# Using p-channel mosfet on high side of h-bridge, do I really need a driver chip?

I know there are fancy h-bridge ic's to drive high side of h-bridge but I guess that is when I have to drive n channel mosfet.

I am planning on using p channel mosfets on high side and n channel ttl mosfets on low side. I will driving the high side with n channel mosfets.

My Vbatt is 12 V and I will be controlling a wheelchair motor which can conduct upto 30A under load.

my concerns:
1. How do I calculate gate drive current for the p-mos since the gate capacitance has to be fully charged for maximum Id? do I just use i=Q/t, where t is charge time and Q is gate charge. But how does the pwm frequency come into play? say if i am using 31khz
2. How can I calculate the values of the resistors (connected from Vbatt to Gate of n-mos)
3. Do I have to worry about gate resistors?

Shoot through won't be a problem since I will put a delay when switching direction. Any help would greatly be appreciated.

• As Andy says, high side gate drive will be very slow and suited only to very low speed PWM.One option is to turn say top left PFET hard on and PWM only bottom right NFET to get variable power control in one direction. Reverse to Top-R and bottom -R for other direction. Even bottom side gate drive from MCU will be inadequate for higher speed PWM as gate capacitance chg/dischg rate is limited by mcu pin drive mA. | V simple current driver uses 2 x bipolar cheap transistors. NPN collector to V+. PNP collector to gnd. Join bases and drive as input. Join emitters as output to gate. ... – Russell McMahon Apr 18 '14 at 1:52
• ... Maybe use a few Ohms emitters to gate to LIMIT gate drive current. This cct has no formal overshoot control but dead area in middle due to 2 x Vbe deadband usually makes it work OK. If that does not make sense draw the cct, look at how it works and then read what I said again. | NB this is a current buffer/driver and you still need a voltage shifter to use it on the "high side". – Russell McMahon Apr 18 '14 at 1:55

Firstly, I think you have your P ch and N ch mosfet symbols reversed. Secondly, no, you don't need to use a driver chip if the top mosfet is P channel (providing you are not making extensive use of PWM to control the motor).

If you are using PWM then I'd suggest you use push-pull drivers because the gate-source capacitance on MOSFETs is usually in the order of 1nF to 10nF and "charging" this amount of capacitance from a GPIO pin takes several micro-seconds. Worst still, the resistor that discharges the gate will take significantly longer if the resistor is in the middle kohm range.

So, if you are using PWM then I'd go the whole hog and use a driver AND use both N ch devices - efficiency will be a tad higher than using a P channel device as the top FET.

As an example use the following formula: -

$\dfrac{dQ}{dT} = C\dfrac{dv}{dt}$ = I (current injected into the gate capacitance)

So if capacitance is 3nF and you can inject 1 amp, the voltage rise on the gate is 333 volts per microsecond - you'd be looking for a rise time less than 1us to achieve maybe 10 volts and that sounds to me like a rise time or fall time of about 30 nano seconds.

If you were relying on a 1kohm resistor to discharge the gate, CR time is 3 microseconds and in reality you might need about 10 micro seconds to discharge it properly.

The option is yours.

• cool thanks! Yea I see that n ch devices also have lower Rdson. But if I go the route of driver I still have to figure our how much current the driver can supply right? And would my equation i=Q/t be ok for that? – rashid Apr 17 '14 at 21:13
• @rashid - see the edits to my answer. Your formula leads to my formula by the fact that Q = CV. – Andy aka Apr 17 '14 at 21:20
• @rashid - not a lot can beat this cds.linear.com/docs/en/datasheet/4449fa.pdf - it'll drive 2 N ch mosfets and is able to inject 3 amps + into the gates. It has shoot thru protection built in. – Andy aka Apr 17 '14 at 21:30
• in the datasheet I see two values Td (turn on delay time) and tr (Rise time). I believe I chose the Td right? I guess that is the time required to charge the capacitor? please correct me if I am wrong. – rashid Apr 18 '14 at 16:11