If you have common ground you will be applying 1000V across the drain and gate.

 

See if this circuit works for you. Q2 and Q3 have to be 1000V rated.

I have a question about the way this circuit works:
if the current from the 1000V supply flows to the drain of Q3 (which is connected to the plus side of the power supply) , then the current will exit the source of Q3 and try to enter the source of Q2.
Would that work (current entering the Mosfet from the source)?
Unless you meant that Q2 is PMOS (but the symbol is for NMOS), if Q2 in NMOS VDS will be negative.
Shouldn't the two MOSFETs be in series?
Thank you.

 

Both Mosfets are of the same type (n-channel). I am replying on my IPad and I do not have the circuit in front of me, but both MOSFETs are N channel and require positive gate voltage with respect to ground to turn on. Once you apply positive voltage to both gates, the MOSFETs will conduct current from an external source, even though there're is no common ground between the local gate control voltage source and external 1000V supply. If I understand correctly you are a little suspicious about the circuit because there is no common ground between your 1000v source and the gates control voltage. Let me know if my assumption is correct and I will try to explain that.

 

Both Mosfets are of the same type (n-channel). I am replying on my IPad and I do not have the circuit in front of me, but both MOSFETs are N channel and require positive gate voltage with respect to ground to turn on. Once you apply positive voltage to both gates, the MOSFETs will conduct current from an external source, even though there're is no common ground between the local gate control voltage source and external 1000V supply. If I understand correctly you are a little suspicious about the circuit because there is no common ground between your 1000v source and the gates control voltage. Let me know if my assumption is correct and I will try to explain that.

Actually I am wondering about the orientation of the two mosfets, I think the current will flow from drain to source in Q3, and then the current will enter Q2 from the source going to the drain (oppesite direction) which will cause negative drain to source voltage. I am wondering if the two mosfets need to be in the same direction (current enters both mosfets from the drain and exit from the source).

 

Wouldn't an H-bridge, using drivers for the mosfets be a better way of doing this? Unless I missed it the -1000V and the +10V aren't on at the same time. Each side of the H-bridge would allow the different voltage to return to its own circuit.

A driver like this - http://www.irf.com/product-info/datasheets/data/ir2213.pdf - has its own separate ground for the "logic" side and the high voltage (com) side, that don't need to be connected to each other.

 

Wouldn't an H-bridge, using drivers for the mosfets be a better way of doing this? Unless I missed it the -1000V and the +10V aren't on at the same time. Each side of the H-bridge would allow the different voltage to return to its own circuit.

A driver like this - http://www.irf.com/product-info/datasheets/data/ir2213.pdf - has its own separate ground for the "logic" side and the high voltage (com) side, that don't need to be connected to each other.

Yes would be perfect but would also double the number of FETs necessary for each direction.

I still prefer something like this

but it also requires two isolated gate drivers and therefore two additional power supplies. You could also use transformer coupled gate signals, they are isolated and don't need an additional power supply but this doesn't go well with the non-50% duty cycle.

The circuit from post #12 may cause excessive switching losses as the FETs are turned off through the 4k7 resistor.

I don't know why the OP is concentrating on only the -1000V if it is actually +10/-1000V he wants to switch.

You could actually use the 100kHz gate signal to generate a secondary power supply for an isolated gate driver. Is the duty cycle fixed or variable?

 

Yes would be perfect but would also double the number of FETs necessary for each direction.

I still prefer something like this

but it also requires two isolated gate drivers and therefore two additional power supplies. You could also use transformer coupled gate signals, they are isolated and don't need an additional power supply but this doesn't go well with the non-50% duty cycle.

The circuit from post #12 may cause excessive switching losses as the FETs are turned off through the 4k7 resistor.

I don't know why the OP is concentrating on only the -1000V if it is actually +10/-1000V he wants to switch.

You could actually use the 100kHz gate signal to generate a secondary power supply for an isolated gate driver. Is the duty cycle fixed or variable?

I was the told the duty cycle is about 20% most of the time, but it could vary.

 

Wouldn't an H-bridge, using drivers for the mosfets be a better way of doing this? Unless I missed it the -1000V and the +10V aren't on at the same time. Each side of the H-bridge would allow the different voltage to return to its own circuit.

A driver like this - http://www.irf.com/product-info/datasheets/data/ir2213.pdf - has its own separate ground for the "logic" side and the high voltage (com) side, that don't need to be connected to each other.

I actually considered using that H bridge, but to be honest I only used H bridges once in my life to control a DC motor, I really dont have a lot of experience of using them as power switches. The application circuit listed on the first page of the data sheet looks little confusing to me, because the load is connected between two mosfets, not between one mosfet and ground like I am used to.
Can you direct me to any material that could help me understand H bridges circuits?

Also, it doesn't really matter how many MOSFETs I will end up using, as long as it works.

 

I actually considered using that H bridge, but to be honest I only used H bridges once in my life to control a DC motor, I really dont have a lot of experience of using them as power switches. The application circuit listed on the first page of the data sheet looks little confusing to me, because the load is connected between two mosfets, not between one mosfet and ground like I am used to.
Can you direct me to any material that could help me understand H bridges circuits?

Also, it doesn't really matter how many MOSFETs I will end up using, as long as it works.

I don't see an advantage.

Now what happens is that the Load resistance , the 10V supply and the -1000V have all different reference voltages. This is what I posted a few pages before (maybe in the other thread). It doesn't matter in which direction you apply the voltage to the resistor if it is not related to the power supply reference voltages...
It's kind of confusing what you want to achieve.

Maybe we can start all over again?

1. Are the 10V and -1000V isolated from each other?
2. Does the load have to have the same reference as one of the power supplies?
3. Does the PWM have to have the same reference as one or both of the power supplies?
4. Maybe you should ask for more details of what the customer has/wants.

A few solutions have been given to you. I don't see where the problem is.

 

Wouldn't an H-bridge, using drivers for the mosfets be a better way of doing this? Unless I missed it the -1000V and the +10V aren't on at the same time. Each side of the H-bridge would allow the different voltage to return to its own circuit.

A driver like this - http://www.irf.com/product-info/datasheets/data/ir2213.pdf - has its own separate ground for the "logic" side and the high voltage (com) side, that don't need to be connected to each other.

I like the idea of using that H bridge very much..
can you please show me how to use it to switch two separate supplies (-1000V and +10V). I can get an isolated logic voltage supply for the H bridge, and I can also isolate the TTL signal generator that controls the switching (if it is not already isolated).
Thank you!!

 

Once you apply control voltage to both MOSFETs each MOSFET will establish a conductive channel between drain and source or source and drain. This conductive channel will conduct as long as control voltage is applied (Vg-s > Vth of your MOSFET). One thing to verify though is that the in ternal resistance of your selected MOSFETs has to be 10 times lower then the load resistance. Rds(on) << Rload. If you can get hold of MULTISIM simulation software you can test this circuit on your computer.

 

I don't see an advantage.

Now what happens is that the Load resistance , the 10V supply and the -1000V have all different reference voltages. This is what I posted a few pages before (maybe in the other thread). It doesn't matter in which direction you apply the voltage to the resistor if it is not related to the power supply reference voltages...
It's kind of confusing what you want to achieve.

Maybe we can start all over again?

1. Are the 10V and -1000V isolated from each other?
2. Does the load have to have the same reference as one of the power supplies?
3. Does the PWM have to have the same reference as one or both of the power supplies?
4. Maybe you should ask for more details of what the customer has/wants.

A few solutions have been given to you. I don't see where the problem is.

1- The two voltage supplies (+10V and -1000V DC) are independent (isolated, each has its own ground)
2- The load has + side which needs to see +10V and -1000V (switched between them) in reference to the - side (or load ground).
3- The PWM is a third supply that has its own ground, it doesn't matter if its ground is tied to the two power supplies (+10, -1000V) ground or not.

I have tried (actually built and tested) several circuits that I designed, but as I asked in the beginning of the thread, I didnt know how to switch the negative supply.
I appreciate all of your help guys, and you gave me great ideas, but I dont really need a circuit that switch between the two supplies. I need to be able to switch the -1000V supply using a low side switch (negative supply connected to the MOSFET source, and the load between drain and ground, as in the little circuit that I posted in the beginning of the thread).

Thank you again.

 

Once you apply control voltage to both MOSFETs each MOSFET will establish a conductive channel between drain and source or source and drain. This conductive channel will conduct as long as control voltage is applied (Vg-s > Vth of your MOSFET). One thing to verify though is that the in ternal resistance of your selected MOSFETs has to be 10 times lower then the load resistance. Rds(on) << Rload. If you can get hold of MULTISIM simulation software you can test this circuit on your computer.

Actually, I understand that concept, and I use it to switch the positive supply, its the negative supply that is giving me hard time.
when I connected the negative supply to the source of the mosfet, and connected the load to the grain and then the other side of the load is grounded, I noticed that no matter what you do to the gate (grounded or connected to +5V) the mosfet is conducting, it only opens when you leave the gate floating (like shown in the schematic I posted in the beginning of the thread).
So my question is, how to switch the -1000VDC on and off without leaving the gate floating.
Also, the gate is tied to the -1000V with 1Meg ohm resistor, that negative voltage connected to the gate is what prevents the mosfet from conducting (push the electrons away from the channel so it doesn't conduct), when you ground the gate, there is no force to push the electrons away, so they get pulled up and they form a channel and they conduct when the gate is grounded.

 

3- The PWM is a third supply that has its own ground, it doesn't matter if its ground is tied to the two power supplies (+10, -1000V) ground or not.

I need to be able to switch the -1000V supply using a low side switch (negative supply connected to the MOSFET source, and the load between drain and ground, as in the little circuit that I posted in the beginning of the thread).

Mmh, it didn't seem to be what you wanted in the beginning. If the PWM is isolated and you only need to switch -1000V then just connect the PWMs ground to -1000V.
What about the 10V?

 

Mmh, it didn't seem to be what you wanted in the beginning. If the PWM is isolated and you only need to switch -1000V then just connect the PWMs ground to -1000V.
What about the 10V?

That is exactly what I did to switch the negative supply, but didnt work. I think it takes more than that to do the switching of NMOS with negative supply.
as I pointed out before, when you connect the negative supply to the source of NMOS transistor, the transistor will be on all the time, whether you apply +15V to the gate, or if you ground it. The only way to turn the transistor off is by connecting a resistor between the gate and source and leave the gate open/floating. once the gate is connected to ANYTHING (+15V or even ground) the FET will turn on and conduct.
so I was wondering if there is a way to do the switching of a negative supply that doesn't involve leaving the gate open?
(the reason why this happens with negative supplies is because the negative potential applied to the source of the transistor, specially a large one -1000V kind of induces a channel between the drain and source and makes the transistor conduct, only when you apply an equal potential to the gate (by connecting a gate to source resistor) you will be able to kill this channel and make the mosfet not conduct. but as soon as the gate potential goes higher than -1000V (by either grounding or connecting the gate to higher potential) the FET will conduct again.

 

Actually, I understand that concept, and I use it to switch the positive supply, its the negative supply that is giving me hard time.
when I connected the negative supply to the source of the mosfet, and connected the load to the grain and then the other side of the load is grounded, I noticed that no matter what you do to the gate (grounded or connected to +5V) the mosfet is conducting, it only opens when you leave the gate floating (like shown in the schematic I posted in the beginning of the thread).
So my question is, how to switch the -1000VDC on and off without leaving the gate floating.
Also, the gate is tied to the -1000V with 1Meg ohm resistor, that negative voltage connected to the gate is what prevents the mosfet from conducting (push the electrons away from the channel so it doesn't conduct), when you ground the gate, there is no force to push the electrons away, so they get pulled up and they form a channel and they conduct when the gate is grounded.

That is your problem. A mosfet is bidirectional in regard to polarity. For your negative 1000V side, the -1000V goes to DRAIN and the source goes to the positive "ground" of that supply. This link at the very bottom of the page gives an example of what I'm talking about(even though they show a motor) http://www.electronics-tutorials.ws/transistor/tran_7.html

I say mosfets are bidirectional because they can be used to switch AC voltage.

This all would be easier if you would identify the whole "non-secret" circuit. Or even the component that this switch controls.

 

That is exactly what I did to switch the negative supply, but didnt work.

Then you did something wrong.

Take the two ground symbols out of this schematic. Now it's not a "negative" voltage anymore, still it's the same circuit.

This is working.

 

That is your problem. A mosfet is bidirectional in regard to polarity. For your negative 1000V side, the -1000V goes to DRAIN and the source goes to the positive "ground" of that supply. This link at the very bottom of the page gives an example of what I'm talking about(even though they show a motor) http://www.electronics-tutorials.ws/transistor/tran_7.html

I say mosfets are bidirectional because they can be used to switch AC voltage.

This all would be easier if you would identify the whole "non-secret" circuit. Or even the component that this switch controls.

But If current would flow from the source to the drain of the N- type MOSFET, that would create negative VDS, which the data sheets clearly don't recommend. The circuit shown in the link you sent me has both NMOS and PMOS, and they have opposite polarities (VDS for PMOS is negative, while for the NMOS VDS is positive).
The mosfets can be used in AC applications (like audio) while they operate in the "small signal" region, the mosfets still need to be biased by a DC supply first before they operate in that mode.

I dont think mosfets can be used to switch AC directly, because they are uni-directional (check out this source http://www.irf.com/technical-info/designtp/dt94-5.pdf)

and remember that the Power MOSFETS have a diode between the drain and source that would allow them to conduct fully if the current flows in the opposite direction (if you send current from the source to the drain the diode will let it pass regardless of the mosfet status)
(http://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LITERATURE/DATASHEET/CD00160703.pdf) ..

I dont know exactly how this "power switch" will be used, I only have the specs of the load as I mentioned before. the rest is purely physics that is beyond my understanding to explain. but the idea of switching between two power supplies to deliver power to the load is clear.

 

I dont think mosfets can be used to switch AC directly, because they are uni-directional (check out this source http://www.irf.com/technical-info/designtp/dt94-5.pdf)

Did you read the document you linked to? They describe exacty how and that it can be done. The way it was posted before with two MOSFETs in series, in opposite directions.

 

The only way to turn the transistor off is by connecting a resistor between the gate and source and leave the gate open/floating. once the gate is connected to ANYTHING (+15V or even ground) the FET will turn on and conduct..

There is really something wrong in the way you look at MOSFET switches. First of all if you connect a resistor from gate to source , it is not floating. Having a floating gate means it is not connected to anything.

The basics is really simple:
Apply a positive voltage from gate to source within the Vgs limits and it will conduct.
Apply 0V (low impedance source, not floating) or a negative voltage from gate to source (within Vgs limits) and it will cease to conduct. Unless you apply a voltage to the source that is more positive than the drain voltage in which case it will conduct through the body diode.

How you denominate your 1000V voltage in your circuit doesn't matter, what I just said holds true. (n-channel MOSFET)