Hey there,
I am using an N channel power mosfet (IRFZ44N) to control power to a high current motor with atmega8.
I've connected atmega 8 I/O pin to gate, one motor terminal to source and other to +ve power and drain of mosfet as well as atmega8 ground to a common ground.

Circuit diagram attached

It seems correct but there are certain wierd problems as follows:
1.For gate voltage 5v there is very little current from drain while for 12v at gate current is very high!

2.The mosfet sometimes turns itself on without any gate voltage applied

3.When gate voltage is set to 0 from some +ve voltage it sometimes take time (5-8 sec) for drain current to become 0


How can I solve the above Problems?

 

You need a P-channel, or you need to wire it for the low side.

Also you need a small BJT and a gate pullup (for P-ch).

Works 100% with any microcontroller, all you need is 2.2K and a 2n3904.

For N-ch the control signal is inverted, the BJT is used the same way.

 

That MOSFET needs the 5V output the atmega8 can provide to go gate to source, so the source needs to go to ground. That means the drain will act as a switch to ground, so you need to wire the motor there.

 

That MOSFET needs the 5V output the atmega8 can provide to go gate to source, so the source needs to go to ground. That means the drain will act as a switch to ground, so you need to wire the motor there.

PRESTO worked like a charm ! thanks a TON

one more question

Datasheet rates it as
IRFZ44 55V, 49A N channel Power MOSFET

so does it mean that with proper heat sink the mosfet can supply a 49A load at 55v continuously

 

If the MOSFET burns out, so your microcontroller will. Not with a BJT.
5V is just on the margin to turn on IRF MOSFETs. No use if you have larger currents. Think it could be 4.5V (USB).

 

If the MOSFET burns out, so your microcontroller will. Not with a BJT.
5V is just on the margin to turn on IRF MOSFETs. No use if you have larger currents. Think it could be 4.5V (USB).

What's your point?

 

What's your point?

Nothing. You will see for yourself over time.

I use MOSFET for 2 Amps 12V, and I don't control the gate directly. It's P-ch anyway.

If it works for your circuit, fine.

 

Maybe you have not looked at this.
It's in the datasheet. Not for your MOSFET but the gate voltage is all the same.

How can I know about your motor.
Of course I assume it is a small motor.

The point is to tell you 5V is just at the margin. If a 12V rail exists, normally that is used for the gate voltage.

 

Yes if you expect large currents through the mosfet, larger Vgs is allways better as it gives you lower Rdson.

 

one more question

Datasheet rates it as
IRFZ44 55V, 49A N channel Power MOSFET

so does it mean that with proper heat sink the mosfet can supply a 49A load at 55v continuously

No. The mosfet has "package" limits. All mosfets that are in the To-220 style package, are limited to around 75 Watts of power. By that I mean that the volts times amps can be no more than 75.

This is probably the best/easiest to understand tutorial for choosing and understanding what the data sheets mean in mosfet selection. http://www.mcmanis.com/chuck/robotics/projects/esc2/FET-power.html

 

No. The mosfet has "package" limits. All mosfets that are in the To-220 style package, are limited to around 75 Watts of power. By that I mean that the volts times amps can be no more than 75.

This is probably the best/easiest to understand tutorial for choosing and understanding what the data sheets mean in mosfet selection. http://www.mcmanis.com/chuck/robotics/projects/esc2/FET-power.html

Much less than that. It is not so easy to cool them effectively, you'd need forced cooling.

But normally they are not used in linear mode. The voltage accross the junction counts. With a large sink, you can dissipate about 20W, with a smaller one, about 6 Watts.

If you switch them fully on, depends on the IRF model number, but 2 Amps are possible typically without any cooling.

That is what I mean by over time, OP would experience all that.

The Wattage as such as well Amps and voltage rating are all misleading.

If you approach the voltage limit and they are hot, these MOSFETs have a good chance for explosion.

It is hard to explain all the details without writing a book chapter. So I tried to raise a few topics which are interesting when you use these in circuits.

Turning them fully on is absolutely essential, unless you only have small currents. Remember what you see in the chart is the limit from the increased ON resistance. You need more than 5V even for 1.5 Amps (but that depends on the particular IRF model number).

These IRF MOSFETs are a particulary good technology.

 

one more question

Datasheet rates it as
IRFZ44 55V, 49A N channel Power MOSFET

so does it mean that with proper heat sink the mosfet can supply a 49A load at 55v continuously

To design a reliable circuit you should not operate any component at or near any of it's maximum ratings.
The maximums are listed to provide a basis for selecting reasonable operating parameters.
Components that handle the highest power/voltages in circuits are also the ones that frequently fail if the design is not conservative.

 

There are few components and circuits even capable to deal such currents to a MOSFET.

Most smaller coils and transformers have such a wire resistance currents never raise that much.

One path is to calculate everything.

Another is to know certain MOSFETs, heatsink sizes and things like that. And to know how many Watts they can handle. I mean knowing from experience.

The datasheet is a helping to choose components when you already have that experience.

Nothing can replace solid experience with such circuits.

OPs post/question was solved. These MOSFETs can do many things. It is well advised to make oneself familiar with them.

The question is rather why would you want to deal with 49 Amps at 55 volts.

 

SO there's a problem now! making an H bridge

I have IRFZ44N mosfet and I want to drive it with 5v WITH LOAD ON HIGH SIDE.

but it wont work

any suggestions

 

N-Channel MOSFETs can't be used as a high side driver in an H-Bridge with logic level voltages. You'll need to use a Driver or a P-Channel MOSFET for the high side switching.

There's A similar thread on this that I've put over 2,000 words into, skim through it thoroughly, then Read this one and stop back here.

 

For your original circuit, if the MOSFET gets hot, you can increase the gate voltage with a few components. The second circuit on this page shows how to do it (the first doesn't work):
http://letsmakerobots.com/node/4285
If you wanted to do speed control for the motor using high frequency PWM you probably need a few more components.

 

heres a simple fet op stage....

 

It's a little misleading to say "limited to around 75 Watts of power. By that I mean that the volts times amps can be no more than 75."

That's true, but the "volts" in question are across the transistor, not the power supply voltage. If the transistor were turned fully on, the power dissipated would be I squared times the channel resistance, which for the IRFZ44 is listed as .028 ohms at room temperature. Generally the way we operate a MOSFET is with some load taking most of the power supply's voltage, and some much smaller voltage driving the current through the transistor, so effectively the MOSFET acts as a resistor in series with the load.

If someone needed to run huge current through a MOSFET, there are ways to put multiple parts in parallel, which looks electrically like a smaller resistance, and also spreads the thermal loss out over more heat sink area.

 

@ takao21203 & John P, I may be wrong(often am) but think you guys are confusing "heat watts/temp" due to Rds on ohms, with actual transfer Watts allowed by package (TO-220) design. A mosfet that was/is limited to the watts of Rds would be pretty much useless. And there would only need to be one part number instead of the many different ones that are available in all the volt and amp there are.

The internal leads to the mosfet die are the limiting factor. And the rateing is set from the mosfet being turned full on, not in the linear region of operation. In the linear region of use the transfer Watts are de-rated according to the SAO graph in the data sheet.

 

I don't get what you mean by "heat watts/temp due to Rds on ohms" even assuming you wanted to say "in ohms" not "on". The heat rise caused by Rds is the minimum you can get away with for a given current; for a large current, it would be the limiting factor. I assume that Rds is actually made up of resistance of the silicon and bonding wires combined, and you can't separate them. If the device overheated, it's not clear which would actually fail.

In the linear region, a smaller current would obviously cause a lot of heating. But you could be sure that any failure then was in the die, as the increased voltage drop would be there rather than in the wires.