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New to electrical engineering. I am following an introductory Cambridge university course but not entirely clear on what the lecturer is saying, and have nobody to consult because I am self-teaching this part of the course under huge time constraints.

The diagram here has RG. I understand that the gate acts as a capacitor in a MOSFET, due to a metal oxide layer, which gives it essentially infinite input impedance. So, why not just connect the gate to ground with a wire? This ensures it's not floating, as charge buildup doesn't occur. Furthermore, the lecturer says RG sets the value of the input impedance from infinite to a finite value (and so should be suitably large for use in an amplifier), which I don't understand, because there is still the capacitive property of the MO layer to consider. I also don't understand why an infinite resistance is a problem in a practical scenario.

I tried to find similar posts: Why is this MOSFET's "pullup" resistor necessary? seems like it might have relevant points (especially on the idea of not being able to just connect VDD and VDS). However I feel like I'm missing lots of basic details. Question about mosfet gate resistor says a high value resistor 'avoids capacitive coupling driving the transistor when it is otherwise not connected', and 'It is common practice to place a resistor... from the gate to ground, just to be sure the MOSFET will be off if the thing driving it... 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'. What does it mean to let the output float, and what is capacitive driving?

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  • \$\begingroup\$ The transistor turns on when the voltage from gate to source is high enough (for N-channel) or low enough (for P-channel). A key part of designing a mosfet circuit is to make sure you control that voltage to keep the mosfet in the desired state at all times. The only reason to use a resistor is because it aids in that control. This happens pretty often, that you want that resistor there. But sometimes the resistor may not be needed. If the gate to source voltage is adequately controlled at all times without the resistor then you don't need the resistor. Just go with the flow for now. \$\endgroup\$ – mkeith Aug 17 at 4:09
  • \$\begingroup\$ Wait... they call the drain-to-source voltage V_DS and draw the arrow from source to drain? \$\endgroup\$ – zebonaut Aug 18 at 11:11
  • \$\begingroup\$ Since you mention it is a high-value resistor, the 99 % correct answer is: You need a weak-ish pull-down resistor to keep the MOSFET off as long as the gate is left floating. However, and because this might be fairly theoretical (academic/textbook question), you could also consider a 1 % chance that the gate input might be current driven, and R_G translates the current from the driving source to a voltage, i.e. V_G = I_in * R_G. This is also a helpful idea if you want to find out what maximum value you can allow for R_G for a given undesired current into the gate node (think: interference). \$\endgroup\$ – zebonaut Aug 18 at 11:19
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Maybe you’re overthinking this.

The diagram you show implies that there is a gate signal connection that is open. Rg is to ensure that the gate has a DC path to source (GND) in the absence of a gate signal. As you noted, this is needed due to the FET gate impedance being practically infinite.

Rg isn’t needed if there’s a ground-referenced gate signal present.

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  • \$\begingroup\$ Perhaps I am overthinking and also very new to this, but I don't understand why a wire wouldn't work. \$\endgroup\$ – debrevitatevitae Aug 17 at 1:12
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    \$\begingroup\$ If you connect the gate to Ground with a wire, you can't drive it with another signal - the MOSFET would be permanently held in the Off state. \$\endgroup\$ – Peter Bennett Aug 17 at 1:15
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    \$\begingroup\$ A wire would work to ground the gate, but then how would you apply a signal? What they're not showing you - it's implied - that to the left of that little diagram is where you would attach a signal source. \$\endgroup\$ – hacktastical Aug 17 at 1:16
  • \$\begingroup\$ If you connect gate to source with a wire, that guarantees that the mosfet will always be OFF under all conditions. Logically, if you want the MOSFET to always be off under all conditions, you can just remove it from the circuit altogether. \$\endgroup\$ – mkeith Aug 17 at 4:06
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So, why not just connect the gate to ground with a wire?

How then would you switch it on and off? If the gate is permanently tied to 0V, you would never be able to tell this transistor to conduct.

David Normal already discussed the value of the pull-down resistor. When there is no other input to the MOSFET gate, Rg ensures that it is sitting at 0V, and thus it will NOT conduct. However, you need the resistance there because you want to be able to overcome its effect by applying an external signal to the gate to tell the MOSFET to turn on.

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If you tied it directly low, as you suggest in the above quote, then if you apply voltage to the gate it will be immediately shorted to ground and your MOSFET would never be able to switch on. In short, Rg "holds" the gate at 0 volts (a known state) until you force it high. There is no in-between.

A MOSFET with a floating gate can cause all sorts of problems. Because its input impedance is so high, any voltage fluctuation on the gate can cause it to partially turn on. This often leads to more fluctuations, and the MOSFET will begin to oscillate. This will, obviously, cause your circuit to behave erratically and, in some cases, will lead to heating of the transistor and potentially permanent damage. It is crucial for the gate of a MOSFET (or other high-impedance input) to be held in a known state to eliminate the possibility of this type of behavior.

Besides the pull-down effect, Rg may serve another purpose in high-speed applications. Because the gate of a MOSFET is effectively a capacitor, if you are switching at a high speed the gate will take some time to discharge and turn the transistor off. Suppose the MOSFET is a 2n7000 with an input capacitance of 50 pF and no Rg in the circuit. The impedance between the gate and ground could be, say, 50 MΩ. The RC delay would then be R x C = [50x10^(-12)] x [50x10^(6)] = 2.5x10^(-3), or 2.5 milliseconds. If you're trying to switch the transistor at 100 kHz (period of 10 microseconds) then the MOSFET will not be able to turn on or off fast enough. A resistor connected between the gate and ground will discharge the gate capacitance significantly faster, allowing you to switch the MOSFET on and off much faster.

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  • \$\begingroup\$ Thank you, this makes sense :). I did think of the grounding problem, but didn't really follow the train of thought to the end haha \$\endgroup\$ – debrevitatevitae Aug 17 at 1:26
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Just like transistors like open collector pull up resistors, MOSFETs also require a pull up only if the gate pin is left floating at any point. When the MOSFET gate is connected to a power source or a microcontroller pin, the gate then has a known state (either high or low). It is also a good idea for the gate to have a pull down resistor Rg to keep the mosfet gate at a known state, in case of a loose connection perhaps, this would keep the gate at a low potential. This would keep the MOSFET's Rds resistance low. In case of any fault and the gate is left floating, the MOSFET's Rds goes high and the MOSFET turns into a glorified heater. This is an N-type MOSFET, it does not require the resistor Rs, especially when you connect an Inductive load such as a motor. Rs comes into play when a resistive load is connected. This is not a common practice for N-type.

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    \$\begingroup\$ or in case someone with an electrostatic charge walks across the room. it doesn't take much charge to turn a mosfet on or off. so don't leave the gate unconnected. \$\endgroup\$ – Jasen Aug 17 at 1:17
  • \$\begingroup\$ electrostatic, or even .. radiowaves or other noise, all depending on how good an antenna is the wire or pcb trace connected to the gate \$\endgroup\$ – quetzalcoatl Aug 18 at 13:49
  • \$\begingroup\$ Both comments are absolutely correct. Especially high frequencies travelling through the board causes intermittent issues. So just as a rule of thumb, use pull down resistors \$\endgroup\$ – David Norman Aug 18 at 19:37
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In short; when the input source has a very high impedance (i.e. is off), RG provides a current path to discharge the gate, while when the input source is high it can only supply a limited current. The value of RG is a compromise between these two requirements; low enough to discharge the gate in a short time, high enough not to overload the source.

If the driver is off (high-impedance) and there is no path to anywhere else, electrically isolating the input and circuit element (in this case the gate) from any direct voltage, they are said to "float". When a component floats, it can pick up static charge or stray fields which will either create a spurious signal or damage the component.

The capacitance of the gate is only significant at high AC frequencies and hence influences its on/off switching time. It does not influence the DC behaviour in any other way.

Capacitive coupling passes the signal through a capacitor. It blocks any net DC bias is the signal but allows fast switching transients through.

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