H-Bridge Design suggestions

Hi , i designed this circuit and want to know your opinion, any errors ? before printing it as PCB , it worked in the simulator , but want to know your suggestions for improvment.
thanks

What does the diode bridge do? You don't seem to have flyback diodes to protect the MOSFETs from the inductive flyback of the motor (the voltage spike produced from shutting off the current to an inductive load).

Also, please make the image a GIF and make it bigger.

yea that what does the rectifier do , it covers the mosfet from the back emf from the motor .

can this circuit work in real life ?
values of resistors , current rating of the BJTs ..etc ??

It covers the MOSFET from the back EMF of the motor? That doesn't make sense. DO you mean:

1. It covers the MOSFET from the inductive flyback of the motor?

2. It measures the back EMF of the motor?

yes
1. It covers the MOSFET from the inductive flyback of the motor?

I see. The BACK EMF of the motor is not what causes the voltage spike. It's the voltage generated from the motor acting as a generator. THe inductive flyback is generated from the inductance of the motor. THey are two different things.

And, I also think that your BJT transistor types are all backwards for Q5, Q6, Q15, and Q16.

i didnt understand this , please explain it

I'm going to delete most of my previous posts and stare at your circuit more before I start saying things that could be wrong. But my concerns right now lie with the diode bridge and the gate driver circuit, mainly because I'm not entirely sure how they work yet. How were you planning on switching the transistors? Were you only going to PWM Q1 and Q2? Or only PWM Q3 and Q4? Or all of them?

PWM Q1 and make Q2 off .

I am having a hard time tracing the flyback path using the bridge rectifier because it seems to be assymetrical. I'm still working on it, but regardless of whether it works or not, what bridge rectifier are you using? I am guessing that the bridge rectifier uses regular diodes and not the very fast, low forward voltage kinds that you need to supress the flyback voltage spike.

Is there a reason you did not place the diodes like this:
https://www.modularcircuits.com/pic/simple_bridge.gif

I have not gotten to the gate drive circuit yet.

FLYBACK DIODES:
So it seems that the diode bridge will probably work IF the diodes are very fast and low forward voltage (like schotkey diodes). The current path for the flyback is also very strange in some cases. I am unsure if you can find a schotkey diode bridge, but you shouldn't use regular diodes. You need at least high speed diodes. But you should know that your circuit relies on a transistor being on to give the flyback current a path to flow. This means that if all transistors are off (like when you just decide to shut everything off) and flyback exists, then something will be destroyed.

I still suggest you switch to this for the diodes just because it's more common and easier to analyze.
https://www.modularcircuits.com/pic/simple_bridge.gif
It also allows a path for the flyback current to flow if all transistors are off for whatever reason.

You also seem to be using the diode bridge though for the voltmeters. What are those for?

GATE DRIVER:

D1 and D2 are not needed becuase the NPN BJTs already have such an intrinsic diode in them (double check me on this). But this is if your circuit does work (and I don't think it does).

Let's just look at the left side driver circuit because they are the same and everything that applies to the left driver also applies to the right driver:
-If Q7 turns on, both Q5 and Q6 stay off because Veb ~= 0 for Q6 and Vbe < 0 for Q5.
-If Q7 is off, then Q6 stays off because Veb < 0 for Q6 and I am uncertain what Q5 will do- it may or may not turn on.

The problem? Q5 and Q6 are of the opposite BJT types so the base-emitter voltages don't match up properly to turn them on with the way you have them connected. If you change Q5 to be PNP (since it is on the high-side, cloest to +V) and change Q6 to NPN (since it is on the low-side) it should then become a push-pull configuration that can be turned on. For example, you seem to realize that PMOS transistors are easier to switch when they are on the high-side, and NMOS transistors are easier to switch when they are on the low-side since that's what you did with Q1, Q2, Q3, and Q4. It's the same thing with NPN and PNP transistors. PNP is easier to switch on the high-side and NPN is easier to switch on the low-side, but for some reason you flipped it all around in your gate drive circuit.

I don't know what D1 and D2 are there for but I do not think they are needed (unless you can explain to me why you have them there).

THat is all I have time for now. I have to get back to studying, but you probably also want large capacitors across the power rails. I don't know if you included those in your circuit or not, and you may need some base resistors for your driver circuit in the same way you have them for the MOS gates. Gate resistors will slow down the turn on time for MOSFETs, but you might cause short-circuits if you don't have base resistors with BJTs.

The bridge is connected OK except for the speed implications noted. Because there's always a MOSFET connection path, you might get away with no diodes. The motor might need a snubber to absorb really fast transients, though.

The gate driver needs to have the PNPs flipped. And 10K is much too large (too slow) for efficient switching. Diodes D1 and D2 should be removed. The gate drivers ("Q5C, Q15C and Q1S") should be at the same potential to properly drive the gates. This must be no more than 12V (the maximum gate voltages on Q1 through Q4.)

Your MOSFETs have no shoot-through protection. Depending on the supply voltage, this might be handled with gate thresholds or the fairly high RdsON, but that's really not dependable.

mneary said:

Your MOSFETs have no shoot-through protection. Depending on the supply voltage, this might be handled with gate thresholds or the fairly high RdsON, but that's really not dependable.

Click to expand...

What is shoot through protection? Is this like dead-time (preventative)? Or is this something that kicks in once shoot-through has occured (like current limiting though I don't think that would work fast enough or frequency enough to deal with shoot through)?

Shoot through protection limits the current when both MOSFETs are on. This can be based on dead time (both MOSFETs never on at the same time) or RdsON. No active current limit (except resistance or inductance) could be fast enough to eliminate shoot-through damage.

thanks for the reply

why The gate driver needs to have the PNPs flipped ?

ahmedragia21 said:

thanks for the reply

why The gate driver needs to have the PNPs flipped ?

Click to expand...

THe answer is in my last post about how PNP and NPN transistors turn on. Read it again carefully.

You can also google the internet or read a textbook about how to use an NPN and PNP transistor as a switch. You don't seem to entirely understand what it takes to turn them on and off (though you seem to understand that the way you placed the PMOS and NMOS transistors in your H-bridge...or do you? Did you copy that part of the circuit from somewhere else? If you did you should almost learn how NMOS and PMOS transistors are used as switches. It is a similar to the difference between PNP and NPN).

I didnt understand what's meaning by flipping the PNPs ?
i know how a BJT acts a switch ,as Vbe >0 NPN is a switch closed hence VCE=0.2V, for PNP VBE <0 is a closed switch , hence VCE=0.2V.

When we say flip PNP and NPN, we mean change all your NPNs to PNPs and change all your PNPs to NPNs. Do you not see why your BJTs won't turn on?

i dont see why really ...
it works in the simulator and i measured the voltage at the VC of each BJT and its working ..
what's your point of view ?

Do you see why you think your circuit will work the way you have it now and why you think it won't work our way? Or did you just randomly pick one of the two and make your circuit like that?

That the NPNs and PNPs will not turn on, or if they do erratically and unpredictably. Build it and find out- as long as you choose NPN and PNP transistors in the same package you can always swap them in your circuit when it doesn't work.

Take a look at your PNP transistors- your emitter is connected to ground so clearly if you apply +5V to the base the voltage from emitter to base (Veb) is -5V, not the postive voltage you need to get current to flow to turn it on.

Now take a look at your NPN transistor. The emitter is floating (if you assume all BJTs start as off!) So if you make the base a positive voltage to send current into it to turn it on, you have no idea if it will turn on or not because the emitter is indeterminate so you don't know what Vbe is because Vbe = Vb-Ve and you don't know Ve because the emitter isn't connected to any voltage (and your base voltage/current is not referenced to Ve either).

At the very least, this is one way to think about it:
THe voltage required to send current in the right way to turn on the BJT is between base and emitter. YOu send signals into the base, so the signals are referenced to the emitter. You have to connect the emitter to a know reference so you know exactly what to send into the base. YOur first problem is your emitter isn't always connected to a know voltage (if it's not then you can use a floating drive that is referenced to the emitter, but you don't have that either).

Don't always trust a simulator. You have to understand how the circuit works yourself- or else you can't use the simulator correctly. Why are you measuring Vc anyways? Vc of one Q5 and Q15 will always be +V since you connected it to the positive supply and Vc of Q16 and Q6 aren't connected to voltage (they are floating). YOu should be measuring current, not voltage. Did you try measuring the current? Are you sure you're using the simulator properly?