Hi,I'm wanting to drive a high current load using a 12V PWM controlled by an Arduino. Since the load requires commonality with ground, I'm forced to use a P-Channel mosfet to drive it. In order to eliminate acoustic rattling, the PWM frequency has to be quite high. This means that I have to use a Gate Driver to reduce the phase transition time so the fet doesn't get too hot due to the time it spends in ohmic mode switching on and off.
I think a suitable driver would be the IR4426 which is an inverting gate driver.the data sheet it shows an example schematic (it's for a non inverting application using N fets but the principle is the same). It has resistors between the FET gates and the IR442X outputs. Wouldn't putting resistors there totally negate the point of using the driver in the first place as it would introduce a delay for the gate to charge/discharge? So naturally I'm a little confused now.
How to deal with negative transients. Layouts and other guidelines. How to isolate the logic from the power circuit. How to boost the output current of an MGD to drive modules. How to provide a continuous gate drive. HIGH VOLTAGE RAIL. Power MOSFET in high side configuration www.irf.
Could anyone clear it up as to why those resistors have to be there? Now that is what I'm not getting. The output from the driver is inverted, so when I apply 0V to the input, I should get VCC out of the output (open circuit low side is off) which turns off the mosfet, and when I apply a logic voltage to the input I should get 0V at the output (low side is on so connects to ground) which turns the mosfet on. Or am I getting all this back to front?I have bread boarded this without using the resistors (output to gate directly) and it does seem to work.I may be wrong in my assumption, so please correct me if so. A low side driver switches a an open voltage to ground and a high side switches an open ground to a voltage?Basically the gate driver has to do a better job than thisThis is a low side inverting driver isn't it? Thanks.I do understand how the circuit works, I'm just not up on the jargon. I thought low-side meant the switch (Q2) was on the low (ground) side of the load and inverted because the gate of the mosfet is at an inverted logic to the base of the transistor.
It's one thing to understand circuits and another to know the terminology.I guess, it seems very counter-intuitive to me. So High Side would be clearer to understand if they said the LOAD is on the High Side rather than the SWITCH being on the high side as the switch by definition is actually on BOTH, the load can only be on one. I still don't get why it's non-inverting though, as the logic is being reversed by the transistor. Is the term is relative to the device it's driving?
If I used the same circuit to drive an N-Channel would it then become an inverting high side?I'm still however a little confused, now I understand the resistors are there to protect from these high currents, but again, wouldn't they mess up the driver's function? The gate would then have to charge and discharge through those resistors and introduce a delay making it no better than using the transistor circuit above? Also if the spiking current is high wouldn't those resistors burn out if they were regular 0.25W ones?
Lets clear this up. Gate resistors are not normally required when usinga MOSFET driver driving one MOSFET. You can select the driver thatoutputs enough current to switch your MOSFET as fast as you want.You normally have to add gate resistors if driving two or more MOSFETs from thesame driver, because there is a differential oscillation mode in this configurationthat can occur and needs damping down. Forget it, the terminology is too ambiguous and counter-intuitive for me. I'll stick to programming, it makes more sense.This is the problem in this field for beginners. Everyone automatically assumes the beginner is going to understand every bit of jargon and terminology.
I can tell you on a physical quantum scale why a mosfet works, I understand doping etc. But because I don't know the definition of 'high side' or 'low side' and/or mis-interpret it, I'm buggered because no where there is a real unambiguous definition.
Like in this example, the SWITCH is the mosfet, where as I thought it meant the transistor, the load is the final load whereas I thought it meant the pullup resistor. That kind of detail is never covered just the terms 'switch' and 'load'. I mis-interpretted because when I imagine a driver, I don't see the final load, all I see is the transistor and pull-up, that to me IS the driver and so I took the parts of the driver itself as what switch and load referred to.But anyway, thank you both for clearing it up.
Now I actually do understand.
Motor DrivingThe most common type of motor you might come across in hobbyist circles for low power applications is the 3V DC motor shown below. This kind of motor is optimized for low voltage operation from two 1.5V cells.And running it is as simple as connecting it to two cells – the motor fires up instantly and runs as long as the batteries are connected.
While this kind of setup is good for ‘static’ applications like a miniature windmill or fan, when it comes to a ‘dynamic’ application like, more precision is needed – in the form of and torque control.It’s obvious that decreasing the voltage across the motor decreases the speed and a dead battery results in a slow motor but if the motor is powered from a rail common to more than one device, a proper driving circuit is needed.This can even be in the form of a – the voltage across the motor can be varied to increase or decrease speed. If more current is needed, this circuit can be built discreetly with a few bipolar. The biggest drawback with this kind of setup is the efficiency – just like with any other load, the transistor dissipates all the unwanted power.The solution to this problem is a method called.
Here, the motor is driven by a square wave with an adjustable duty cycle (the ratio of on time to the period of the signal). The total power delivered is proportional to the duty cycle. In other words, the motor is powered for a small fraction of the time period – so over time the average power to the motor is low. With a 0% duty cycle, the motor is off (no current flowing); with a duty cycle of 50% the motor runs at half power (half the current draw) and 100% represents full power at maximum current draw.This is implemented by connecting the motor high side and driving it with an N-channel MOSFET, which is driven again by a PWM signal.This has some interesting implications – a 3V motor can be driven using a 12V supply using a low duty cycle since the motor sees only the average voltage. With careful design, this eliminates the need for a separate motor power supply.What if we need to reverse the direction of the motor? This is usually done by switching the motor terminals, but this can be done electrically.One option could be to use another and a negative supply to switch directions.
This requires one terminal of the motor to be permanently grounded and the other connected to either the positive or negative supply. Here, the MOSFETs act like an SPDT switch.However, a more elegant solution exists.The H-Bridge Motor Driver CircuitThis circuit is called H-bridge because the MOSFETs form the two vertical strokes and the motor forms the horizontal stroke of the alphabet ‘H’. It is the simple and elegant solution to all motor driving problems.
The direction can be changed easily and the speed can be controlled.In an H-bridge configuration, only the diagonally opposite pairs of MOSFETs are activated to control the direction, like shown in the below figure:When activating one pair of (diagonally opposite) MOSFETs, the motor sees current flow in one direction and when the other pair is activated, the current through the motor reverses direction.The MOSFETs can be left on for full power or PWM-ed for power regulation or turned off to let the motor stop. Activating both bottom and top MOSFETs (but never together) brakes the motor.Another way to implement, which we discussed in previous tutorial.Components Required. Nice idea for learning H-bridge circuits, your logic is right on; some major things you may have missed.Shoot thru occurs when both the upper and lower MOSFETs turn on at the same time.
With your circuit as shown this will happen at the transition where one is turning on the is turning off. Note: the higher the VCC the worse it will get, causing excess heating of the MOSFETs and eventually failure to occur.Examine the Vgs of both upper and lower devices over the full gate voltage swing.
Gate Driver Circuit For Mosfet Pdf File
You will fine you have them both in the partially enhancement mode where both are trying to conduct. The slower the rise and fall time of the gate the worse this will be. This will create some huge current spikes.What is shown on the breadboard will work because of the limiting impedance of the interconnections. If this is built for say 20A 40V you will be blowing MOSFETs. The primary failure will be either shoot through or VGS over voltage.The Vgs of both devices is +- 20V which is way short of the max 40V suggested.
At approximately 3V Vgs both devices are turning on. There is also the Miller Capacitor effect which will also slow down the switching time however the TC4427 (4.5 - 18V) is a good choice to solve this problem.Do continue with this, you have made a great start.Gil.