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How do drive a power mosfet gate

BobK

Jan 5, 2010
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twister,

You do not understand power MOSFETs. The gate is basically a capacitor. In order to switch it quickly, you need to charge / discharge this capacitor quickly, which means lots of current. Typical gate drivers can source and sink 1 A for the hundreds of nanoseconds that it needs to charge / discharge the gate capacitor.

Bob
 

bonedoc

Dec 21, 2011
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Ok, I am putting together that circuit that Resqueline posted. I dont have a 100V zener, but I have the rest. So far, I have added the 15V zenner at the gate, the 10 ohm resistor to the gate, and the 100k pull down. My temps have actually dropped a little. The max temp I have hit is 120C, down from 150+. I just noticed something interesting though. The Mosfet heats when it is NOT receiving the square wave. When it is charging and receiving the square wave, the temp drops. Does this make sense? I have NOT added the 100V zener or the uf4007 yet. I have the uf4007 but need to go by some 100v zeners.
 

twister

Feb 12, 2012
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twister,

You do not understand power MOSFETs. The gate is basically a capacitor. In order to switch it quickly, you need to charge / discharge this capacitor quickly, which means lots of current. Typical gate drivers can source and sink 1 A for the hundreds of nanoseconds that it needs to charge / discharge the gate capacitor.

Bob

Wow! one amp of current! I wonder how this guy is driving a mosfet with just a 555 if it needs so much power? :)
I do see that I was wrong about the diode across the transistor. I believe that with that high voltage transistor that he is using, he doesn't need a diode anyway. Some mosfets have a zener built in.
I wonder if his driver goes to ground in the off state? If not, he needs a gate bias resistor to ground to turn the transistor off. I wish I could see his schematic. That may be why his transistor is getting hot. The pic was probably grounding the gate, whereas the driver is not.
The oscilloscope photo is very interesting. It shows the on time of the transistor and the voltage is very low. I wouldn't think the transistor would even be warm, but I have never experimented with a boost converter. It sounds like he is building a flyback converter anyway.
http://www.dos4ever.com/flyback/flyback.html
 

bonedoc

Dec 21, 2011
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Yes, I do have a 100k pulldown from the gate to the ground. I do think that the mosfet is latching to the on state, but I think it is do to the fact that I need the 10V zener and uf4007 on the drain.
 

twister

Feb 12, 2012
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Ok, I am putting together that circuit that Resqueline posted. I dont have a 100V zener, but I have the rest. So far, I have added the 15V zenner at the gate, the 10 ohm resistor to the gate, and the 100k pull down. My temps have actually dropped a little. The max temp I have hit is 120C, down from 150+. I just noticed something interesting though. The Mosfet heats when it is NOT receiving the square wave. When it is charging and receiving the square wave, the temp drops. Does this make sense? I have NOT added the 100V zener or the uf4007 yet. I have the uf4007 but need to go by some 100v zeners.

I was wrong about putting a diode across the transistor and with your high voltage transistor it probably is not needed.
Some thing is wrong with your driver I think. Is it turning completely off? Have you checked the voltage at the gate with no drive? I should be 0. If the transistor is on with no drive and a gate bias resistor to ground, sounds like the transistor is bad. You can check it with an ohm meter. It should be infinite resistance from Drain to source.
Sounds like your driver is on all the time. You should only need one transistor to drive it, and the base bias should be a little less than .7V.
If your pic puts out more than 5V it should drive it without a driver. Try it again since you put the gate bias resistor in. How much current will you pic put out? You need to put in a large enough resistor in the gate so it won't take out your pic if the transi shorts out.
Here is a link that I found that might help.
http://www.dos4ever.com/flyback/flyback.html
 

BobK

Jan 5, 2010
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The driver should be push-pull. If it is not, that is why your MOSFET is staying on. Have you looked at it with a scope?

Bob
 

bonedoc

Dec 21, 2011
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Ok, so I scoped it and found something out. I think it is possible that I am misunderstanding the function of the driver all together. I am using the TC427. Gonzo recommended it. It is the 2 channel, non inverting type. Here is the datasheet:

http://ww1.microchip.com/downloads/en/DeviceDoc/21415C.pdf

The driver is putting out an excellent square wave. BUT....I notice that the driver is ALWAYS putting out the Vdd voltage. When I send the square wave to the input, it behaves perfectly. Shut the square wave off, and the output pin goes to Vdd. I have it hooked EXACTLY as the datasheet says. I am decoupling Vdd with a 1.0uF and 0.1uF cap. I am missing something...
 

Resqueline

Jul 31, 2009
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The 1000µF should be a low impedance type. It's imperative to decouple these pulse circuits.
The gate pull-down resistor and the 15V zener should not be needed if all is ok, permanently mounted, and you drive it with less than 15V.
The 100V zener might not be needed either. It's there just to make sure, and the need depends on the transformer and circuit layout etc.
The 0.1Ω resistor is there just to be able to scope the current ramp. TP = Test Point. Ground the scope at +12V. The resistor could also be put in the source lead.
Edit: do you have any kind of capacitor in series with the input to the driver circuit? What is the input voltage when the square wave is off?
 
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bonedoc

Dec 21, 2011
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LOL...I am going to look even more stupid, but I think I know what is going on. I wanted my microchip be be pretty safe. So, I am sending my square wave to an optocoupler, and the optocoupler is passing this to the driver. Maybe all of this is not necessary. Maybe the driver is enough immunity to the chip. I dont know. Anyway, I am using the HCPL2630 opto. It is an open collector.

http://www.fairchildsemi.com/ds/6N/6N137.pdf

So, in order to get my opto output signal, I had to put a 10k pullUP on the optocouplers output pin. So, this i the problem. The output pin of the opto is pulled up, but also connected to the input of the driver. So, when it is sending the square wave, all is great. When the wave is off....it is just holding the driver output high and burning up the mosfet.

So....suggestions? Trash the opto...microchip is already safe. Change setup of current opto. Change optos?
 

Resqueline

Jul 31, 2009
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Hehe..
When you say wave is off I suppose you mean that the PIC output stays low (not going high impedance)?
If so I suppose you have the IR LED connected between PIC and ground? If so you just need to move it between +5V and PIC.
Unless you have a very bad grounding scheme the PIC should be safe enough behind the driver. I'd ditch the opto.
Supplying a complete diagram (or a description) would have had this case solved by the second post btw..
 

bonedoc

Dec 21, 2011
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You are correct as to the pic hookup. I think I am going to ditch the opto. I believe my ground to be good. I have the grounds separated on the pcb, all the way back to the ground wire to the battery. I'm bad about going from breadboard to pcb cad. If I do schematics, it's pencil and may cause seizures. I'm still going to use that circuit u posted, minus my opto use :) ian
 

Resqueline

Jul 31, 2009
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Big lasers use optocouplers extensively to protect the computer boards from transients that can take unexpected routes.
Without looking at the full picture I can't say for sure what can take place in your circuit or not.
You just created an inverted drive signal by the way you connected that optocoupler.
 

bonedoc

Dec 21, 2011
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Ok, here is my total circuit. It is broken into 3 parts, which are connected to a microchip.They are:

1. The charger
2. The monitor
3. The de-charger

I did not include my power supply or microchip decoupling. I will have a tank cap on the main power. I will have a 0.1 uF decouplers on the inputs and outputs of the 7805 regulator, and I will also put a 4.8V zener on the regulator, just in case.

For an explanation:

-the first pic is the charge circuit, without the optos.

-the second pic is how I monitor the charge levels. I am using a voltage divider. The power resistor is actually divided into two 1M Ohm resistors to break up the voltage and power dissipation. The lower portion of the divider is a 33k resistor in series with a 10k wiper resistor. This allows me to tune the divider. I have a BAT43 schottky on the divider to the 5V rail to protect the ADC from over voltages.

-the last pic is of what I am using to discharge the caps for safety. I use it in case the voltage goes out of the desired range, and to automatically discharge when unpowered. I am achieving this by using a Normally CLOSED relay. No power = discharge. I also have microchip control of the relay with a 2n222 NPN. I am using a 25w 5K resistor to discharge slowly through the relay. I also have the relay poles connected into series to help limit sparking.

Let me know what you think. I would be happy to pay you for reviewing it. Just PM if his is the case. I dont know what protection things I have left out for devices. For example, if I need a diode on the potentiometer.
 

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bonedoc

Dec 21, 2011
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Resquline, I was testing with the circuit you showed and noticed a problem.. The uf 4007 that goes from drain to the 100V zener shorts out the system. Is this an error? Keep in mind that I dont have the 100V zener...just the uf4007 going from drain to sink.
 

Resqueline

Jul 31, 2009
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Oh yes, that'll create a short circuit condition.
A transient absorber diode would be better than a zener btw.
I'm sorry I haven't found the time yet to analyze the rest of your circuit.
 

bonedoc

Dec 21, 2011
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No problem at all! I was reading about transient absorber diodes, and according to wiki, those are also zeners. LOL. I am a little confused as to what I am doing wrong. What should I be using to protect the mosfet from the inductor? Someone said the 100v zener may bot be needed, so that is why I removed it.
 

Resqueline

Jul 31, 2009
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Well, I said it might not be needed, but that is no reason to short it out instead. The normal thing to do with superfluous parts is to just remove them.
Of course removing the zener also removes the need for the UF4007. There's no need for protection if the spikes never exceed 100V. Did you have a 'scope?
You said yourself that according to the transformer ratio and the output voltage that the transistor shouldn't see much more than 30V, so you have a triple safety.
 

bonedoc

Dec 21, 2011
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Sorry for the dumb questions ;-) I think I am getting closer to pulling it all together. I think it would be smart for some protection, since I am dealing with an inductor. In that case, let me see if I understand this.

The DS rating on this mosfet is 100V. Just as the gate has a 15V, zener and that is also the rating of the GS voltage, I should reverse bias a 100V zener from the ground to the drain? I can omit the uf4007 and I am probably be safe. Correct? In the case that region sees 100V, the zener will conduct it to the ground. Am I correct?

I dont see the purpose of the uf4007 being on there. Can you tell me? I wish I could sit down and get a tutorial in person. I could ask a billion questions.
 

Resqueline

Jul 31, 2009
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The purpose of the rectifier diode is somewhat subtle, and while you're right that a zener would work if connected directly - there are some drawbacks.
Zeners are not made for this kind of pulse duty, and they have some capacitance that will be taken "on a ride" if connected directly. The diode isolates this.
The zener could even benefit from an added parallel capacitor. There are combined diode + zener components made, called ZenBlock by Philips for example.

The 100kΩ + 15V zener are only useful if ever the drive circuit is disconnected. If the drive circuit is run on 12V there should be no need for a zener though.
The zener + 10Ω will also protect the drive IC if there's a d-s punchthrough in the MOSFET. The 10Ω may protect against oscillations.

If there's perfect coupling in the transformer + rectifier then there will be no extra spikes on the primary side. It's only the leakage inductance that's causing this.
If the leakage inductance is small then the spike energy will also be small and the MOSFET may handle it all on its own.
It doesn't take much to quench small spikes down in voltage. Have a look at this discussion for snubber examples and explanations.
 
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bonedoc

Dec 21, 2011
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Thanks for the info (and being patient, ;-P)

One *last* question. As you can see by my schematic I posted, I have a slow discharge of my 300V cap bank through a large resistor and a relay. Its working pretty good thus far. I would also like to figure how to do a safe, but single pulse, quick discharge of the bank (through a inductor/solenoid load). So, I wanted to use my IRF840. It is rated at 500V and has an RDS of .850. I prefer not not have this directly connected to my pic. I have an extra mosfet driver channel. Is it stupid or beneficial to use this driver? If it is not impractical, can you tell me:

-with a 500V Vds AND Vgs, is there ANY reason to add protection? If so, what type of diode? I have not looked into zeners at that rating, but the UF4007 is rated here. Would this be fine?

-with the power mosfet, I understand applying more gate voltage/current may be better, so a smaller base resistor may be better. In the case of a irf840, is this true? It seems as though I would want a larger resistor on the gate for damage immunity to the driver. Also, since I am also going for a single pulse, would a lower resistor value mean that the discharge time be significantly different?


EDIT- I overlooked that the GS voltage is 20V, so I should clamp it. I guess the questions is then....do I clamp at 20V or 15V? Base resistor size needed from driver output? I suppose the uf4007 can then go reversed biased on the inductor leads.
 
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