Will this mosfet circuit work?

[liquibyte]

I have been looking for a solution to a problem I have and after coming up with the circuit in the picture I was reading down the new threads and came upon this discussion which kind of touches on a few concerns I have about VGS and such.

I've got an over voltage situation happening and was wondering if I could just put this circuit in line before the power supply circuit and have it delay the turn on time to the power supply by around 850ms or so and eliminate the 60+ volts I'm getting now due to the transient surge.  I've been looking into solutions and found this app note and while trying to calculate the values in figures 5a and 5b came across the attached circuit and then this calculator / equation.  I thought it might be on the simple side but since I prefer a simple solution I thought I should ask before attempting it.

[Falcon69]

correct me if I am wrong, but I think most mosFETs can only handle +- 20 volts on the gate. That means You'll need to place a 30volt zener or so between the gate and source.  Anode to the gate.

Also, if the mosFET doesn't have one (most do), you may also want to put a Diode in there.  Anode connected to the Source, Cathode to the Drain.

[liquibyte]

correct me if I am wrong, but I think most mosFETs can only handle +- 20 volts on the gate. That means You'll need to place a 30volt zener or so between the gate and source.  Anode to the gate.

Also, if the mosFET doesn't have one (most do), you may also want to put a Diode in there.  Anode connected to the Source, Cathode to the Drain.

I actually have a couple of 30V zeners, so that's good to know.  I did know about the diode and just mocked this up quick so I could visualize the circuit a bit better.  I guess I could have chosen a part that had one on the schematic but then I wouldn't have had that issue verified.  What I'm worried about is if this will keep power off for the time calculated or am I just wasting time with something so simple.  I've never worked with mosfets before other than using one once for an electronic load and I didn't come up with that circuit, I just verified the mosfet I used was close enough to the one they used because I didn't have the exact part.

Edit: How's this one?

[Falcon69]

now you switch to a p channel mosfet. You need to switch the drain and source on it. and move the diode to the other side and connect the anode between the resistor and the gate.  I am by no means an expert, in fact a beginner myself, but I think the resistor needs to be connect to ground instead, and maybe the cap connected to the supply?

[Dave]

Also, if the mosFET doesn't have one (most do), you may also want to put a Diode in there.  Anode connected to the Source, Cathode to the Drain.

MOSFETs don't actually have separate diodes in the package. The construction of the transistor itself makes a PN junction between the bulk (which is in most cases connected to the source) and the drain. See the picture attached below.

As for the circuit above, you are going to get a exponential ramp on the output, which in its steady state is going to be a couple of volts lower than the supply voltage. If the losses don't bother you, you are going to be fine.
As Falcon69 suggested, add a zener between gate and source (15V or so). You aren't going to have a problem when your circuit will be powered, but you will have a problem when you cut the power. The potential of your drain and source will suddenly drop, while the capacitor on the gate will still be fully charged, holding the potential of the gate high.

[Falcon69]

Dave is correct,  I just simulated it, and a 15v zener will work when the supply voltage is 43.7v?  Wont that put ~28 volts to the gate?  could blow the mosFET won't it?

Here's the pic of the simulation.  Dave is right, when the supply is removed, the cap holds power and slowly drains as the LED dims to nothing.

[Dave]

You might want to put that zener between source and gate, not drain and gate. Also, that cap should be in parallel with the zener and you should keep the resistor the way you have it now. Add another resistor (higher value) in parallel with that capacitor to drain it when the circuit is powered off.

[Falcon69]

fixed

[Dave]

nope

[Falcon69]

This simulation program is fun, but the way they have their mosFETs drawn confuses me.

Here's another one.

[rs20]

This simulation program is fun, but the way they have their mosFETs drawn confuses me.

Here's another one.

You've got a diode straight from input to output there, that's not going to stop any overvoltage...

[Falcon69]

Sorry, don't mean to hijack your thread liquibyte

I'm confused, it looks the same as this circuit. 

[T3sl4co1l]

Sorry, don't mean to hijack your thread liquibyte

I'm confused, it looks the same as this circuit.

This is a reverse protection circuit, not overvoltage.

You're looking for something that looks more like a linear regulator or LDO.

You can't use an N-channel MOSFET here because you lose at least Vgs(th) (and it's not "low dropout).  An NPN emitter follower might be marginal (> 0.7V drop).

You can also get purpose made protection chips that include drive circuits to use N or P channel MOSFETs correctly (without losing supply voltage in the process).

Tim

[Falcon69]

That's why I said the circuit simulator is confusing.  That's a P-Channel Mosfet in the circuit I drew.

And yes, it's a reverse Polarity Protection Circuit.  But, it does work in simulation. remove the supply voltage, and the cap keeps the LED lit for a time being.  I think that is what liquibyte was after?

[liquibyte]

This is a reverse protection circuit, not overvoltage.

You're looking for something that looks more like a linear regulator or LDO.

You can't use an N-channel MOSFET here because you lose at least Vgs(th) (and it's not "low dropout).  An NPN emitter follower might be marginal (> 0.7V drop).

You can also get purpose made protection chips that include drive circuits to use N or P channel MOSFETs correctly (without losing supply voltage in the process).

Tim

The other guy working on this solution came up with the circuit below.  Is this the kind of thing you're talking about?

I don't really need low dropout because, as I said, the op amps are being powered by the full 43.7 volts and they are rated at 44 so it's always seemed like I'm too close to the tolerance for the part anyway.

[liquibyte]

That's why I said the circuit simulator is confusing.  That's a P-Channel Mosfet in the circuit I drew.

And yes, it's a reverse Polarity Protection Circuit.  But, it does work in simulation. remove the supply voltage, and the cap keeps the LED lit for a time being.  I think that is what liquibyte was after?

Not really, no.  I've got a transient surge of around 61-62 volts happening at power on that I'm trying to delay.  I figure maybe I can use a mosfet as a delay until the transient passes.  If there's a few volts drop as a by product then I've actually solved a couple of issues.

[Falcon69]

Oh!  Can't you just put a TVS diode in there to suck up the transient voltage?  I think TVS diodes reset once the transient is passed and circuit should operate as normal then.

[liquibyte]

Oh!  Can't you just put a TVS diode in there to suck up the transient voltage?  I think TVS diodes reset once the transient is passed and circuit should operate as normal then.

The problem is that the transient changes at the output of the power supply based on the voltage setting so what value am I shooting for?  I've measured it at different voltage settings but If I clamp it at 43.7V then it will still allow more than I want when I've set the supply at a lower voltage wouldn't it?  I'll give you guys a run down of basically what I'm seeing.  The first voltage is where the power supply is set at after things settle down.  The second voltage is what I measure as the spike if I turn off the power and let the filter caps drain and then turn it on again.

30V - 36.7V
25V - 33V
20V - 29V
15V - 23.5V
12V - 19.7V
10V - 16.2V
5V - 7.95V
3.3V - 5.49V
2V - 3.6V
1.5V - 3.2V
1.2V - 3V
1V - 2.7V
.8V - 2.2V
.5V - 1.6V
.2V - .9V
.1V - .36V

My thought was to delay the power to the supply circuitry until any transient has passed and the problem would go away.

[Falcon69]

so you are looking for something that would connect the circuit to the power supply after a set time amount, say 5 seconds for example?  Dunno, as your power to supply the 'something' would also undergo those transients as well, unless you had an external power supply for that.  Maybe a button battery, 555 timer, and some caps/resistors?

[liquibyte]

I've looked into using a relay as a soft start on the primaries of the transformers using a separate 12V supply but this adds quite a bit in complexity.  What I want to do with this is connect it between the filter caps and the input to the power supply so that it delays turn on for just under a second until the filter caps have had the time to fully charge after power on.  Basically it's a wedge in solution because the boards are made and the supply (dual actually) works fine but for this one issue.

[rs20]

I'm so confused by this thread. If you want the switch to turn on x seconds after power is applied, that is straightforward to do with a PFET, resistor and cap (and maybe a zener if your PFET can't withstand enough Vgs, although I'd prefer to just get a strong enough MOSFET if possible). You were all going down that route before, what happened?

[liquibyte]

I'm so confused by this thread. If you want the switch to turn on x seconds after power is applied, that is straightforward to do with a PFET, resistor and cap (and maybe a zener if your PFET can't withstand enough Vgs, although I'd prefer to just get a strong enough MOSFET if possible). You were all going down that route before, what happened?

I'm confused by this.  Why would a PFET work better than an NFET? The only PFET I saw was here and I thought using one would require a low side configuration.  I thought using an NFET was where this was going by adding the zener to control VGS which I wasn't really aware of for the most part because I've never really used mosfets before.  The rest of the calculations were done but I wasn't sure if they were right because I've never tried this before and thought I'd ask those that know more than I do.  The app note I linked had a filter cap configuration for dealing with inrush current and while I was doing research on the calculations for that I stumbled on this circuit and because it seemed simpler I was curious if it would work.

[Falcon69]

PFET are normally used as a High Side Switch.  NFETs are for Low Side switching.

[Falcon69]

Oh, I also want to say liquibyte. NEVER use Google images for circuits. I did that once, and screwed up an order of circuit boards.  The picture I was going off of, had the source and drain switched.  Really messed up the entire circuit. However, if I had gone to the link where the picture was, and read the thread, I would have known it was wrong, as it was mentioned in the thread.  Lesson learned.

[rs20]

The rest of the calculations were done but I wasn't sure if they were right because I've never tried this before and thought I'd ask those that know more than I do.  The app note I linked had a filter cap configuration for dealing with inrush current and while I was doing research on the calculations for that I stumbled on this circuit and because it seemed simpler I was curious if it would work.

The process goes like this:

1. Draw a PFET on your piece of paper. Make sure the body diode is pointing against the normal flow of current.
2. (Optional) add a zener from G to S, to prevent Vgs from getting too high.
3. We want the MOSFET to stay switched off initially; capacitors tend to hold voltage steady. Since the capacitor will be discharged initially, it'll have zero volts across it. We want the FET to be off initially, Vgs = 0 volts. Therefore, we should connect the capacitor from G to S (in parallel with the optional zener).
4. We need the FET to eventually turn on. A PFET is turned on by a negative Vgs, S is fixed at the positive supply, so this is great because we just have to pull the gate to ground to turn it on (if you used an NFET, this is where you would get stuck). Therefore, we connect a resistor from the gate down to ground.

Design done, with simple reasoning only. Now, the only issue is that the FET might turn on sort of gradually, and might burn a lot of power while it's only partially turned on. If you want to avoid that, you'll need some more active components.

Oh, I also want to say liquibyte. NEVER use Google images for circuits. I did that once, and screwed up an order of circuit boards.  The picture I was going off of, had the source and drain switched.  Really messed up the entire circuit. However, if I had gone to the link where the picture was, and read the thread, I would have known it was wrong, as it was mentioned in the thread.  Lesson learned.

I think surely the lesson here is, understand and test your circuits rather than blindly accepting schematics from ANY source. If you got circuit boards fabricated without ever testing or thinking "What's the VGS on this FET here?", then you're orders of magnitude short of due diligence. Google images is just fine (wonderful, even), as a source of inspiration.