I have n mosfet gate connected to 4043 logic and Id is about 100mA. Both 4043 and mosfet has +5v. I have plan to use 2N7000 mosfet

Questions are: How large gate resistor i need between 4043 and mosfet? Logic output is sometimes put on rapidly. How fast? Motherboard hdd led controls that. Do i need to place pull down resistor from logic to -0v, between 4043 and mosfet?

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    \$\begingroup\$ The answers by jippie and PhilFrost are good, but if you would like to get a more quantitative view of gate resistance in MOSFETs you could look at this post (electronics.stackexchange.com/questions/60427/…). \$\endgroup\$
    – gsills
    May 11 '13 at 20:20
  • \$\begingroup\$ It also depends what the MOSFET's load is: irf.custhelp.com/app/answers/detail/a_id/215 \$\endgroup\$
    – Fizz
    Nov 8 '15 at 13:45
  • \$\begingroup\$ If the logic never gets abruptly disconnected from the MOSFET, such as when having two instruments connected with a cable, you can do without any resistors here. The CMOS 4000 line is push-pull, so no pull-up/down resistors are needed. And unless your MOSFET is used on the edge of its capabilities, you don't need to worry about the added cable capacity either. \$\endgroup\$
    – Zdenek
    Nov 10 '17 at 20:35

It is generally a good idea to include a gate resistor to avoid ringing. Ringing (parasitic oscillation) is caused by the gate capacitance in series with the connecting wire's inductance and can cause the transistor to dissipate excessive power because it doesn't turn on quickly enough and hence the current through drain/source in combination with the somewhat high'ish drain-source impedance will heat the device up. A low ohm resistor will solve (dampen) the ringing.

As @PhilFrost mentiones, a high value resistor to ground is a good idea to avoid capacitive coupling driving the transistor when it is otherwise not connected.


simulate this circuit – Schematic created using CircuitLab

At all times keep the wiring between logic output, transistor gate, transistor source and ground as short as possible. This will ensure fast turn on/off.

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    \$\begingroup\$ A trivial observation: I would put the pull-down resistor before the gate limiting resistor - that way, the two resistors do not form a voltage divider (however minor), and thus the input voltage is fully expressed at the gate. \$\endgroup\$ May 10 '13 at 5:52
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    \$\begingroup\$ @AnindoGhosh This was discussed in an answer to one of my questions. supercat noted that putting R2 before R1 creates a voltage divider when the G-D is shorted in case of MOSFET failure and thus, protect the driver a little. Of course, the values should be selected appropriately and there is a power consumption trade-off. \$\endgroup\$ May 10 '13 at 6:31
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    \$\begingroup\$ @abdullahkahraman The driver would be protected in either case, R2 before or after - the protection is courtesy the R1. \$\endgroup\$ May 10 '13 at 6:40
  • \$\begingroup\$ @AnindoGhosh never really considered an alternative location for R2 and although the voltage division is minimal I guess you are right. For protection of the microcontroller when D-G shorts out, a 5V1 zener can be placed in parallel to R2. Not sure how effective it is, but at least you took a shot a protecting the controller. \$\endgroup\$
    – jippie
    May 10 '13 at 6:56
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    \$\begingroup\$ @Anindo I actually prefer the series resistor near the source and let it double as source terminator. The voltage divider is a non-issue (100/1M = 0.01%, not even 1mV for a 5V driver). \$\endgroup\$ May 12 '13 at 6:14

You do not strictly need a base resistor. Not only do MOSFETs not have bases (they have gates), but the gate is (very) high impedance. Except when the MOSFET is changing states, the gate current is essentially zero.

Sometimes, a gate resistor is prudent to reduce ringing, especially if the trace driving the gate is long, or if you are concerned with generating electromagnetic interference. A higher value dampens ringing but also slows switching times: the appropriate value depends on how bad the ringing could potentially be and what switching times are required.

It is common practice to place a resistor (the value isn't terribly critical -- anything between \$ 1k\Omega \$ and \$1M\Omega\$ will do) from the gate to ground, just to be sure the MOSFET will be off if the thing driving it (the 4043 in your case) is letting the output float. Otherwise, very small currents from your finger, capacitive coupling, inductive coupling, or other things you'd rather not worry about can change the gate voltage of the MOSFET, resulting in unintended behavior.

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    \$\begingroup\$ I think to limit the inrush current, one should put a series resistor, when working with microcontrollers. Or is it just overkill ? \$\endgroup\$ May 9 '13 at 17:15
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    \$\begingroup\$ @abdullahkahraman I/O ports are generally not stiff sources and sinks, so I don't see a need for a series resistor. \$\endgroup\$ May 9 '13 at 17:20
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    \$\begingroup\$ Sometimes you do put a small (100-200 ohms) resistor, or maybe a ferrite bead, in series with a MOSFET gate in order to kill the gain at high frequencies (RF) if it is showing signs of instability. \$\endgroup\$
    – Dave Tweed
    May 9 '13 at 18:16
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    \$\begingroup\$ I'd like to remind everyone that we are talking about driving a 2N7000 with a 4043. Neither is the 2N7000 a high gate charge device nor is the 4043 capable of high drive currents. I seriously doubt this combination requires a gate resistor any more than all the MOSFETs in a CPU require gate resistors. \$\endgroup\$
    – Phil Frost
    May 9 '13 at 19:12
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    \$\begingroup\$ @Madmanguruman -- Your conclusion is correct given your assumptions, but your assumptions don't apply in most consumer-grade MCU's. (1) You can (and will) exceed rated I/O current if you short the output driver. (2) The driver will behave as if driving into a short circuit if the switching frequency is high -- the input can switch at high frequency independently from the output. The output won't climb much in response resulting in a sustained short-like condition. However, Phil is correct, that isn't the case here, but your generalization deserved this qualification. \$\endgroup\$ May 10 '13 at 7:06

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