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NC switch with time delay


blek
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i'm looking for a component that is normally closed, but when voltage/current is applied to the gate/trigger, it will:

1) open the load (turn it off)
2) keep it open while the current/voltage is applied to the gate at *around the same amount* of trigger current/voltage
3) keep it open for a few milliseconds after the gate/trigger current/voltage drops lower than the holding threshold.
4) after a that immediately closes the load again.


i've read that the most reliable way to do this is with a depletion-mode mosfet, but (correct me if i'm wrong), hardly any of them has time delay in the milliseconds range.... so would adding a capacitor and resistor change it??:

would it work even if there was no resistor (just the capacitor)

post-64076-14279144331763_thumb.png

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A depletion mode MOSFET is a normally open switch but it needs a negative voltage to fully turn off.


waitta... i thought, thought i did, enhancement mode mosfet are normally open and depletion mode are normally closed  ???
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ok.... how about..... THIS:

[take a'look at attached image!!]

tada!


more details:

oad voltage: ~6V
load current: ~500mA
... but it opens it, so i guess breakdown doesn't matter that much

gate/trigger: ~6V
gate/trigger: ~500mA

nyah

post-64076-14279144333361_thumb.png

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The MOSFET will not fully turn off because the gate voltage ill not be negative with respect to the source, as I said earlier.

Why not simply build an astable multivibrator?

The only time a depletion mode MOSFET is required for switching is when it's in a solid state relay as a normally open contact and the load is isolated from the switching signal.

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alright, would it actually help that there WON'T INITIALLY BE current or voltage difference between the source and drain, as it will be open all that time and the depletion of the gate is simply a way to keep it such open, when there would be a short period in milliseconds that the path from drain to source would be closed.

post-64076-142791443337_thumb.png

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I'm not sure what you're trying to say.

Putting the load in series with the source rather than the drain won't help because as I said before, in order for no current to flow the gate voltage needs to be below the source, not equal to but a couple of volts below the source.

Take a look at a datasheet for a typical depletion mode MOSFET. The threshold voltage ranges from -2.1V to -1V, any higher (i.e 0.5V, 0V or 1V) and a significant current will flow. If the gate-source voltage is 0V and the drain is 10V, 140mA will flow. If you connect a load in series with the source and the gate to 0V, when the power is applied, the current through the MOSFET will go down until an equilibrum is formed between the gate voltage and the draind current, much like connecting an enhancement MOSFET's gate to the drain, the source to 0V  and the load between the drain and +V.
http://www.infineon.com/dgdl/BSP149_Rev1.2.pdf?folderId=db3a304412b407950112b408e8c90004&fileId=db3a304412b407950112b42ef8fa4acf

This is one of the reasons why depletion mode MOSFETs are rarely used as switches.

As I asked before, why can't you just use an ordinary astable?

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alright... this is what i'm trying to say...

under normal conditions, the circuit is closed for the path to Va, but sometimes, (now this is the unavoidable behavior of the components), a voltage will "SPIKE" towards Vb (well that's not actually what's happening but for a simplified model, we'll think it as such). now, to stop that "spike", we'll have to disconnect the path to Vb temporarily.

at the same timeframe of the spike, connection is absolutely opened to Va, therefore, we need a time delay.

post-64076-14279144333887_thumb.png

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I don't think you've given this enough thought. Va will be short circuited to 0V.

Attached is a cricuit showing how to bias a depletion mode MOSFET off, on and somewhere in-between.

The MOSFET has a threshold voltage of 2.5V.

1) The MOSFET is fully one with 0V between the gate and source.

2) The MOSFET is half way on. With the gate voltage 2V below the source, the voltage across the load is also 2V so the lamp glows very dimly.

3) The MOSFET is fully off. With the gate 3V below the source, no current flows through the drain so there's no voltage accross the load.

Note that the only way to completely turn the MOSFET off is to use a negative power supply which in this case is a 3V battery with the +V terminal connected to 0V and the -V to the MOSFET's gate.

Why can't you use a monostable such as a 555 timer, a couple of logic gates, a comparator etc?

Sorry, about my previous questions, I got astable and monostable back to front.

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