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I always see a very high resistance between the gate and the ground in n-channel MOSFET common-source amplifier. What is it's role? I think it is to prevent the signal input from bypassing the MOSFET and going to ground directly, but I am not sure. Am I right or wrong?

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    \$\begingroup\$ It is not clear whether you are asking about the external biasing resistors (usually, voltage divider), or about internal parasitic resistance. Clarification or schematic will be appreciated \$\endgroup\$ – Vasiliy Aug 4 '13 at 19:33
  • \$\begingroup\$ Sorry if I wasn't being clear. I was talking about an external resistor between the gate terminal and the ground. \$\endgroup\$ – Douglas Edward Aug 4 '13 at 20:55
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    \$\begingroup\$ @DouglasEdward Please could you share any of the schematics you see this high resistance in, and specify which resistor(s) on that schematic you are referring to? \$\endgroup\$ – Anindo Ghosh Aug 5 '13 at 11:35
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If I understand you correctly, you're asking why \$R_1\$ (and \$R_2\$) in the following schematic is usually taken to be very large, right?

enter image description here

In summary:

if you take big resistors for Gate's biasing you get low power waste by these resistors + high input impedance of the amplifier.

Explanation:

The main function of these resistors is to form a voltage divider which defines the DC portion of Gate's voltage. This action is known as "biasing".

However, this does not answer why take such a high value resistors. You can form voltage divider with any resistors, as long as their proportion is correct. There are (at least) two reasons why you want these resistors to be big:

  1. Note that the function of these resistors is to just bias the Gate of the MOSFET with correct voltage. However, there is current flowing from the supply to the ground through this voltage divider. The magnitude of this current is \$I_{DC}=\frac{V}{R_1+R_2}\$. The bigger these resistors are, the lower the current drawn from the power supply by this biasing divider.
  2. The input impedance for small AC signal (as seen from \$V_{in}\$ pin) of this amplifier is \$\sim R_1||R_2\$. The higher the values of these resistors, the higher the input impedance of the amplifier.
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    \$\begingroup\$ OP provided a somewhat limited clarification that the resistor referred to is one between Gate and ground. This suggests that it isn't biasing but pull-down, else two resistors would be mentioned, one from gate to supply and another from gate to ground. However, it's best to ask OP to edit the question and add a schematic with part references. I'll add that as a comment to the question. \$\endgroup\$ – Anindo Ghosh Aug 5 '13 at 11:34
  • \$\begingroup\$ @AnindoGhosh, the answer applies to both. Even when the resistor is there for pull-down only, its value still affects power consumption and input impedance. \$\endgroup\$ – Vasiliy Aug 5 '13 at 11:45
  • \$\begingroup\$ Not debating that there is some effect. Just asking OP for more detail. A pull-down will have lower power consumption, since it will conduct only when the gate drive is high, as compared to a bias resistor pair, which conducts all the time. \$\endgroup\$ – Anindo Ghosh Aug 5 '13 at 11:56
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The resistance you see when you look into the gate of a MOSFET is extremely high, usually several orders of magnitude larger than the gate resistor you refer to. Withouth the gate resistor it will pick up static easily and the charge build up on the gate may easily drive drain-source to conduct or even worse it may build up such high voltages that the metal oxide insulator between gate and drain-source will break down. Once that happens, your transistor is dead.

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  • \$\begingroup\$ +1 @Douglas -- This is the correct answer to the question most students mean when they ask what you did. \$\endgroup\$ – DrFriedParts Aug 4 '13 at 19:46
  • \$\begingroup\$ @Douglas, this answer is correct in general, but has almost nothing to do with the initial question. You don't want the Gate to be floating - its true. However, the reason why the resistance across the Gate is high is due to other factor. \$\endgroup\$ – Vasiliy Aug 5 '13 at 10:27

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