# Will the large gate current when turning on/off power mosfets with gate drivers cause problem to limited current power supply?

I have a idea using Arduino to control power mosfets(CSD18532). After considered some problems about wiring power mosfets directly to Arduino pin, like considerable slow switching time, I turned to using a gate driver IC(LM5114) between Arduino and the power mosfet. I think I could use the picture below taking from LM5114 datasheet page 10 to used as my schematic. The VDD on LM5114 will wire to Arduino's +5 volts power supply directly, while the PWM wiring to Arduino pin.

I learned from a TI's appnote that when turning on/off power mosfet, the gate current is considerably large(I roughly searched the internet, and got answers like 1 amp or so...). So my basic question is that will it be ok for a USB powered device like Arduino to power the gate driver IC's VDD through its +5 volts power supply directly? I know that USB port can only source 500mA...

With this question, I found TI's LM5114 Evaluation Board's schematic on page 8. Although I don't know the schematic clearly, I found that LM5114's VDD directly coming from a 50mA LDO. I used the equation from previous appnote(page 9):

So the IG is roughly: (5 - 1.4)/2 + 1.5 + 1 = 800mA. Will the relatively large gate current cause damage to the 50mA LDO(LP2982)?

• How often are you switching it on and off? – Andy aka May 28 '14 at 7:26
• @Andyaka maybe Arduino's pwm frequecy, 500Hz... – iouzzr May 28 '14 at 7:32
• What MOSFET are you using? – Andy aka May 28 '14 at 7:35
• @Andyaka Sorry for the unclear information for you... It is CSD18532 ti.com/lit/ds/symlink/csd18532q5b.pdf – iouzzr May 28 '14 at 7:40

800mA peak current for a short amount of time will not damage the LDO. The gate driver should be accompanied by a capacitor on the Vdd pin. The capacitor will supply the peak current and not the LDO. A 10uF capacitor is much larger than the gate capacitance.

The LDO has to provide the net (average) current for the gate driver.

If you are switching the MOSFET slowly and using the MCU pin, the current you pump into the gate will be limited to (say) 20mA. Using Q = CV as the starting point and differentiating we get: -

$\dfrac{dQ}{dt} = C\dfrac{dv}{dt}$ = current = 20mA

Therefore the rate of change of voltage you can get driving 20mA into a gate capacitance of 5nF is

$\dfrac{20mA}{5nF}$ = 4 million volts per second or 4 volts per micro second.

It looks like 4.5 volts on the gate is going to turn it on adequately and if your MCU can supply this (i.e. it is powered from 5V) then the MOSFET will turn on in about 1 micro second. If you want to operate this 500 times per second (2 ms), then 1 us every 2 ms is 0.05% of the time that the device is being turned on or, 0.1% of the time it is being turned on OR turned off.

If your MCU can supply 5V from a GPIO pin I wouldn't worry about the driver chip.

• Thanks Andy... Actually, my boss is worrying about have MCU IO directly wires to MOSFET's GATE. I searched a little and got the saying that a series resistor will protect the CMOS pin of MCU from SCR or any others... Is that reliable? Anyway, it is another question... – iouzzr May 28 '14 at 8:08
• Oh, I forget one thing. On some situations, I will control four power MOSFET. Will the 4*20 mA current source from IO pins become too much for an MCU? I remember that atmega328 has a MAX 200mA limits. Or it is just a peak current which only last for a short amount of time and that's just gonna to be OK... – iouzzr May 28 '14 at 8:19
• @iouzzr this might become a problem but given the low time spent (0.1%) switching, you could typically put 470 ohm current limit resistors to prevent no more than ~10mA flowing - on/off times would double of course but is this a big deal? – Andy aka May 28 '14 at 8:46