!

@eirik:
1.5us switching time is quite ok, and the inductor that I have given are just about the minimum values that would ensure CCM, and continious inductor current.
I hope you are not using this for a battery charger! The bootstrap capacitor, with high output voltage from the battery or on light load conditions, cannot charge to provide enough voltage for gate drive!
Your MOSFET current will be Io+2A during the on time.
Regarding the heating of the MOSFET,
Pd = Rds(on) * Id^2 * Duty Cycle +
(rise time + fall time) * Vin * Id * Switching frequency / 2.
The second factor is switching losses, approximated.
What are the test conditions when the heating takes place?
What is the load current, heatsink, MOSFET, etc.,

 

The math appears to be very incorrect.
What will reach 150V? The simplified circuit consists of Voltage Source, MOSFET (switch), Inductor and the load.
The voltage across the MOSFET, when it is on, is Rds(on) X Id, and Pd is Id^2 X Rds(on), without considering the switching losses.
The 50% 150V, is not the MOSFET!

Hey RamaD, can you please help me find the mistake I made?

Rds(on) X Id was already calculated using the circuits total resistance value at a specific point in time. (in other words, the inductor and the mosfets resistance are equal at a given point in time, thus having equal potential difference, 150v)

 

Never Mind! I see it!

I calculated voltage with Rds(on) * Id then added I again on top with this statement:

"1 amp means 150w of heat. 3 amps mean 450w of heat. your mosfet is rated absolute max 400W."

My Bad!

Just to clarify, the current would be excessive for the mosfet to be @ 150v. but the general concession of the lesson is understanding the relationship between potential difference on an inductor in series with a resistor and time.

an example of time of 1.5us means the voltage difference on the mosfet would be about 1.7v @ about 45A for a 10uh inductor. (which is why RamaD is recommending 85uh!)

 

Hi again guys,
work has been slow the last weeks but now I hope to get some real progress. Because the design has not worked properly until now I am thinking about redesigning parts of the circuit. I want to move the mosfet to low side in the buck regulator so I can get rid of the gate driver. Doing this means I have to modify my voltage sense loop. But I need some tips on how to do this. I know I need a transistor and some resistors, but how to make sure I sense the correct voltage? So does anyone know about a good article on this or have some tips?
As always all response is appreciated!

 

Have you seen this site? You put in the parameters you want to use and it shows you what components will work.

http://www.ti.com/ww/en/simple_switcher_dc_dc_converters/tools.html