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This is my H-bridge circuit without considering Back EMF. In the real circuit I would put four fast recovery diode to eliminate the Back EMF.Maybe schottkies? enter image description here

Before I came up with this design , I googled "DIY H-Bridge". And I only found out H-bridge design that is composed of two p-channel mosfet and two n-channel mosfet , as follows. My question is: Will this four n-channel mosfet H-Bridge circuit really work? If it works. Why the 'TWO N-CHANNEL TWO P-CHANNEL' plan is more popular? I mean. It is said that N-channel mosfet can handle high current much better than P-channel mosfet.I have concerns that the top N channel Fets are not getting enough drive .Will they run hot ?My proposed circuit is simple but do you think it needs refinement? Do you think that shoot through could be a problem?

enter image description here


At first , I didn't know what is high-side switching.This thread might help for people who also have question about what is high-side and low-side switching.

http://www.edaboard.com/thread182857.html

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  • \$\begingroup\$ Take a look at this question that I answered few weeks ago. This might give you some idea. \$\endgroup\$ – Whiskeyjack Jan 3 '16 at 10:59
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No. it won't work.

there's nothing to turn the top two mosfets all the way on, you need to get gate voltage higher than the drain voltage, else the mosfet will overheat.

for a solution search on high side MOSFET driver

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  • \$\begingroup\$ Gate voltage higher than the drain voltage to turn on N-channel mosfet? I mean . I only know that gate voltage should be higher than source voltage to turn on the n-channel mosfet. And why dose the simulation work? \$\endgroup\$ – Frank Jan 3 '16 at 10:31
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    \$\begingroup\$ when the upper mosfet is on the difference between source voltage and drain voltage should be less than 1 volt. if it's not you'll cook the mosfet. so gate voltage should be higher than both drain and source. \$\endgroup\$ – Jasen Jan 3 '16 at 10:40
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Why the 'TWO N-CHANNEL TWO P-CHANNEL' plan is more popular? I mean.

The complementary design with N and P type FETs is more popular because it can be driven from a single supply voltage. With All N-channel FETs you need a Gate drive voltage higher than the motor supply voltage to turn the upper FETs on fully.

Your simulation 'works', but if you check the numbers you will see that the motor only has 7.75V across it (1.55A*5Ω). The other 12.25V is dropped across the upper FET, which will get pretty hot trying to dissipate 19 Watts!

It is said that N-channel mosfet can handle high current much better than P-channel mosfet.

This is true, but it's not important unless you want the highest possible current handling. The IRF540 is nowhere near 'state of the art' and many P-channel MOSFETs can match it. They may have higher Gate charge, but this is only a problem if you need to drive them with high frequency PWM (one reason for only applying PWM to the low-side FETs).

The main advantage of using all N-channel FETs is that you only need one part instead instead of two, so you can get a better bulk discount. Also P-channel FETs with the same specs tend to be a bit more expensive.

The FETs in your second circuit will suffer shoot-through if their combined Gate threshold voltages are lower than the supply voltage. I suspect this is what the "Vcc < Vgs!" is referring to. Note that threshold voltage can differ widely between individual FETs of the same type, so finding FETs that are guaranteed to work properly could be tricky.

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