Buck Converter Driver

Thread Starter

gusmas

Joined Sep 27, 2008
239
You want the cap to be large enough so that the required gate charge (from the data sheet) does not lower the cap voltage below 10V. You place C1 in parallel with the gate capacitance to determine the drop in voltage as the charge on C1 is now shared between C1 and the gate capacitance.

25V rating is fine for the cap.
Ok, I understand what you are saying... but how do i calculate the farad value for the capacitor?
 

Thread Starter

gusmas

Joined Sep 27, 2008
239
OK so I went and looked for formulas for IR driver bootstrap calculation and found the following from the application notes:
Document: Application Note AN-978

Note: Changed my mosfet driver IC to IRS2186.

Now:

Using the formula on page 6:

Qg = 73nC
f = 30kHz
ICbs (leak) = Not using electrolyte cap
Iqbs (max) = According to driver IC is 150uA. (Page 3 of IRS2186 datasheet)
VCC = 15V
Vf = Max 1.1V (Bootstrap diode)
VLS = This is where my problem is, if my Vsupply to my drain is 60V, and let ignore Rdson value, and my load resistance is connected at the source pin of the mosfet to ground. then the voltage drop across my load will be the full 60V, thus my answer will be negative??
VMin = 10V (unsure about this value).
Qls = level shift charge required per cycle (typically 5 nC for 500 V/600 V MGDs and 20 nC for
1200 V MGDs)
 

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shortbus

Joined Sep 30, 2009
10,045
@ gusmas - You need to use Ohm's law to figure the voltage drop of the load. That is then the value to use. As an example if you figure from your values the load is dropping 1V that is the number to put in the equation. V= I x R of Ohm's law.
 

crutschow

Joined Mar 14, 2008
34,285
VLS is the voltage drop across the transistor when it is on (don't know why they also say "or load"). Just multiply the maximum load current times the maximum ON resistance of the MOSFET (from the data sheet). That should be no more than a few tenths of a volt.
 

Thread Starter

gusmas

Joined Sep 27, 2008
239
VLS is the voltage drop across the transistor when it is on (don't know why they also say "or load"). Just multiply the maximum load current times the maximum ON resistance of the MOSFET (from the data sheet). That should be no more than a few tenths of a volt.

Cool thanks man!

and the VMin = 10V, that is the 10V diff required to switch mosfet correct?
 

Thread Starter

gusmas

Joined Sep 27, 2008
239
VLS is the voltage drop across the transistor when it is on (don't know why they also say "or load"). Just multiply the maximum load current times the maximum ON resistance of the MOSFET (from the data sheet). That should be no more than a few tenths of a volt.
Quick question, u say tenths of a volt, where my answer is way higher..
Final product will be a buck converter, which should be able to supply 10A. Thus using the Rds(on) = 0.38 ohm, value from the STP14NK50Z datasheet alongside the maximum amount of current the transistor can handle, Id (@25 degrees) = 14A. This results:

Vls = Rds(on) = Id = 0.38 * 14A = 5.32V.


I changed my driver IC to a High Side and Low Side Driver: IRS2186 , but I am only using the High side driver hopefully that wont be a problem. please comment on this.

Using my original values:

Qg = 73nC
f = 30kHz
ICbs (leak) = Not using electrolyte cap
Iqbs (max) = According to driver IC is 150uA. (Page 3 of IRS2186 datasheet)
VCC = 15V
Vf = Max 1.1V (Bootstrap diode)
VLS = 5.32V
VMin = 10V
Qls = level shift charge required per cycle (typically 5 nC for 500 V/600 V MGDs and 20 nC for
1200 V MGDs)

with a Vcc of 15V results in a negative answer:

Cboot >= -219.7183099nF

However using Vcc = 20V (max safe operating capabilities of the IRS2186):

Cboot >= 87nF

So IF this is all correct, I can finally order my components and because I used all the extreme values for all calculations, if I decrease the Vdd and just add a 10kohm resistor at the source of the transistor, the whole circuit should still work fine correct? I just want to see if the driver is doing its job. IF successful I will use the exact design for my buck converter, thats going to be a whole new thread on its own :).

Note: The reasons I am asking so many questions is because I could not find a clear cut document that explains everything regarding buck converters, MOSFET'S and MOSFET drivers. So hopefully this can serve as a reference material to future threads regarding MOSFET drivers.
 
Last edited:

crutschow

Joined Mar 14, 2008
34,285
Quick question, u say tenths of a volt, where my answer is way higher..
Final product will be a buck converter, which should be able to supply 10A. Thus using the Rds(on) = 0.38 ohm, value from the STP14NK50Z datasheet alongside the maximum amount of current the transistor can handle, Id (@25 degrees) = 14A. This results:

Vls = Rds(on) = Id = 0.38 * 14A = 5.32V.
...................................................
Your calculations are correct. I said "should be no more than a few tenths of a volt" because you normally pick a MOSFET with a low enough ON resistance to keep the ON voltage low at the operating current to minimize transistor dissipation and maximize converter efficiency. Your transistor will be dissipating 38W when conducting 10A.

I suggest you find a MOSFET with a lower ON resistance to keep the ON voltage below a volt and the dissipation below 10W.

Edit: I would use a larger Cboot capacitor, since the value you calculated is the minimum.
 

Thread Starter

gusmas

Joined Sep 27, 2008
239
Your calculations are correct. I said "should be no more than a few tenths of a volt" because you normally pick a MOSFET with a low enough ON resistance to keep the ON voltage low at the operating current to minimize transistor dissipation and maximize converter efficiency. Your transistor will be dissipating 38W when conducting 10A.

I suggest you find a MOSFET with a lower ON resistance to keep the ON voltage below a volt and the dissipation below 10W.

Edit: I would use a larger Cboot capacitor, since the value you calculated is the minimum.
Ok thanks!

I found a alternative: IRF540NSPbF

Thus my Vls = 0.044 * 33A = 1.452V

Thus

Pd = Idd(buck)^2 * Rds(on) = 10^2 * 0.044 = 4.4W.

Thanks for all the help/tips/equations!! I will order components soon and post my findings in this thread!
 

Thread Starter

gusmas

Joined Sep 27, 2008
239
If you peak current is 10A then Vls = 0.044 * 10A = 0.44V.
YIP i totally agree, I just went with the absolute worst case, to see if everything is still fine. From what I could see in the MOSFET gate driver datasheet is that the high side and low side are connected to one another through the PULSE Generator block (Vcc passing through) on the high side... Now I hope I can just use the high side output without connecting anything to the low side output.

Driver is the IRS21864 HIGH AND LOW SIDE DRIVER.
 
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Thread Starter

gusmas

Joined Sep 27, 2008
239
HIHI

its been a very long while since we talked!! I finally tested my circuit and I am very pleased to announce it works!! well the first part anyway now it is time to do the buck converter part!!
 
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