Mosfet DC to DC converter problem

Hi guys,

I have a problem regarding a DC to DC converter circuit.
Firstly, the circuit is collecting the signals from 2 coils. The voltage on them is somewhere around 1000VAC. It's a sinusoidal form and the 2 signals when they're add look okay. I've used there some diodes, to get 1000VDC.

The problem is at the MOSFET. This one takes the 1000VDC signal and makes it around 0-17VDC depending on the potentiometer. The idea is that I get mostly no current at all at the output. The potential between the gate and source is 3.2Volts. In the datasheet it says that it should be greater than 4Volts.
Just for the record, I've tried measuring the output current using a resistor. What I got was 0.2mA at 17Volts.

I have here the circuit: **broken link removed**
And this is the MOSFET datasheet: https://datasheet.octopart.com/IRFBG30PBF-Vishay-datasheet-57413.pdf

Could somebody help me regarding this circuit?

Thank you in advance,
RobertEagle

RobertEagle said:

The voltage on them is somewhere around 1000VAC. It's a sinusoidal form and the 2 signals when they're add look okay.

The problem is at the MOSFET. This one takes the 1000VDC signal and makes it around 0-17VDC depending on the potentiometer. The idea is that I get mostly no current at all at the output. The potential between the gate and source is 3.2Volts. In the datasheet it says that it should be greater than 4Volts.
Just for the record, I've tried measuring the output current using a resistor. What I got was 0.2mA at 17Volts.

Click to expand...

The gate to source voltage (3.2V) or 4V is dependent on temperature and output current and varied from part to part.

The temperature is around 25 degrees, just like in datasheet.
But why the current isn't bigger? I should have had 200mA.

The 4 volt threshold voltage is for a current of only 250 micro amps. It needs to be higher for 200 ma.

Robert,
I don't know why it is not working. With a G-S voltage of 5 or 6 volts the MOSFET should be very much on.

With no load measure the gate voltage. Now apply the source load. What is the gate voltage now? The gate voltage should not change with load. I am looking to see if the gate is damages.

Maybe I'm misreading the schematic, but I don't see how you could be getting the GS voltage you are reporting.

As I see it, your schematic shows a 47V zener. From that voltage you have a 180K resistor feeding a 10K pot.

With the wiper at the top of the pot, the highest gate voltage you should be able to get is 2.47 volts. And that is with respect to ground. You will need to take the gate voltage up to about 20V to see 17V at the source.

I suggest changing R2 to 10K, and see what that does for you.

I've changed the R2 resistor with one of 1.1kOhm.
Now, with no load, I get 0.79V and with load I get 2.80V.

The load is a 2.2kOhm resistor. The voltage across the resistor is 9.55V. From my calculus it would mean that the current is around 4.34mA, which is extremely low.
And by the way, the MOSFET gets very easy hot.

Now, with no load, I get 0.79V and with load I get 2.80V.

Click to expand...

Perhaps the FET is oscillating? Check with a scope if you have one.

And by the way, the MOSFET gets very easy hot.

Click to expand...

Not surprising; even at a few mA current it's dissipating a few Watts! At 200mA you will be asking it to dissipate ~ 200W

BTW if the coil voltage is 1000VAC (RMS) the peak of the sine wave is ~ 1400V.
The 1N4007 diodes are being overstressed, so is the FET.

The temperature is around 25 degrees, just like in datasheet. But why the current isn't bigger?

Click to expand...

And by the way, the MOSFET gets very easy hot.

Click to expand...

"around 25 degrees" in the data sheet is referring to the die temperature, which is very high. Not the room temp!
That part (without a heat sink) is not good for more than 1 watt. You may have over heated the part until it died, or partly died.

Now, with no load, I get 0.79V and with load I get 2.80V.

Click to expand...

What voltage?

What are you trying to do? There has to be a better way.

ronsimpson said:

What are you trying to do?

Click to expand...

Agreed. Give us the requirements; not make us infer them from a partial, non-working design.

Ronsimpson, I've looked too now in the datasheet and without a heatsink it dissipates about 1W at every 62 degrees, so theoretically it would dissipate about 2W.
Now, the voltages:
With the load I have:
- 60.2V between Drain-Source
- 57.4V between Drain-Gate
- 2.75V between Source-Gate
- 9.75V at the output with the potentiometer at the lowest value.

Now, there's something I've got to say. The signals from the input are working at 12.5kHz, are AC, therefore sinusoidal, and reach around 2kV.
After they go into the diodes (1N4007, 1000V rated), the signal drastically steps down to 650VDC(with the diodes left in air, with them connected to the circuit it reaches 250VDC). Here the signals are a little bit sinusoidal, that's why there are the capacitors.
This measurements were taken with the scope with no load, except the circuit itself.

I didn't know what was the actual power that I was getting from those signals so I calculated. I've took 2 resistors of 459kOhm rated at 2W. I've connected them to the inputs. I got 937.5V, meaning that the power per channel is 1.91W. From here we can say that the current would be 2,04mA.
Keeping in mind that circuit is a linear regulator, the absolute maximum current I could get from the MOSFET would be 4.08mA. Right? Would this explain the low GS voltage?
So power getting from the circuit is 0.00408*9.75, meaning 1.04% efficiency, am I right?

I didn't become aware of the fact that my circuit is like a 7805, dissipating all the rest of the energy till now.

Do you know a method, a type of circuitry, which it would be able to manage these voltages and simultaneously have a high efficiency (>90%)? Excluding a transformer, because it radiates EM waves, and loses much power by this way.

Thank you again,
RobertEagle

Sorry, I've posted the same thing twice. My bad. Apparently, I can't delete a post. Look up for what I have written.

RobertEagle said:

...The signals from the input are working at 12.5kHz, are AC, therefore sinusoidal, and reach around 2kV.
After they go into the diodes (1N4007, 1000V rated), the signal drastically steps down to 650VDC(with the diodes left in air, with them connected to the circuit it reaches 250VDC). ...

Click to expand...

If the sources are ~2000V, then the minimum requirement for the rectifiers is that they have a PIV rating of >5kV. 1N4007s wont hack it.

RobertEagle said:

Now, the voltages:
With the load I have:
- 60.2V between Drain-Source
- 57.4V between Drain-Gate
- 2.75V between Source-Gate
- 9.75V at the output with the potentiometer at the lowest value.

Click to expand...

D-S is only 60V???Why? If D is 1000V how can D-S=60V

Do you know a method, a type of circuitry, which it would be able to manage these voltages and simultaneously have a high efficiency (>90%)? Excluding a transformer, because it radiates EM waves, and loses much power by this way.

Click to expand...

I think transformers are efficient. Certainly better than 1.04%. Maybe 90%

I measured again, and I got the same results.
D isn't 1000V. Without load it gets to 250V, and with load it gets around 67V. What I said is that when I measure the diodes with their pins not being connected to the circuit I get 650VDC (down from 2kVAC).

Could you show me some circuits which corresponds with my situation that includes a transformer? Ain't there a more a different approach? Couldn't I use something else than the transformer? And if there's a possibility, how would it look like?

After all, all that I want is to convert to lower voltages, with power minimal loses.

RobertEagle

What actually puts out the 2000kV 12.5kHz out-of-phase signals? Is that from a center-tapped transformer?

RobertEagle said:

I measured again, and I got the same results.
D isn't 1000V. Without load it gets to 250V, and with load it gets around 67V. What I said is that when I measure the diodes with their pins not being connected to the circuit I get 650VDC (down from 2kVAC).
RobertEagle

Click to expand...

I don't understand. The diodes are not connected?

I said that diodes were connected to the coils, but not to the circuit. In that situation I got 650VDC.
But when diodes are connected to the coils and circuit it gets to 250VDC with no load, and with load it gets 67V.
Tell me if you don't understand.

Yeah...a sort of center tapped transformer.
Look. Here is a snap from the scope with both the channels. 250V/div. **broken link removed**

RobertEagle

One more time.
Diodes are (all the time) connected to the coil.
When the diodes are not connected to the MOSFET the voltage on the transformer is 650V.
When the diodes are connected to the mosfet the voltage on the transformer is 250V.
If there is a load the voltage on the coil drops to 67 volts.

Yes. The diodes are all the time connected to the coil.
But when the diodes are not connected to the MOSFET the voltage on the diode's pins that are heading into the circuit is 650VDC. That's why it's DC, it's after the diode.
When the diodes are connected to the MOSFET, the voltage on the diode's pins (Drain-GND) is 250VDC.
And if there's a load the voltage between Drain-GND is 67Volts.