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Complementary transistor MOSFET driver.

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cavemen

Wed Sept 08 2010, 03:55PM

Registered Member #2008 Joined: Tue Mar 03 2009, 05:11AM
Location: USA, Frederick, MD
Posts: 118

Hello.
I am back to work and struggling with the same questions.

I am trying to build a universal two transistor MOSFET driver, sometimes called totem pole driver. It uses push-pull output of the microcontroller to control powerfull applications.

It is a great module to use in solid state Tesla coils, radio control systems and CNC machine control modules that I may be developing in the future.

As far as I heared there are adapted modules that utilize both MOSFET and bipolar transistor functions but I am not sure how those components are called and if those components can stand high current.

Here is a problem. (I am not in a rush to blame my Multisim.)
Why WRONG circuit works better than RIGHT circuit and why my pair of complementary transistors doesn't want to sense the 3.3V voltage?

It looks like hFE is not theissue here since I placed darlingtons and ordinary transistors into the circuit.
RIGHT circuit does sense +-5.3 input.

The resistor that is used in the WRONG circuit is definitely noa a solution but an adjustment made to get the result in Multisim.

So what do I need to change?

("RIGHT" circuit is the way it is supposed to work)
Have a great day.

V.T.

Mattski

Wed Sept 08 2010, 07:48PM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

What is the gate threshold voltage of the MOSFET? The output of the NPN when V_in > 0 will be about V_in-Vbe(on) ~=Vin - 0.7V. It acts as an emitter follower, so there is no voltage gain, just a DC drop. So 2.6V can be a little low for turning on power MOSFETs.

Then when on the negative swing of the signal generator, the NPN will be off and the PNP's output will be V_in + 0.7V, or about -2.6V. So this is fine as it will turn of the FET easily.

It's worse for the darlington case because now you have two base-emitter drops to contend with, so the gate will only be charged to 1.9V, which is certainly not enough.

If your signal generator was for example 0 to 6.6V instead of -3.3 to +3.3V that should probably fix it.

The "wrong" circuit works because each bipolar can pull the FET gate near the power supply rails - enough to turn it fully on and off. However you will get shoot-through where both bipolars are on at once and just conducting current straight through, dissipating a lot of power and causing a spike in the power supply.

Avalanche

Wed Sept 08 2010, 07:57PM

Registered Member #103 Joined: Thu Feb 09 2006, 08:16PM
Location: Derby, UK
Posts: 845

Yes Mattski has summed it up with the gate voltage being too low. Threshold voltage (typical) for an IRF540 is 3v IRRC and even that is WAY too low to be reliable. If your drive signal really can only be 3.3v, then connect another NPN transistor in a common emitter configuration with a pullup, and use the output from the collector to drive your output stage. It's not ideal, and will also invert the signal, but then your output stage will follow the voltage on the pullup resistor and switch +-12v (minus vbe for the top)

cavemen

Thu Sept 09 2010, 03:21PM

Registered Member #2008 Joined: Tue Mar 03 2009, 05:11AM
Location: USA, Frederick, MD
Posts: 118

Is there any way to use the "wrong" type of circuit and make sure there is no shoot-through current?

I added some resistors to help compensate for the Vbe of the NPN transistor in the "right" circuit. That caused the circuit to do the switching but MOSFET doesn't open fully.

Vbe is a voltage drop that has to happen for transistor to work? Right?

The problem is that Multisim virtual oscilloscope doesn't work at khz. It shows straight lines.
My circuit will have to PWM at 5000hz. The MOSFET may not have time to charge/discharge through a transistor.But i did the calculations for this and have the exact numbers.

I am not sure how another NPN transistor can be used to make "right" circuit more effective. Can you explain this please.

I am surprised that noone came up with an easy solution to this problem before.
Maybe I will stick with the wrong circuit.

Avalanche

Thu Sept 09 2010, 04:56PM

Registered Member #103 Joined: Thu Feb 09 2006, 08:16PM
Location: Derby, UK
Posts: 845

Here's what I meant...

The problem is caused by your input signal only being 3.3v. The emitter follower stage on the output can only follow this 3.3v - it cannot amplify the voltage (as soon as your output voltage as seen on the gate of the IRF reaches 0.7v less than the base of the top transistor, no base current can flow so it stops there!)

The extra transistor just amplifies the switching signal from a 3.3v signal to a (nearly) 12v signal before the emitter follower on the output.

As I said, it isn't ideal because it inverts the duty cycle of the signal, but that might not matter.

Mattski

Thu Sept 09 2010, 06:21PM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

wrote ...

Is there any way to use the "wrong" type of circuit and make sure there is no shoot-through current?

The problem is when Vin is between 11.3V and 0.7V both of the transistors will be on. So to alleviate the problem you would need to make Vin nearly rail to rail, and you will still have shoot through while Vin transitions from its high to low value.

wrote ...
Vbe is a voltage drop that has to happen for transistor to work? Right?

Yes. The base emitter junction is a diode, for an NPN transistor if points from base to emitter, for a PNP it's the opposite. You need current to flow through this diode to turn the transistor on. As with a normal silicon diode, pretty much no current flow until is has a voltage of 0.7V across it. So in order to get a silicon bipolar to conduct you need 0.7V across the base emitter junction.

The circuit Avalanche provides should solve your problem.

cavemen

Fri Sept 10 2010, 02:00PM

Registered Member #2008 Joined: Tue Mar 03 2009, 05:11AM
Location: USA, Frederick, MD
Posts: 118

Thank you for explaining.

One thing I don't get is why my mosfet doesn't open when I connect it directly to the signal source.
Yet two transistors that only work as emmiter followers (in the "right" circuit) open it partially even though they don't amplify the signal by voltage.

Avalanche, what simulator software are you using?

Avalanche

Fri Sept 10 2010, 03:49PM

Registered Member #103 Joined: Thu Feb 09 2006, 08:16PM
Location: Derby, UK
Posts: 845

It's the free version of simetrix

cavemen

Sun Sept 12 2010, 11:51PM

Registered Member #2008 Joined: Tue Mar 03 2009, 05:11AM
Location: USA, Frederick, MD
Posts: 118

OK
Thank you for your explanations.
Just to fill all my understanding gaps:

Why my mosfet doesn't open when I connect it directly to the signal source.
When two transistors that only work as emmiter followers open and close it even though they don't amplify the signal by voltage.

A ULN2001 transistor array chip requires 2Vce. Does it mean that this chip requires 2 volts applied to the input pins to turn something on?
Ii is the input current required to operate the circuit?

Just found an ultimate bug in multisim11.

Thank you.
V.T.

Mattski

Mon Sept 13 2010, 07:33PM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

cavemen wrote ...

Why my mosfet doesn't open when I connect it directly to the signal source.

Is this still the 3.3V source? When the gate-source voltage is below the threshold voltage current will be very small. I think for your device threshold was 4 or 5V. But you must be careful not to just get it working in simulation, because the model in simulation may use one value, but threshold voltage is not well controlled for power mosfets, so the device you build with may be worse. You need the gate-source voltage to be sufficiently above the maximum specified threshold for your device.

wrote ...

When two transistors that only work as emmiter followers open and close it even though they don't amplify the signal by voltage.

By "open" I assume you mean "turn on" but for best terminology "open" typically means "open circuit", and that is when the FET is off. Offhand I'm not sure why it's happening. More information is needed, in particular it would be good to see the plot for the "right" circuit with units or volts/div of the y-axis, plus a plot of gate voltage (labeled axis), and maybe current through the collectors of the BJT's.

It could be as simple as an artifact of simulation where that output voltage is so small that it's negligible. I've seen such things in simulations where you think something interesting is happening but it's really so small that you're just seeing tiny errors from the numerical solver. That's why we need the scale of the axis.

wrote ...

A ULN2001 transistor array chip requires 2Vce. Does it mean that this chip requires 2 volts applied to the input pins to turn something on?
Ii is the input current required to operate the circuit?

Not sure here, can you link to the datasheet you're looking at and the page? Maybe someone else here already knows. But Vce refers to collector-emitter voltage difference, not 2V.

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