here's the final version , many thanks for replying

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If you are going with the single-pin control, just mind the shoot-through. It may become a problem. You can reduce it's effects by adding in one resistors at the drain of each transistor, but of course, that will decrease efficiency.

Also, use bigger than normal MOSFETs that can handle the shoot-through current without overheating and beign destroyed. ANd don't use a PWM frequency that is too high. THis will also reduce shoot-through since it will happen less frequently since you are switching less.

(You can also try doing what I described in my last post).

 

Hi , i changed the circuit and used 4 transistors with inverter , i think its now can work 100% ?

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Place each inverter in front of the BJT base resistor for each PMOS so that the inverters do not affect the NMOS gate drivers. This will make it so that a HI signal will turn the PMOS rather than a LO signal (so the PMOS will behave just like the NMOS for the same control signal).

Then take the top-right PMOS and the bottom-left NMOS and connect their gate drive circuits together.

Do the same thing with the other two MOSFETs.

Do you see how this works?

EDIT: I added a schematic of what I mean. You can move the inverters from the PMOS to the NMOS. You probably should actually, since right now you have to apply a LO to the forward (backward) control pin to get the motor to move. If you move the inverters to the NMOS, then the logic isn't inverted anymore, and applying a HI to the forward (backward) pin will make the motor move.

(My PMOS's are upside down and so they are connected the wrong way. THe ones in your schematic are connected the RIGHT WAY. So don't change them.)

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no really , would you wire it please ? thanks alot .. i didnt get it

 

It is a bit hard to describe the changes I mean. I added a schematic to show you what I mean on how you wire diagonal FETs to be controlled by the same signal in my previous post.

But now that it's there, do you see how you use the PWM to control it? And what I mean when I say one control signal makes the motor go forward and the other one makes the motor go reverse?

 

is this what you mean ?

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Yes that's what I meant.

It makes no difference, but it might be easier if you move the inverters to the NMOS from the PMOS. Since the way I put the inverters (on the PMOS) it makes it so a HI is STOP and a LO is GO. If you changed it then a LO is STOP and a HI is GO which makes more sense.

Do you understand how the circuit works though? One pin will always have the FETs off, while you PWM the other pin to have it move with variable speed in a certain direction. And when you change directions, turn all the FETs off and wait a bit to make sure they are turned off and then apply the PWM for the new direction and you will never have shoot-through.

 

many thanks =)), hey i would use a 5V for the NPNs and i think the VGs is higher than 5V cuz this source is avaliable to me ... would be any problem ?

 

Vgs if a number of MOSFET gates, not NPNs. THe number you are interested in is the voltage drop across the NPN's base emitter diode. It's probably around 0.7V but it's in the datasheet.

So as long as your voltage is higher than 0.7V (to turn on the base-emitter diode), you size the resistor so that the extra voltage above 0.7V (ie. 5V-0.7V, if you use 5V) drives enough current into the base of the NPN. If the voltage is less than 0.7V, the you will not be able to turn on the base-emitter diode inside the NPN won't let any current flow into the base. THe calculation is the same as the one you use to size a resistor to control the current in an LED.

 

I mean 5V for the VCC for the NPNs so that the MOSFETs turns on using a 5V source ....

Rbase= 5-0.7/25ma (I source or sink from the inverter) = 172 ohms min resistor so i choosed 1kohms.

 

Your Rbase calculations is correct since your microcontroller (or inverter) outputs ~5V into the base of the NPN. But you are talking about Vb, not Vcc. You already connected each NPN to Vcc=12V through a resistor.

There are some things you should know. Vcc on your NMOS's NPN transistors must be a voltage that will turn on your NMOS when the gate is pulled high and not damage their gates, Your motor voltage must also be a voltage that is high enough so that your PMOS transistors can turn off when you pull their gate low. This places a minimum voltage that your bridge will work at (probably around 10V depending on your MOSFETs).

Also, your motor voltage can't be too high because of the PMOS transistors. When you pull their gate low, you can't exceed the maximum gate voltage or it will be damaged.

If you want to run your motor at lower than this voltage, things get tricky and more circuitry is needed. By far, the simplest thing to do is change your MOSFETs that need lower gate voltages to turn on and off.

I don't see any more problems in your circuit.

 

Im going to use a 5V for all the npns vcc cuz i've it instead of 12V, from the datasheet of the mosfets VGS max is 20V, why i cant then use a 5V !! ? and 24V for the upper mosfets ...
how do i calculate the Rc of the NPNs ? Rc= Vcc-Vce(sat)(0.4V)/150ma ?

from 2N2222 data sheet,VCEsat =400mv,at ic = 150ma, ib=15ma..

thanks really for being very helpfull ! wish you a good day

 

You can't use 5V because it is not high enough to turn on the NMOSs. If you go beyond 20V then you damage the NMOS. Maybe you can use a linear regulator or something but it might get too hot for Vcc.

The 24V for the upper MOSFETs might be a little bit too high (is it within the range of the maximum gate voltage for the PMOS?) Think about what Vgs is for the PMOS when you pull the gate low if your battery voltage is 24V. Vgs ~= 24V. Is it higher than the max gate voltage? It probably is by about 4V.

Rc has to be big enough so that your NPN won't burn out when it is on, but it has to be low enough to source enough current into the MOSFET gate to switch it quickly (to turn on the NMOS, or to turn off the PMOS). I don't know how much current this is, but my guess is that you want the current to be as high as possible without burning out the NPN. So if your NPN can handle 150mA, then size the resistor for 150mA:

Rc= Vcc-Vce(sat)(0.4V)/150ma
(just like you did)

I wouldn't worry about it too much- after all when you get the PCB you can always change the size of the resistor until you find one that works well. Other people on this forum know more about how to size pull-up resistors to drive MOSFET gates than I do. Maybe someone will tell us how to size it.

 

By the 24V i mean by The Vdrain of the PMOSs, and for the VGS for NMOSs i dont see why 5v cant turn it on ,the problem is in the datasheet it doesnt show the VGSth to know how much voltage needs to turn it on .. anyway im going to use 12V for it , but i need to know how to calculte the Rc of the NPNs