N channel high side switch

[BlinkingLeds]

Hi.
I have been searching for diy solution (no ic) for a N-channel high side switch at 600v+ (rectified 400v ac) and it doesn't going very well.
I need to boost the Gate Source voltage somehow so i figured i could use a capacitor voltage doubler or an inductor but both have huge voltage drops so i end up with big capacitors or a huge inductor.

Also i have never used inductors before and i don't know much about them. Do they have max voltage ,current the ones i have are rated at 100mH but they are very small about the size of a 400v 1uf electrolytic cap. How much those can withstand?

I suspect those Gate driver ICs have some sort of voltage boosting internally.
I don't like to use ICs that won't be around for many years so if i must use an IC i would prefer one that i could change afterwards if it runs out of stock or something that is expected to have many years of service (like the 555 for example)
Thanks

 

[(*steve*)]

A high side switch is typically a P channel device.

Is there any reason you want to use an N channel device?

You'll need to drive the gate higher than your positive rail. This will entail floating another power supply on top of your +ve rail, or employing some sort of boost circuit to create this voltage.

There are ICs that have this built in, but you say you don't want to use one.

 

[BlinkingLeds]

A high side switch is typically a P channel device.

Is there any reason you want to use an N channel device?

There isn't any p channel for more than 600v i know it would had solved my problem very easily

You'll need to drive the gate higher than your positive rail. This will entail floating another power supply on top of your +ve rail, to create this voltage.

How do i do that? Should i connect my 12v PS - to source and + to gate? can i do that?

 

[BobK]

You can get high-side driver chips that will do this. A DIY solution would be difficult.

Bob

 

[BlinkingLeds]

Yes but most of them are 600v max
Steve's idea looks good just like this one
http://tahmidmc.blogspot.gr/2013/02/n-channel-mosfet-high-side-drive-when.html

 

[BlinkingLeds]

Will a bootstrap capacitor work?

http://engineeringconnections.com/blog/2012/04/05/pulling-up-your-gate-drive-by-its-own-bootstrap/

If so how many uF it must be
What is the bootstrap diode?
Thanks

 

[BobK]

That would put 600V on the gate of the MOSFET. Not a good idea.

Bob

 

[BlinkingLeds]

That would put 600V on the gate of the MOSFET. Not a good idea.

Bob

I don't think so because the gate voltage should not exceed 20v Gate to Source so it might be even 100kv Gate to Ground as long as Gate source voltage is less than 20v
So if the voltage on the Drain is 700v to ground Gate's voltage should be 700+20v = 720v Gate - Ground
I read that elseware and i couldn't believe it at first i thought that the Fet will bolow up if i give it more than 20v to the gate. I tried to give it 300v and it survived because Gate-source voltage was only 3v and it was acting as a resistor so i need to boost up that voltage to 18v
Thanks

 

[KrisBlueNZ]

BobK is right. If you use a bootstrap capacitor to make a charge pump to generate the high positive rail, its voltage will be 600V above the top MOSFET's source voltage and that will blow the snot out of the MOSFET.

Also a charge pump will only work if the MOSFETs are being switched regularly. You have not said anything about this.

If you want any useful suggestions, describe what you're trying to achieve and upload a schematic of what you have so far. We would need to know what you want to drive the MOSFETs from and what the MOSFETs will be driving. Any other information would be helpful. Without giving this information, you would just be wasting everyone's time.

 

[BobK]

Here is a high-side low-side driver that can handle 1200V.

http://www.irf.com/product-info/datasheets/data/ir2213.pdf

Bob

 

[KrisBlueNZ]

BobK is right. If you use a bootstrap capacitor to make a charge pump to generate the high positive rail, its voltage will be 600V above the top MOSFET's source voltage and that will blow the snot out of the MOSFET.

Oops I was wrong about that. As long as the bootstrapped voltage rail is capacitively referenced to the source of the high-side MOSFET and the capacitor is charged from a suitable ground-referenced supply (e.g. +15V) you will be OK. But of course you need a high-side driver that's either fully isolated or coupled with a circuit that's tolerant to high voltage, to create the gate drive signal for the top MOSFET using the bootstrap supply rail. I suppose you could use a high-speed optocoupler with output buffer built in, such as a 6N137, if you really want to avoid using an IC (and if you don't consider an optocoupler to be an IC), but a high-side driver is really the best solution.

 

[BlinkingLeds]

Here is a high-side low-side driver that can handle 1200V.

http://www.irf.com/product-info/datasheets/data/ir2213.pdf

Bob

That's good but i'm really fond of diy solutions (where possible).

BobK is right. If you use a bootstrap capacitor to make a charge pump to generate the high positive rail, its voltage will be 600V above the top MOSFET's source voltage and that will blow the snot out of the MOSFET.

Also a charge pump will only work if the MOSFETs are being switched regularly. You have not said anything about this.

If you want any useful suggestions, describe what you're trying to achieve and upload a schematic of what you have so far. We would need to know what you want to drive the MOSFETs from and what the MOSFETs will be driving. Any other information would be helpful. Without giving this information, you would just be wasting everyone's time.

Thanks but can you please explain what you mean, for the beginners (particularly about the 600v) ?


The image shows what i was trying to do and failed because the fet only got 3v Gate source so it was acting as a resistor and was heating up allot with a 15w test lamp (ofcourse this was done at 250v and not 700v)

I want to test this circuit as a dc dimmer so it won't be less than 100hz but i also plan to use it as a long period switch if possible

it will just drive resistive loads like lamps heaters and maybe leds

 

[BobK]

Going back to Steve's original suggestion, the simplest way to do what you want is to use a separate floating power supply that connects to the source of the MOFSET, and an opto-isolator to switch the gate based on that power supply.

Bob

 

[KrisBlueNZ]

OK, so you understand why the MOSFET will not fully turn ON with your circuit? It's acting as a source follower; the source voltage follows the gate voltage with a few volts' drop (the Vgs voltage) and the MOSFET never saturates. You need two things to make your circuit work: a supply rail for the gate drive that is referenced to the MOSFET's source, and a way of isolating the drive signal.

As I said in post #11, my comment that you can't use a bootstrapped power supply was wrong. You need a capacitor with its negative terminal connected to the MOSFET's source, which will be charged up to a suitable Vgs voltage (e.g. 12V) when the source is at 0V (i.e. when the MOSFET is OFF) and will hold its voltage while the MOSFET is ON (when its source is at a high positive voltage) so that the driver circuit can always provide a source-referenced positive voltage source to drive the gate.

You charge the capacitor from a low-voltage 0V-referenced power source through a diode. The diode obviously needs to be rated for at least the positive rail voltage, and should be a fast switching type such as a UF4007. You should use a current limiting resistor in series with it; for the UF4007 (1A rating) a 15 ohm resistor is appropriate. The capacitor is typically in the range 10 nF to 1 uF.

This bootstrap circuit is ONLY USABLE if the MOSFET is being switched regularly. The capacitor cannot hold the charge indefinitely while supplying current to the driver circuit. You may be able to fulfil this requirement by using a drive signal with a duty cycle of either 99% (for nearly always ON) or 0% (for OFF). Of course this fast switching of a high voltage will generate a lot of electromagnetic interference.

If you have the bootstrapped power source you can transfer the control signal using an optocoupler. If you use a regular optocoupler such as a 4N32 I would connect it between the bootstrapped positive rail and the gate, with a pulldown resistor, so that when the optocoupler is not energised, the MOSFET will also be OFF. Alternatively you could use an optocoupler with a digital output, such as the 6N137 I mentioned or the H11L1/MOC5007. In this case you should use an inverting stage between the output and the MOSFET gate for the same reason - when the optocoupler is OFF, the MOSFET is OFF.

If you want to avoid switching the MOSFET continuously (which is probably a good thing to avoid if you don't need it), then you need a drive circuit that transmits voltage. Some semiconductor devices will do this; I think a stack of photodiodes in series can generate enough voltage to turn on a MOSFET (especially if you use one with a logic level gate voltage threshold) or you could probably even use a stack of small solar cells! Another option is a transformer with a rectifier and smoother on the secondary; applying AC to the transformer will turn on the MOSFET. Google isolated MOSFET voltage drive for other suggestions.

 

[BlinkingLeds]

i have tried this thing with the inductor but it has a huge voltage drop.
and i could try this as shown here http://tahmidmc.blogspot.gr/2013/02/n-channel-mosfet-high-side-drive-when.html and i think that's also what steve suggested in his first post?

 

[KrisBlueNZ]

Your first schematic is so incomplete I can't tell much about how, and whether, it would work.

Regarding the second circuit, the isolated power supply is normally generated locally and it's probably a bad idea to use a separate mains-powered adapter for it; at the least, the adapter may not be rated for 700V isolation! Another big problem with the second design is the choice of R1, because the voltage across R1 ranges from 700+ volts to 0V.

R1 must be high enough that it doesn't dissipate excessive power when the driving MOSFET is ON and the main MOSFET's gate is pulled down to 0V (when R1 has 700+ volts across it), but this means that it can supply very little current in the opposite state. This is why it's normal to develop an isolated supply that's relative to the source of the MOSFET and switch the gate voltage within the positive and negative limits of that supply rail.

If you're not switching the MOSFET often, perhaps your best option would be a small transformer to generate the gate bias voltage relative to the source. You would need to feed a signal into the transformer to turn the MOSFET ON; when the signal stops, the capacitor on the MOSFET gate would discharge and the MOSFET would turn OFF.

You need to ensure that the MOSFET will switch quickly enough that it won't be damaged by power dissipation during switch-on and switch-off. If you use a transformer driver, you may want to use some kind of threshold circuit after the rectifier/smoother, to make sure the signal to the MOSFET gate is clean and switches quickly.

That article you linked to is very good. He shows a number of schematics though; I don't know which one you're referring to.

 

[BlinkingLeds]

I'm referring to the last schematic on the page.
I need something small and something that will insure that the mosfet turns on and off as quickly and as accurately (to the applied signal) as possible.
Can you draw up what you mean because i have a hard time understanding it?
Thanks

 

[KrisBlueNZ]

OK, if you want immediate control of the MOSFET, my suggestion of starting and stopping the drive to a transformer that generates the gate bias will not work.

The last schematic on the page you linked to is a good idea, but it requires an isolated power supply that can be referenced to the MOSFET's source. This point is switching over a wide voltage range, so you have to be careful what you connect to it, since anything connected to the source can radiate interference and any capacitance to ground may affect the MOSFET's switching. But you can use a small transformer to generate the isolated power supply to use with the schematic you linked to.

 

[BlinkingLeds]

Ok i will use an isolated power supply.