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JMLStamp2p
04-24-2008, 10:21 PM
Good morning,
I am·driving the Gate of·a 2n7000 N-Channel FET via a 5V output pin. Someone on the Forum told me that I should pull the gate down via a resistor because the gate of a FET acts like a capacitor. Could someone give me a correct value for this resistor, would a 1K be sufficient. My circuit will be installed in a noisy electrical environment.

Thanks,
JMLStamp2p

Philldapill
04-24-2008, 10:33 PM
I don't think you REALLY need a resistor if your 5V ouput pin swings between 5V and ground(0V). The gate on a fet does indeed act as a capacitor due to the nature of fets. The reason mosfets are voltage dependent instead of current dependent, like BJTs, is because the amount of charge on on the gate is what widens/contracts the channel for current to flow. Because of this, you will have a resistance across the source and drain instead of a distinct voltage drop like BJTs.

Back to your question. I don't know what your 5V output is from, but like I said, if it can source and sink current from 0-5V, you should be ok while the pin is active(i.e. the device is ON). You may want a pulldown resistor to just keep the gate low(off) if the pin becomes a high impedence, such as when the device is off or something and you want to make sure the fet is off too. If this is the case, you could use a resistor of 100K or higher. Just don't make it too low, then you'll start having a voltage drop from the 5V pin and you may only get your gate up to 4V or something where the channel isn't fully open.

TomS
04-24-2008, 11:22 PM
I think JML is refering to a series resistor to limit current flow into and out of the driving device. The maximum value of the resistor would depend on how fast you need the FET to switch as the resistor and capacitor form a RC time constant. If you're switching high currents you don't want to make the switching time too great or power dissipation would increase. Too low a value of resistor and you might as well leave it out. Anything from 100 Ohms to 10K Ohms would probably work.

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Tom

Philldapill
04-25-2008, 12:12 AM
Tom,
You are right about using a series resistor to limit the inrush current, but this is usually for a power fet since they have a fairly large capacitance. A 2N7000 has a typical input capacitance of 60pF, while a power mosfet may have something on the order of 5000pF. A big difference.

As for the resistor, I'm pretty sure he was talking about a pulldown resistor. In that case, 10K is a bare minimum, but the series resistor TomS mentioned is correct.

JMLStamp2p
04-25-2008, 12:28 AM
Thanks guys,
And Yes I was talking about a Pulldown on the gate. I will try a 10k and see how it reacts ...

JMLStamp2p

Philldapill
04-25-2008, 12:38 AM
JML,
You can try a 10K resistor, but remember, this is only there to keep the gate low. Device driving the pin is on, the pulldown resistor will serve no purpose, and will infact, increase power consumption when the pin is high. At 5V with the pin high, the pulldown will be drawing 2.5mW just sitting their. a 100K will pull 1/10th of that. If power isn't an issue in your design, a 10K will be fine, but again, it is ONLY to ensure the gate stays low while the device is off.

Paul Baker
04-25-2008, 02:46 AM
Bias resistors for this application are typically 10k, and if the gate is always actively driven (pin is never input, including at startup), you don't need the resistor.

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Paul Baker (mailto:pbaker@parallax.com)
Propeller Applications Engineer
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