I Have only used N and P channel enhancements before and thought I could just pretend it was a P channel mosfet. It worked fine and lit the LED when the gate was connected to ground, but when I switched the gate to +3.8v it started smoking.
I'm unsure if it was smoking because I was 0.8v over the gate limit and should be applying positive voltage Or if I really be applying negative gate voltage and am 6.8 volts over the limit.
Can anyone help me? I'm guessing it's a very obvious and stupid mistake, but I'm just getting confused by the datasheets.
Whoops. I believe so. Am I missing something else or will this circuit work?
It would only work until you flipped the switch. Then you are sending 3.8V-.6V= 3.2V right to ground through the lower diode. In order to keep those diodes reverse biased, G<S<D
Whoops. I believe so. Am I missing something else or will this circuit work?
No the problem isn't that you need to add those two diodes. The problem is that you need to take those two diodes away, but you cannot. Those two diodes are internal (*see note) in that particular mosfet. This means that you MUST NOT forward bias the gate on that particular device.
*NOTE. From the data sheet: "The body of the mosfet is connected to the gate pin". This is what forms those two internal (parasitic) diodes. One from the "p" body to the "n" drain, and the other from the "p" body to the "n" source.
BTW ComputerLag. I'm just wondering why you have chosen to use a depletion mode mosfet, (by far the less common type), for that particular application?
I'm unsure if it was smoking because I was 0.8v over the gate limit and should be applying positive voltage Or if I really be applying negative gate voltage and am 6.8 volts over the limit.
The datasheet quotes-
Absolute Maximum Ratings
Parameter Value
Drain-to-Source voltage BVDSX 9V
Source-to-Drain voltage BVSDX 9V
Gate-to-Source voltage -12V to +0.6V
Gate-to-Drain voltage -12V to +0.6V
Vgs (off) -.8 to -3V
Vsg (off) -.8 to -3V
Tells me you can swap the d/s pins, but the gate must be driven -3V relative to either to turn it off.
Seems easier to set it up as a high side switch to keep out of negatives when connecting ground to either the drain or source
stuart0:
BTW ComputerLag. I'm just wondering why you have chosen to use a depletion mode mosfet, (by far the less common type), for that particular application?
I'm working on a circuit that charges a capacitor and discharges it if power is lost or logic level low is set. So I need the N channel specifically to be On with 0v gate voltage.
I can't use a P channel because I need it to source ground.
stuart0:
No the problem isn't that you need to add those two diodes. The problem is that you need to take those two diodes away, but you cannot. Those two diodes are internal (*see note) in that particular mosfet. This means that you MUST NOT forward bias the gate on that particular device.
*NOTE. From the data sheet: "The body of the mosfet is connected to the gate pin". This is what forms those two internal (parasitic) diodes. One from the "p" body to the "n" drain, and the other from the "p" body to the "n" source.
Thank you for the clarity. So If I invert my logic level high to a negative voltage it should be able to turn the transistor off. I smoked it because It was sorting the battery directly to ground through from gate to source.
ComputerLag:
I'm working on a circuit that charges a capacitor and discharges it if power is lost or logic level low is set. So I need the N channel specifically to be On with 0v gate voltage.
Ok, the enhancement mode device will work there. But now you introduce a new complexity in that you require both a positive and a negative supply. Specifically you need 0 volts on the gate to turn the fet on, but you need a negative voltage (about -3 volts) to turn the fet off.
This requires more than just inverting your logic. It requires an auxiliary negative supply (albeit low power, so could be just a small capacitive "charge pump") and an appropriate level shifting driver (which you can build with a small BJT, or various other ways).
I smoked it because It was sorting the battery directly to ground through from gate to source.
ComputerLag:
I'm working on a circuit that charges a capacitor and discharges it if power is lost or logic level low is set. So I need the N channel specifically to be On with 0v gate voltage.
I can't use a P channel because I need it to source ground.
That sounds like a strange circuit. Can you describe better what your intention, maybe draw a sample schematic of the idea you have?
ComputerLag:
I'm working on a circuit that charges a capacitor and discharges it if power is lost or logic level low is set. So I need the N channel specifically to be On with 0v gate voltage.
I can't use a P channel because I need it to source ground.
Wouldn't a pulldown resistor work with a lot less headache?
Jiggy-Ninja:
That sounds like a strange circuit. Can you describe better what your intention, maybe draw a sample schematic of the idea you have?
Yep Jiggy. I'm thinking that when ComputerLag realizes the amount of extra complexity involved in needing dual power supplies and level shifting, he might post the actual problem so that we can see if there is an easier way.
It might be that the presently proposed method really is the best way to go, but without knowing the full problem requirements we will never know.
stuart0:
No the problem isn't that you need to add those two diodes. The problem is that you need to take those two diodes away, but you cannot. Those two diodes are internal (*see note) in that particular mosfet. This means that you MUST NOT forward bias the gate on that particular device.
"forward biased" really applies to pn-junctions of the body diodes, and I'm sure you can forward bias
them within limits, the datasheet doesn't tell you what the limit is, but implies you should not exceed 0.6V
I am not sure whether 'forward' or 'reverse' is meaningful terminology for the gate of a depletion-mode FET,
since it acts to decrease current. The simple rule to remember is that for n-channel more positive gate
means more current in drain-source circuit (since more electrons are attracted to the n-channel)
It is confusing that the datasheet's test circuit incorrectly shows a positive-going pulse generator...
Note that the gate and substrate (body) are connected for this device (its unusual), the diodes are
simply a representation of the interface between source/drain wells and the substrate layer. They
are not deliberately created, just a fact of life that certain MOSFETs have a substrate layer
(high power enhancement MOSFETs do not, note, since the substrate is the drain for those).
Jiggy-Ninja:
That sounds like a strange circuit. Can you describe better what your intention, maybe draw a sample schematic of the idea you have?
I'm working on creating something similar to an H-bridge but I am using it to control a latching valve so I only need a small pulse of current. So I'm trying to both save power for the battery and always close the valve even if the battery somehow dies or is disconnected or the program crashes.
The circuit will work with enhancements and high logic for closed and low logic for open but the capacitors path to ground will by blocked by the N channel if power is lost or the battery is disconnected and will be unable to discharge to close the valve.
MarkT:
It is confusing that the datasheet's test circuit incorrectly shows a positive-going pulse generator...
No, the test waveforms on the datasheet are correct. They show a pulse starting from -3 volts and going up to zero, exactly what this device requires.
Note that the gate and substrate (body) are connected for this device (its unusual), the diodes are
simply a representation of the interface between source/drain wells and the substrate layer.
Yep I know. That's why I said exactly that in my first reply.
ComputerLag:
I'm working on creating something similar to an H-bridge but I am using it to control a latching valve so I only need a small pulse of current. So I'm trying to both save power for the battery and always close the valve even if the battery somehow dies or is disconnected or the program crashes.
Ok, thanks for elaborating. Depending upon the amount of current required to latch off the valve, then the way you're trying to do it sounds reasonable. Just remember that you can't drive the depletion mode mosfet in the simple way shown in your diagram. As I said before, you'll need a negative bias of about -3 to -4 volts and some kind of level shifting driver to turn off that bottom device.
I'm working on creating something similar to an H-bridge but I am using it to control a latching valve so I only need a small pulse of current. So I'm trying to both save power for the battery and always close the valve even if the battery somehow dies or is disconnected or the program crashes.
The circuit will work with enhancements and high logic for closed and low logic for open but the capacitors path to ground will by blocked by the N channel if power is lost or the battery is disconnected and will be unable to discharge to close the valve.
Is that how your valve is latched? Positive DC current to turn it on, negative DC to turn it off? What's the datasheet for that valve? There's probably a better way. Since you couldn't even spot the body diode configuration in the datasheet (it was the first thing I looked for), I can't really trust that you even have the right idea in designing this circuit.
stuart0:
you'll need a negative bias of about -3 to -4 volts
BTW. If you do stick with the n-channel depletion mode design, here is a very crude charge pump that will get you about a -4 volt bias voltage. In case you can't tell from the rough freehand sketch, those diodes I used where schottkey.
You can buy ICs to do this, or for a low power bias like you need you can even run the circuit below straight from an Arduino output pin (switching at about a 10 kHhz square wave). For a 0 to 5V input waveform the output will be about -3.5 to -4 volts, provided you don't put too much load on it (we're talking about tenths of a mA here).
