Driving Dual MOSFETs From A NPN Transistor

"Michael"

** You must supply more info.

Are the MOSFETS wired in parallel to increase current ?

Are the MOSFETS switching or linear types ?

Is you circuit switching or linear ?

Etc.

........ Phil

what you describe suggests that you have the NPN transistor manipulating the voltage on the gates of the mosfets, that effects their resistance.

not much more can be said without seeing the types of transistors and how it's all hooked up.

plenty of rules, the more info you give the better.

--

Bye.
   Jasen

Hi,

Sorry for not getting back sooner,

I've drawn you a schematic and uploaded it to:

I'm trying to work out a suitable transistor (Q3) to use and it's base resistor (R1) value... From what I've read in the mosfets datasheet

2-4V needs to be applied to the gates so that's where I get the 3V value from and the 36V figure is from the onboard batteries (it's for a robot).

Where should I go from here?

Thanks,

Michael

Looking at this, you could do the following:

1. Replace Q3 with a logic level MOSFET (the venerable VN2222LL would be perfect), and change the sense completely. As shown, the circuit really won't work that well.

Either use a PNP with emitter to positive rail or use a P-channel MOSFET (Source to positive rail). If you need to drive with 3V or 3.3V logic, then there are solutions. The simplest fix here (assuming 3.3V logic) and the fact you need 5V for gate drive (see 3) is to

a. Remove Q3 emitter from the gates and tie it to ground b. Remove Q3 collector from supply, and pull it to the supply via a 10k resistor c. Connect a 2N3096 with collector to gates, base to Q3 collector via a

1k resistor, emitter to 5V

1. Add a pulldown from the MOSFET gates of 10k or so. There has to be a DC return, and a resistor to ground is the simplest.

2. You need to provide a minimum of 4V to the gates of Q1 and Q2. This is the only way you are guaranteed to exceed Vgs(th). Better would be

5V, so change the power supplied to the collector [drain if you take the advice in 1] to that level. Note your TTL circuitry could be operating on 3V and the VN2222LL would work fine as it would if it were a bipolar device.

1. I assume M1 and M2 are motors. As inductive loads, you need to put diodes across them; anode to the MOSFET drain, cathode to positive power for the motor. If you don't, the MOSFETS will operate once, and at the first turn off, they will be toast (literally).

Cheers

PeteS

Ok thanks Pete....

Just to make sure I understand you correctly...are these right?:

OR

Michael

In

1. Change Q4 to a 2N3904 or a VN2222LL

1. Remove R2 or R3 (only one needed)

2. D1 and D2 should be across the motors, not the FETs

Then we're starting to get there.

Cheers

PeteS

Thanks, so this ok then?:

Cheers,

Michael

Your 2N3904 shows a PNP - it should be an NPN :)

Apart from that, you'll need some decoupling - put a 0.1uF cap from the

5V rail. The amount necessary on the 36V rail depends on the motors.

The DC ratings:

Use 10V or better rated for the 5V rail Use 63V electrolytics on the 36V rail.

As a first cut, I would use 47uF on the 36V rail.

Cheers

PeteS

PS : I hope you are learning something :)

Hi, Michael. I'd guess you're overthinking this a bit.

First, according to the data sheet (ST page 4/18), the gate threshold voltage of the MOSFET is between 2V and 4V. That means that, with a gate voltage of less than 2V, 100% of test devices will be OFF. At a gate voltage of 4V, 100% of test units will be ON. It gets a little more complex than that, though, because it kind of depends on what you mean by "ON". You'll have to look at the graphs for that info.

If you look at the graph titled, Figure 8. Static drain-source on resistance (ST p. 7/18), you'll see that the juicy Rds(on) spec of 0.02 ohms is with a Vgs of 10V. Since you've got a 36V supply to play with, that should be easy. You can do this with a couple of resistors like this (view in fixed font or M$ Notepad):

36V + 36V 36V 36V 36V | + + + + .-. | | | | 10K| | 36V | | | | | | + / \\ - / \\ - '-' | ( M ) ^ ( M ) ^ | |< \\_/ | \\_/ | o-| | | | | | |\\ | | | | .-. | 2 X STB55NF06 o----' o----' 10K| | .-. | | | | | |6.8K ||-+ ||-+ '-' | | ___ || | | GND | GND | .-. | ___ | | | | '----|___|----------' === | |2.7K 100 GND '-' | === GND (created by AACircuit v1.28.6 beta 04/19/05

This is somewhat easier than your original circuit, because you don't have to supply another regulated voltage. It would be even easier if you could live with an active low TTL signal (TTL low = ON) instead of the active high signal shown in your application, but there it is.

When the TTL signal is high (typically 3.5V unloaded), it will source

1/2mA or so into the base of the first transistor, turning it on. That will turn on the second transistor, making about 10VDC available to turn on the two MOSFETs. The 100 ohm resistors are typically used to isolate the gates and help avoid oscillation. Be sure to use diodes rated for the motor current, so they don't smoke to save the MOSFETs on turn-off. Be sure to use transistors rated for the voltage (such as the 2N4401 NPN / 2N4403 PNP).

Hope this has been of help. Not the most elegant thing, but it will work very well.

Good luck Chris

lol, I appreciate the help :-)

This better?:

Where do the values of 10K and 1K for the resistors come from? I presuming they aren't 'rigid' values.....

Cheers,

Michael

Looks good, but to know the proper values for the decoupling/mass storage for the motors I would need more information about the motors.

You are correct, they are not rigid values. Let's take R4 / R5

When Q5 turns on, there will be a few 10s of milliamps of current to charge the gates, and assuming a worst case hfe of 10, a milliamp or so of base current into Q5 during this time. It will stabilise at about

500uA, with a base current of 1/10 of that max.

R4 + R5 provides that current into the base. We could probably increase the values significantly, but the trick of 'make it work' errs on the side of plenty of current for bipolar devices.

Hope that helps too!

Cheers

PeteS

Ok thanks Pete, that makes sense...

The motors are pretty heavy duty and rated at 1.8Kw (I'm thinking the cap value will need to be increased a fair amount...?)

Could you (or Chris) explain why your circuit has decoupling caps and Chris' doesn't?

Cheers,

Michael

Perhaps because we were trying to look at the basics first, and the practicalities afterwards :)

Cheers

PeteS

Hi, Michael. Wow. A 50 amp, 36V motor. Quite a load there.

First off, your motor current will exceed the MOSFET rating. You might want to consider running each motor off two or three of your MOSFETs in parallel. Since they're ohmic when fully on, they'll share current quite nicely. The driver circuit I showed can run four or six MOSFETs as easily as two. You see, you just can't run the motor off one MOSFET in the real world. That's because it's rated to 50A at a junction temp of 25C, but is derated to 35A at a junction temp of 100C. Even with the magical 18 milliohms (it is to dream), you'll still have nearly 50 watts boiling off one MOSFET at 50A. It's gonna be very difficult to even keep the die temp down to 100C, without a continuous spray of freon. Use at least two MOSFETs -- I'd recommend three in parallel for each motor.

Second, you're going to have to be very careful about layout. Try to remember that any wire has some resistance as well as inductance. The best thing to do might be to keep the line between the MOSFET source pin and GND as short as possible, and have the GND for the logic power and the motor power meeting at one and only one point.

Use a big heat sink. Make sure you include the big honker diodes (rated for 50 amps repetitive surge). The avalanche diode internal to the MOSFET can handle a meager 340 millijoules -- and they won't run in parallel. Your motor can exceed 340 millijoules on turnoff with both hands tied behind its back, hopping on one foot and humming "Dixie" while typing the Gettysburg Address from memory on a Palm PDA with the stylus gripped between its teeth. You turn it off and depend on the internal diode, you let out the magic smoke. And if you run MOSFETs in parallel, it will destroy them one at a time until, within a second, they're all toast.

Borrow a storage scope if you can, to actually see what's happening on turn-on. Aristotelian theorizing only takes you so far, and it's depressing to pay a buck apiece times 2 or 3 just to have another WAG at what's wrong.

Now, if you can do this, you will have learned something. You might want to start out switching at a lower voltage. This will give you the advantage of lower power and lower current. As you get a handle on the switching, you can bump up the power supply until you're riding at 36V.

Best of luck, and Godspeed Chris

Michael, all things considered, it might be best just to use a couple of contactors if you're not switching too fast or frequently (or one contactor with two N.O. contacts).

If you place the honker diodes across the motor, you won't have to worry about arcing welding the contacts.

Suitable special purpose contactors with DC coils are available from many sources. You might just want to use a 24VDC coil (they're pretty common), and put it in series with a power resistor that has half the resistance of the coil. This will divide up the 36V pretty well. Be sure to use a diode across the contactor coil to protect your power transistor or MOSFET.

Actually, your MOSFET might be a good choice to drive the contactor.

Good luck Chris

Thanks Guys,

Chris,

It's going to be used on two 'sets' of motors - one 12V 200W motor pulsed at

33% max and the 1.8kW one at the full 100%. But even so I'm happier with the 'mosfets only' way Could you explain this please:

AND

Do you mean the internal avalanche diodes won't survive if I havn't got the 'big honker diodes' in place?

Cheers,

Michael

Yes. The internal diodes of your MOSFETs are rated for 340 millijoules non-repetitive. This is far less than the back EMF generated by the

1.8KW motor when it turns off -- they won't survive. And diodes in parallel tend not to share current very well. The one with the lowest forward voltage will hog the current.

No matter what you do to turn off the motor (contactor, solid state, whatever), this back-EMF needs to be dissipated. Placing a big diode across the motor will cause the energy to be dissipated through the motor itself. Probably a good idea no matter what.

Good luck Chris

Thanks Chris,

Actually I was playing with the circuit today and hit something that's puzzling me (I wired it as a h-bridge)...

Could you explain why when one the inputs (the little squares at the base of Q1 and Q17) is high, the motor turns as expected but only 5.51V reaches it with the 30.49V being dropped accross the top mosfets?

I'm guessing it has something to do with the load connected to the source as opposed to the drain.....

Schematic:

Schematic5.png

Cheers,

Michael

Oops, apologies:

(I realise the diode isn't there as well)

Cheers,

Michael