How to design a H bridge with N-MOSFETs and control them?

Your circuit, as shown in the image, is creating a continuous rail to rail short via M3 and M4. The bottom Mosfets are also inverted.

The very best way to learn and properly design an H-bridge: read manufacturer's datasheets and app notes. They provide an incredible wealth of information which simply cannot be summarized on a single post.

International Rectifier, now part of Infineon, has been the king of H-bridge drivers for many years. Their drivers are literally used everywhere.
So I would start there. But TI, OnSemi STMicro and other also have drivers.
 

crutschow

Joined Mar 14, 2008
23,365
You have the MOSFETs wired back-to-back and the gates wired to a fixed voltage so the bridge is permanently in one state.
Do you have a question? :confused:
 

GopherT

Joined Nov 23, 2012
7,983
Thanks for your reply.

My main question is, in which way i can control the for MOSFETs?
H-bridges are just a tool, they don't need to be controlled. The thing connected to the output of the HBridge is getting controlled. And the HBridge sin smart enough to control anything either so you'll need some Microcontroller or logic chip to connect to the ?HBridge inputs.

What do you want to control? A motor, a solenoid / linear motor, ...?

What is your experience level for the input side. That is, how complex (on/off, microcontrollers, ...)
 

GopherT

Joined Nov 23, 2012
7,983
Hello,

I need for my project a H bridge rectifier only with n-channel MOSFETs.
I attached a curcuit simulated with LTSpice and a pic.

thanks
Your picture is confusing. Flip the MOSFETS on a vertical plane (y, z axis) Then the load should connect to the MOSFETS without crossing any wires. Connect the inputs from the left and right side of the H. Power top to bottom (you got power right).
 

Thread Starter

Jimpes

Joined Jan 12, 2017
8
Ok new try.

I think I'm on the right way.

But now I have two little tolerances.
1. voltage is little bit break down
2. nearly the zero line....

How can I fix these two things??
 

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Thread Starter

Jimpes

Joined Jan 12, 2017
8
Thanks for this hint.
But I should design it, only with simple components.. (as FETs, resistor, capacitors, single OP permitted too)
 

GopherT

Joined Nov 23, 2012
7,983
Ok new try.

I think I'm on the right way.

But now I have two little tolerances.
1. voltage is little bit break down
2. nearly the zero line....

How can I fix these two things??
That is actually a good thing near the cross-over. An EXACT match is impossible so making is shorter is possible. The thing you DONT WANT is known as "shoot through" where both P-channel and N-channel on a given side of the H are on at the same time. Instead of your power curve dropping to zero for a fraction of the waveform, the power curve spikes to infinite power for that cross-over period.
 

GopherT

Joined Nov 23, 2012
7,983
Thanks for this hint.
But I should design it, only with simple components.. (as FETs, resistor, capacitors, single OP permitted too)
PS, your drawing is still a mess. Here is a common and clean way to represent an H-bridge - the motor is in the center of the image.

Inputs come from left and right sides, power from top (+) and bottom (-).
(NOTE: this version is using 4 N-channel MOSFETs)

upload_2017-1-13_15-24-42.png
 

MrAl

Joined Jun 17, 2014
6,500
Ok new try.

I think I'm on the right way.

But now I have two little tolerances.
1. voltage is little bit break down
2. nearly the zero line....

How can I fix these two things??
Hello there,

Apparently you want near perfect rectification, perhaps for a measurement application?

Anyway, it is unclear what A1 and A2 are.

The reason for the loss at the peak is because the MOSFETs have some internal resistance. To minimize that effect you'd have to get new mosfets if the drive is the correct level, or if the drive is not correct for the mosfets then you'd have to improve the drive level.

The reason for the loss at the zero line is because the A1 and A2 do not change state at exactly zero input. That's because they need some finite non zero level input in order to change their output to a logic HIGH. Since they are driven by the input sine and the sine does not get to that level at zero degrees, they do not switch to a high output at zero degrees but instead some time later.
To correct that you'd have to get a gate or make your own that detects smaller voltages and thus changes state sooner.
Be aware though that you always need some dead time because the mosfets do not switch perfectly. That is, each mosfet does not turn off at the very instant that the drive signal says it should turn off. It takes some time, and if during that time the other mosfet of the bridge turns on, then the input AC gets short circuited, and even for a short time this is a very bad condition which could blow the mosfets. So some dead time is always needed in any bridge. Dead time is just the time that all four mosfets are OFF.

To minimize the amount of dead time needed, you have to increase the drive current to the mosfet gates, especially during turn off. This makes the mosfet turn off as fast as they possibly can and thus the needed dead time decreases. This kind of drive can only come from a good mosfet driver chip, so if you use other means you may just have to settle for a little more dead time. Lots and lots of circuits make out just fine like this.

Again, we dont know what A1 and A2 are so it's not possible to evaluate this design in full.
 

crutschow

Joined Mar 14, 2008
23,365
For a MOSFET AC bridge rectifier bridge driver such as the LT4320 I mentioned, there obviously is no MOSFET gate drive around the AC voltage near zero volts. The current is carried by the MOSFET substrate diodes in this region until the AC voltage increases to above the MOSFET's threshold voltage, at which point the appropriate MOSFETs turn on.
(Note that the normal conduction direction for the bridge is in the reverse direction from source to drain when the MOSFET is on, but that works because a MOSFET conducts equally well in either direction when on).
Thus there is not a particular concern about dead-time or shoot-through..
This does give about a 0.7V increase in the bridge output voltage when the conduction changes from the MOSFET substrate diode to its channel after the MOSFET turns on.
But since the current is low in this region of the waveform (or zero if the output is a capacitor filter), the losses are low in the substrate diode.
 
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MrAl

Joined Jun 17, 2014
6,500
For a MOSFET AC bridge rectifier bridge driver such as the LT4320 I mentioned, there obviously is no MOSFET gate drive around the AC voltage near zero volts. The current is carried by the MOSFET substrate diodes in this region until the AC voltage increases to above the MOSFET's threshold voltage, at which point the appropriate MOSFETs turn on.
(Note that the normal conduction direction for the bridge is in the reverse direction from source to drain when the MOSFET is on, but that works because a MOSFET conducts equally well in either direction when on).
Thus there is not a particular concern about dead-time or shoot-through..
This does give about a 0.7V increase in the bridge output voltage when the conduction changes from the MOSFET substrate diode to its channel after the MOSFET turns on.
But since the current is low in this region of the waveform (or zero if the output is a capacitor filter), the losses are low in the substrate diode.
Hi there Carl,

I sort of agree for the most part, but there are some circumstances that could be of concern.

For a couple examples:
What if all four mosfets are on at the same time?
What if there is a typical capacitor filter at the output of the rectifier?
 

crutschow

Joined Mar 14, 2008
23,365
What if all four mosfets are on at the same time?
The control circuit must be designed so that does not happen.
What if there is a typical capacitor filter at the output of the rectifier?
The control circuit is also designed to allow for that, to make sure no MOSFET is on that would allow backward current through the bridge..
All in all, it requires some sophisticated control circuits.
That's why a control chip such as the LT4320 is normally used for that task.
 
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MrAl

Joined Jun 17, 2014
6,500
Hi again,

Well, that's why i mentioned the dead time :)

But i think you bring up a good point here about the control chip because there's more to this then we are seeing in his schematic.
If we have a capacitor output filter then we need to be even more careful about when we turn the MOSFETs on. The capacitor will have some voltage across it, and that means we dont want to turn on a mosfet when the sine voltage is lower than the cap voltage. This means monitoring the voltage across the mosfet, which isnt being done in his schematic yet. Of course he just has a resistor load right now, and that would work like that, but that's not usually how a bridge rectifier is used as there is almost always a cap filter. Granted if there was an inductor also it could work, but who wants to use an expensive inductor. IF we dont monitor the mosfet voltage then we should at least monitor the sine voltage and the output voltage and dont allow a turn on if the sine is less than the output voltage. In a diode bridge the diodes do that automatically.

I guess it is a different story if he really wants a resistive load only, but that seems unlikely except maybe for some special purpose.
 

crutschow

Joined Mar 14, 2008
23,365
Well, that's why i mentioned the dead time
Since only two MOSFETs are on at a time and the changeover to the other pair of MOSFETs occurs at the AC zero crossing I don't see that dead time to prevent shoot-through is a significant factor in the design. :confused:
 
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