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 DC
C 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