Hello,
I'm trying to drive high side n mosfet with highest current peak possible ( lowest on/off time ) by using low value resistors and npn transistor.

The 22v source will come from 12v to 22v boost converter.
When PWM is 0% it drives 100% speed and when it's 100% it drives at 0% speed. ( stopped )
I managed to reach around ~700mA current peak as you see but I think it has something to do with proteus PWM configuration.
I have set 100hz low frequency for helping Proteus to simulate but I'll probably use around 5khz in real life.
The PWM peak voltage is 3.3V

Does this seems correct ?
Thanks

 

Notice that your BJT current is 22V/10ohm = 2.2A.

Try one of this circuits


Or this one

 

A few comments - good mosfet drivers are VERY difficult to design. I've spent a good portion of my career designing and redesigning them. My suggestion is to purchase a mosfet driver that does this and has many years of design and redesign behind the driver.

Otherwise - you're on the right track... Option 1) you need to increase your resistance of R1 to maybe 4.7k, and add a base resistor... about 10k... this will slow down the turn-on time of your switch, but keep the turn-off time very fast. Option 2) You can also put a 4.7k resistor on the emitter of the transistor and remove the base resistor. This will help provide more even turn-on/turn-off times, but will impact power dissipation. It's always a good idea to have a resistor on the order of a few 10's of ohms to a few 100's of ohms on the gate of the FET to provide stability to the FET.

The resistance values may need scaled down to be able to run at 15kHz.

 

Thanks for your answers,

I've just found this circuit from "Tahmid's blog":
http://4.bp.blogspot.com/-o-EOdJF0GMY/UNYUHlNJCbI/AAAAAAAAAR8/37_J58aLK6o/s1600/driver04.PNG

I adapted it to high side with +V_DRIVE = 20v and it works fine on Proteus with high current peaks on the gate.
According to the blog " both turn-on and turn-off are very quick".
http://tahmidmc.blogspot.fr/2012/12/low-side-mosfet-drive-circuits-and_23.html

 

Here's THE app note to reference any time you are working with mosfets - all other appnotes are second-rate...
http://www.irf.com/technical-info/appnotes/mosfet.pdf

Using the calculations there, you can actually predict your turn-on / turn-off speeds.

The image you linked to is called a push/pull driver... it will help make your on/off times very fast... usually too fast, causing EMI concerns. The good news is that you can easily change R24 to make it slower... a few quick calculations tells me about 100 ohms is probably about right to get 5% of 15kHz. The calculations are in the app note I referenced (Q=I*t). I usually prefer to make R25 a bit larger (~20k) and clamp the input with a 12V zener. You don't want your total on/off time to exceed 10% of your operating frequency. Usually better to be below 5% if you can. I find that 5% is a good balance of speed, radiated emissions, and power dissipation, at the higher switching frequencies, anyway - which you're pushing.

 

wait if Q = I*t
If I use IRFB7540 : http://www.irf.com/part/_/A~IRFB7540
with Qg = 88nC

with I = 22V/10R = 2.2A
t = Q / I <-> t = 88 nC / 2.2A = 40 ns
f = 1 / 40 ns = 25 megaHertz !!

It means I could drive this with less than 2 milliAmperes ?? (2mA / 88nC = ~22.7kHz )

 

A couple things:
You want your switch time to be at least 10x's your switch frequency... so you want your switch time to be on the order of 6.67us... I was suggesting about 3.33us as your primary rise/fall time. This will put your primary harmonic at about 100kHz (0.35/3.33us)... not 25MHz! I also thought you were using thee IRF1405 based on your first post, so my numbers may be a bit skewed.

There are some tricky things that make that equation - not quite as simple as it seems... the first is that the charge (Q) of the primary on/off time is the threshold voltage to the end of the platau of figure 8 of the 7540 datasheet... which is closer to 40nC. It will take some additional time to finish fully turning on/off, but the power dissipation time of the switch will be completed in this time. so that also means that the voltage differential will be closer to 4V or so. Also the current will be a triangle for the most part - so you have to divide your current by 2. It's all quite complex, but easy at the same time.

So you end up with the following equation
R = t*V/2Q = 3.33us * 4V / (2 * 40nC) = 166ohms... hence my suggestion for 100ohms... anything in that 50-200 ohm range should work nicely... but not get too fast. Too fast can very much be a problem!

 

Thanks for your answers,

I've just found this circuit from "Tahmid's blog":
http://4.bp.blogspot.com/-o-EOdJF0GMY/UNYUHlNJCbI/AAAAAAAAAR8/37_J58aLK6o/s1600/driver04.PNG

I adapted it to high side with +V_DRIVE = 20v and it works fine on Proteus with high current peaks on the gate.
According to the blog " both turn-on and turn-off are very quick".
http://tahmidmc.blogspot.fr/2012/12/low-side-mosfet-drive-circuits-and_23.html

You should be careful because the Q11, Q12 have the dead time, it may kill them, R28, R29 could protecting the bjts from the dead time, but they also will affecting the frequency.

 

A couple things:
You want your switch time to be at least 10x's your switch frequency... so you want your switch time to be on the order of 6.67us... I was suggesting about 3.33us as your primary rise/fall time. This will put your primary harmonic at about 100kHz (0.35/3.33us)... not 25MHz! I also thought you were using thee IRF1405 based on your first post, so my numbers may be a bit skewed.

There are some tricky things that make that equation - not quite as simple as it seems... the first is that the charge (Q) of the primary on/off time is the threshold voltage to the end of the platau of figure 8 of the 7540 datasheet... which is closer to 40nC. It will take some additional time to finish fully turning on/off, but the power dissipation time of the switch will be completed in this time. so that also means that the voltage differential will be closer to 4V or so. Also the current will be a triangle for the most part - so you have to divide your current by 2. It's all quite complex, but easy at the same time.

So you end up with the following equation
R = t*V/2Q = 3.33us * 4V / (2 * 40nC) = 166ohms... hence my suggestion for 100ohms... anything in that 50-200 ohm range should work nicely... but not get too fast. Too fast can very much be a problem!

Thanks for your time!

But I did'nt understand, why "0.35/3.33us" and not "1/3.33us" for the frequency ?

And what is the utility of R25 ?

 

Notice that your BJT current is 22V/10ohm = 2.2A.

Try one of this circuits
View attachment 76120

Or this one

View attachment 76119

That's what I'd have suggested if I wasn't too lazy to draw the schematic and post it.

 

Notice that your BJT current is 22V/10ohm = 2.2A.

Try one of this circuits
View attachment 76120



Or this one
View attachment 76119

I am a newbie in electronics and these circuits seems very complicated to me and I am having trouble to understand them. For example it is the first time I see an input from the emitter of npn transistor. ( second picture Q1)

 

http://www.edn.com/electronics-blog...-The-bandwidth-of-a-signal-from-its-rise-time