Buck Converter Driver

Discussion in 'The Projects Forum' started by gusmas, Oct 2, 2012.

1. gusmas Thread Starter Active Member

Sep 27, 2008
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0
Hi

I am designing a very simple buck converter. Just to test equations that is given in application notes from various electronic manufacturers.

The MOSFET I will be using is STP14NK50Z (N Channel).

Here is the parameters for my buck converter:

Vin: 60V
Fsw: 30kHz
Vout: 48V
Iout: 10A

Before I start the design of the buck converter, I want to to calculate what driver I will be requiring.

Using the app notes from Microchip (AN799 and AN898)

Power Dissapation in Driver:

Power loss Due to charging and discharging gate cap in mosfet:
Pc = Cg * Vgs^2*Fsw
and
Cg = Qg/Vgs

From Figure 11 in the STP14NK50Z datasheet, a Vgs of 10V is chosen and the Qg is read from the datasheet: Qg = 73nC

Thus: Cg = Qg/Vgs = 73nC/10V = 7.3nF
And:
Pc = Cg * Vgs^2*Fsw = 7.3nF* 10V^2 * 30kHz = 21.9mW

Power loss Due to cross over distortion mosfet:

Ps = CC * Fsw * Vin,
where CC is read from STP14NK50Z datasheet, Table 7, "tc" typical value
thus
Ps = 20nS * 30kHz * 60V = 36mW

Calculate Current required from Gate Driver
Idriver = Qg/dT, where
Qg = 73nC
and according to the AN799 app note dT = Turn-on/turn-off time, now I dont know if they mean Turn-on OR turn-off time or Turn-on Divide by turn-off time. So i took the absolute worst case, took td(on). (Table 7)
This resulted in:
Idriver = Q/dT = 73nC / 24nS = 3A.

The current that the gate driver is suppose to supple seems quite high.. Can you guys have a look through and suggestions/comments are welcome!!

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2. bountyhunter Well-Known Member

Sep 7, 2009
2,498
510
I didn't use that particular FET, but I do remember that it took peak currents in the several A range to make them switch fast enough to minimize losses. That's because large power FETs have a lot of gate capacitance and you have to charge it up very fast.

3. gusmas Thread Starter Active Member

Sep 27, 2008
239
0
Can you maybe share/supply the equations u used to determine how much current your driver needed to supply to your mosfet?

4. crutschow Expert

Mar 14, 2008
20,532
5,817
If you can tolerate a slower switching speed and the increase in switching dissipation that would cause, then you can increase the value of dT in your equation which will reduce the required driver current. MOSFETs do require a large momentary gate current to switch rapidly.

5. gusmas Thread Starter Active Member

Sep 27, 2008
239
0
Hi

Thanks, Ill take that into account, but I found a driver that can supply up to 4A
its the IRS21851SPbF. Seems like my formulas and the values I used are correct then? The ton/toff times is around 170nS according to the datasheet of the driver, thus the switching time will be lowered for the mosfet, resulting in a decrease in current required by the gate of the mosfet:
Idriver = Q/dT = 73nC/170nS = 0.4A.

Am i correct?

6. crutschow Expert

Mar 14, 2008
20,532
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Sounds correct to me.

A simple linear estimate of the switching loss is to multiply the rise time by the voltage times the current times the switching frequency. (This assumes 1/2 the loss on the rise and 1/2 on the fall).

7. bountyhunter Well-Known Member

Sep 7, 2009
2,498
510
gate drive current capability is a lot like horsepower: no such thing as too much. The driver is forcing the gate voltage to "snap" to a specific voltage level but to do it the driver's source impedance (ergo, drive current) has to be strong enough. The higher the current, the faster it switches. And faster is always better because it reduces switching losses.

We never worked equations, just cranked up the current and made sure the switching losses stayed low.

8. gusmas Thread Starter Active Member

Sep 27, 2008
239
0
Haha thats a bit confusing mind supplying a formula instead of words ?

9. gusmas Thread Starter Active Member

Sep 27, 2008
239
0
Ok I looked at the datasheet of the IRS21851SPbF, but I am confused.

Problem

The Vho pin from the driver is going to the gate of the MOSFET. (Vgs signal). In the datasheet it says Vho(min) = Vs and Vho(max) = Vb. If i look at the "typical connection" diagram on page 1 of the high side driver datasheet, they connect Vcc through a diode to Vb. Thus the Capacitor sitting between Vb and Vs, will be charge to Vs+ (Vcc-0.7V(diode)), Vs is the reference for the capcitor.

However Vs is connected to the source of the MOSFET, which does not make sense to me. The voltage on the source of the mosfet will be 0V because if my supply voltage to the drain of my mosfet is 50V, it will mean my Vgs must be 60V at least, and the Vgs at that point in time will be:

Vcc for gate driver: 15V

Vgs = Vs + (Vcc-0.7V)
= 0V + (15V-0.7V)
= 14.3V whic is not nearl 60V?

I hope I am wrong but if not what is the use of the typical connection diagram in the datasheet for a HIGH SIDE DRIVER, if the maximum voltage for Vgs is totally dependent on the Vcc for the gate driver IC?

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10. crutschow Expert

Mar 14, 2008
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P= τ x V x I x f where τ equals the rise/fall time, V is the supply voltage, I is the transistor current and f is the switching frequency.

Better?

11. crutschow Expert

Mar 14, 2008
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The charge pump bootstrap capacitor generates a voltage that is 14.3V above the source voltage. As the source voltage rises during turn-on (VB connected to HO), this voltage rise is transmitted to VB through capacitor C1 and then through VB to HO. Since there is already 14.3V on the capacitor to start, the voltage at VB (and thus the gate) rises at 14.3V above the source voltage (ignoring any losses). For a 60V drain voltage, the gate voltage would thus be 74.3V when the transistor is fully on.

Note on the data sheet, that device is not recommended for new design. They suggest the IRS21850.

12. gusmas Thread Starter Active Member

Sep 27, 2008
239
0
Oh my word I am stupid!!!! I dont know what i was talking about!! I somehow confused the drain with the source.. hence it is called Vgs (gate to source) and not Vgd(gate to drain) XD.

You mentioned that VB connected to HO, I assume they are internally connected to one another? and if i connect it up like I did in the schematic i posted previously it should work fine? Calculating the correct value for the bootstrap cap ofcourse!

13. shortbus AAC Fanatic!

Sep 30, 2009
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Did you really mean to say that? The bootstrap capacitor charges to driver voltage when the source is to ground (when mosfet is off) and then as source voltage rises it injects it back to keep the gate at the driver voltage. The bootstrap cap makes a floating ground level, even though it isn't really "ground". This is needed to keep from going over the maximum allowable gate voltage. Correct?

14. gusmas Thread Starter Active Member

Sep 27, 2008
239
0
I wanted to calculate the bootstrap capacitor, but all the formulas i find has variables I cant find values for, anyone got a basic formula I can use?

Much appreciated

15. crutschow Expert

Mar 14, 2008
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VB is connected to HO when the transistor is turned on. When the transistor is turned off VB is removed from HO and HO is pulled to ground.

16. crutschow Expert

Mar 14, 2008
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Yes. The source voltage is 60V when measured to ground, when the MOSFET is fully on, and the gate voltage is 14.3V above that due to the bootstrap capacitor (for a large capacitor) making it 74.3V.

Edit: The maximum gate voltage is the voltage from gate to drain, not gate to ground.

Last edited: Oct 5, 2012
17. gusmas Thread Starter Active Member

Sep 27, 2008
239
0
ok are there typical values for the bootstrap caps? coz i cant find a formula for it.... and the Voltage rating for the caps should be fine 25V? because the potential diff will always be around 10V

18. crutschow Expert

Mar 14, 2008
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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.

19. shortbus AAC Fanatic!

Sep 30, 2009
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Not trying to argue with you, but this goes against everything I've learned. The data sheet shows it as Vgs, not Vgd. The only Vgd shown is the leakage voltage limit, which in most cases is also the maximum Drain voltage.

If the gate voltage was referenced to drain how would it be able to use a mosfet as a low side switch? I must be missing something in this?

20. crutschow Expert

Mar 14, 2008
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You didn't miss anything. Sorry, my brain was scrambled. It is, of course Vgs, the gate to source voltage.

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