H-bridge

[YAN-1]

Hello. I am trying to build an h-bridge for my 12-volt, 11-amp DC motor and I was looking for 2 n-channel and 2 p-channel MOSFETs to do that. But then I was thinking, why can't I build it using 4 n-channel MOSFETs?! That should work, right? Just like using 4 npn BJTs for the same purpose.

 

[Nigel Goodwin]

Hello. I am trying to build an h-bridge for my 12-volt, 11-amp DC motor and I was looking for 2 n-channel and 2 p-channel MOSFETs to do that. But then I was thinking, why can't I build it using 4 n-channel MOSFETs?! That should work, right? Just like using 4 npn BJTs for the same purpose.

You can, but you make much higher losses in the FET's, just as you do with 4 NPN's.

 

[YAN-1]

Well....why is that?!

 

[Nigel Goodwin]

Well....why is that?!

Because a bipolar transistor requires a voltage a fair bit higher than the base in order to provide enough base current to turn the transistor fully ON (and you also lose considerably more volts in common collector than common emitter). Likewise, an FET requires a good bit higher voltage (but not much actual current, unless you're doing it fast!) in order to turn it fully ON - so you would require a gate supply higher than the supply rail.

Using NPN/PNP or N/P channel devices cures this, as you have the full supply voltage available to the base/gate of each of the four transistors

 

[Styx]

Hello. I am trying to build an h-bridge for my 12-volt, 11-amp DC motor and I was looking for 2 n-channel and 2 p-channel MOSFETs to do that. But then I was thinking, why can't I build it using 4 n-channel MOSFETs?! That should work, right? Just like using 4 npn BJTs for the same purpose.

You can, but you make much higher losses in the FET's, just as you do with 4 NPN's.

why?
just use a floating supply for the top two FET's to ensure adequate GATE-SOURCE driving potential
You dont get any extra losses in the FET, it doesn't know what potential it is at??? IF anything the P-type has more losses then the N-type (higher RON, slower switching time,...)


However, the extra complexaty when working at 12V really isn't worth it.
P-type and N-type is ideal

Now IF you went double-voltage to say 24V then yup you would need to use a floating supply to be able to drive the Top FET's so using all N-types makes sense since you will then get better efficiency.


So for your application I wouldn't bother.

 

[Nigel Goodwin]

why?
just use a floating supply for the top two FET's to ensure adequate GATE-SOURCE driving potential

That's basically what I said! - but it's probably going to make life more difficult having to provide an extra higher supply rail.

 

[YAN-1]

But I already have two 12-volt batteries. I was gonna use one for the logic circuit (including turning on the N-channel MOSFETs) and the other for the motor supply and I would connect the grounds together. Is that inefficient? I mean I don't get it. Did you mean that the P-channel MOSFETs will already be using the 12 volts of the motor supply for their switching and I would need to add another supply for the N-channel ones?

 

[Styx]

why?
just use a floating supply for the top two FET's to ensure adequate GATE-SOURCE driving potential

That's basically what I said! - but it's probably going to make life more difficult having to provide an extra higher supply rail.

I was questioning the extra loss statement. In my experience for correctly driven FET's a P-type always has higher losses then an N-typre regardless of position

 

[Nigel Goodwin]

why?
just use a floating supply for the top two FET's to ensure adequate GATE-SOURCE driving potential

That's basically what I said! - but it's probably going to make life more difficult having to provide an extra higher supply rail.

I was questioning the extra loss statement. In my experience for correctly driven FET's a P-type always has higher losses then an N-typre regardless of position

The extra loss I was talking about was because you can't turn the top FET ON properly (just as you can't turn a top bipolar on properly either). 'Correctly driven' is a different story all together! :lol:

 

[Styx]

why?
just use a floating supply for the top two FET's to ensure adequate GATE-SOURCE driving potential

That's basically what I said! - but it's probably going to make life more difficult having to provide an extra higher supply rail.

I was questioning the extra loss statement. In my experience for correctly driven FET's a P-type always has higher losses then an N-typre regardless of position

The extra loss I was talking about was because you can't turn the top FET ON properly (just as you can't turn a top bipolar on properly either). 'Correctly driven' is a different story all together! :lol:

Ahhh well thats different then...

 

[YAN-1]

Well..this is the basic design of the circuit I intend to use. The MOSFET types are not the same though and I will add the freewheeling diodes. But I want to remove N1 and N4 and drive all MOSFETs using a MOSFET driver IC. The IC I have is the TC4427. The N-type MOSFETs are driven by outputting a '1' from the IC (which passes 12 V and up to 1.5 Amps to the gate). But as for the P-type ones, should I send a zero so that it 'pulls' the current and the gate is ON and then send a '1' to turn it off? Thanks.

 

[Styx]

Ahh a H-bridge.

Yup you can replace N1 and N4 by the driver chip to drive the P-type.

IF you were to use N-types at top you would need 24V. Why? you have 12V for you link and to turn-ON the top-FET's their GATE's would have to be raised 12V above the source. Treat the FET as a short when on puts the source at 12V, thus to ensure the gate at a higher potential it needs to be at 24V w.r.t. DC-link LOW

I dont mean to put 24V onto it, you would burn it out, if you were to take a scope probe you would see that it was at 24V.
What you have to do is use a isolated DCC converter and tie the 0V from that to the source of the top FET to provide a floating gate-drive potential

I go over this in the tutorial I am writing.



So as you can see it is alot of complexaty, so if you are working from 12V you might as well just use P-N type phase leg.
That driver chip will still work perfectly in this arranagement and since you are working from 12V you can reference it from 0V making control easier

Remember to turn the P-type ON you need to bring its GATE to the source potential, in this case 12V and to turn it OFF you need to bring it down low enough. Thus you can see why that chip will be fine here.




Basically because you have a complimentary-pair phase leg you send to the upper P-type the inverse signal to the lower N-type


Send "1" to UPPER and "0" to LOWER
P-type gate now at 12V at its GATE and is now conducting
N-type gate now at 0V and is now blocking



Send "0" to UPPER and "1" to LOWER
P-type gate now at 0V at its GATE and is now blocking
N-type gate now at 12V and is now conducting.

--EDIT--
Err I kinda hate P-type N-Type phase leg I have an itching feeling you send the P-type the same signal as the N-type (no inversion) since the N-type has an inherant inversion already



One warning!!!! you are entering into shoot-through territory.
There is two types.

1) Hard shoot-through (what we call it at work)
Basically when you tell an UPPER device and a LOWER device of a leg to turn-ON providing a short across your DC-link.

This is extreamly BAD and will always kill you devices and if you are unlucky your gate-drive. To give you an idea a converter I am working on has a DC-link of 350Vdc and a phase current of 600A. I have done a shoot-through (well actually a DeSat testing) and I was getting upto 3000A in 8us. The only reason my gate-drive and the IGBT survived was because of a DeSat cct

But generally a shoot-through will kill your devices!!! THIS kind of shoot-through (the most damaging) is pretty much always downto bad control (ie you told both to turn on)



2) soft shoot-through (again what we call it at work)
This isn't an intentional shoot-through like the hard shoot-through BUT more due to the fact that a switch cannot be turn-ON or turned-OFF instantly



IF a device is told to turn-OFF at the same time as its compliment device is told to turn-ON then there will be a period where current is flowing through both, and worst still when both are in high-power, high-loss linear region BAD!!!!

The picture isnt that good but you should get the idea. I have IGBT's here but the same is true for FET's and BJT's


Now how you get around this is via dead-time


So as you can see from this, there is a period where both IGBT's are told to be OFF, allowing one to fully turn-OFF before the other is told to turn-ON



Such a technique is CRITICAL to high-power bridges (where IGBT's would be used). Such interlocks/dead-time would be put in in the commutation logic.


For FET's however even though they to also suffer from the same problem they have the advantage that they switch 100x faster then an IGBT (partly due to the fact they take 100x less power...)

What you can do is use a sneeky trick to slow down a FET at turn-ON
You have a gate resistor of abt 18R (from another thread) This gives fast Turn-OFF as well as fast Turn-ON!!

What you want to do is change that resistor to a higher value one (say 100R or higher) THIS higher resistor sets your Turn-ON time since it will take longer for the FET to pass threshold potential

In parallel with that 100R put a diode and a 18R resistor making sure the diode is the right way round. This will provide a lower gate-resistance at turn-OFF allowing fast turn-OFF


The general rule is

Fast turn-OFF
Slow turn-ON



My gateboards that I design and build are effectively like a phase-leg in this H-bridge. A P-type at top and an N-type at bottom. I get shoot-throughs on my gateboard at switching because of this effect. This ages the FET's and draws more power.

I use the diode-trick to ensure that doesn't occur

 

[YAN-1]

Well thanks a lot for this explanation. I really appreciate it. But I don't get something. If I send a '1' (12 V) to the gate of the P-type, then its gate will be at 12 and its source is already at 12. How will that turn it on? (Vgs = 0). And If I send it a '0', this will make Vgs = -12 V. Isn't this supposed to turn it on instead of off?! As for the shoot-through, can't I simply switch it all off, give a small delay, and then turn the other direction on? Thanks a lot.

 

[YAN-1]

If I send a '1' (12 V) to the gate of the P-type, then its gate will be at 12 and its source is already at 12. How will that turn it on? (Vgs = 0). And If I send it a '0', this will make Vgs = -12 V. Isn't this supposed to turn it on instead of off?! As for the shoot-through, can't I simply switch it all off, give a small delay, and then turn the other direction on? Thanks a lot.

Can someone please try and elaborate? :lol: