How exactly does a MOSFET amplify gate voltage?

Discussion in 'General Electronics Chat' started by midnightblack, May 17, 2012.

  1. midnightblack

    Thread Starter Member

    Feb 29, 2012
    31
    0
    Hi,

    I am trying to understand some of the basics of how the mosfet amplifies. Specifically, n channel depletion mode mosfet.

    I understand it happens after the drain voltage reaches pinch off. But I am not entirely sure why after this point it is that there is so much gain for small increases in gate voltage.

    I am not sure if it contributes to the "gain" but correct me if I'm wrong, after the mosfet reaches pinch off voltage and enters saturation, whilst it is amplifying the gate voltage, the channel length is also modulated?

    Is it this effect that is causing the amplification? The width of the depletion layer between the n+ region connected to drain and the channel is being modulated? If so why exactly does it cause gain?

    Or am I wrong about the channel length modulation part?

    I am not looking for equations per se, but rather a sort of physical intuitive explanation.

    EDIT: Following on from this idea (if it is correct), the voltage caused by the depletion layer between the n+ region connected to the drain and the substrate is somehow proportional to the gain?
     
    Last edited: May 17, 2012
  2. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    The area of the pinched off channel in the saturation region determines how much current can pass through the drain-source channel. Variation in the gate-source voltage varies the area of the pinch off region which varies the drain current. Thus it is the variation in drain-source current that causes the amplification (note that the transistor gain is given in transconductance -- variation in channel current versus gate voltage).

    This current variation is converted to a voltage variation by the drain resistor which is how you get voltage gain.

    That's a very simplified explanation but it gives the basics.
     
  3. midnightblack

    Thread Starter Member

    Feb 29, 2012
    31
    0
    Specifically which area are we talking about here? Would it be the same as saying volume of the pinched off channel?

    I am not sure I fully understood this part: The variation in the gate-source voltage varies the drain current correct understood. But then drain source current causes the amplification? I thought the gate-source voltage causes the amplification? If I understand correctly, once Vds>Vgs-Vth, increasing it further has little effect. But increasing Vgs has increases in Id. As shown in such a graph:

    [​IMG]

    Lastly, are you talking about output resistance or an actual resistor connected to the drain?

    Thanks for your reply..
     
  4. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    The area is is area of a cross section of the channel. You could also refer to it as the volume to the channel. It's sort of like the cross section of a pipe.

    I'm trying to say that it is variation of drain current with gain voltage that is the basic transistor amplification mechanism. It is a voltage to current amplification (gm). This current variation is converted to a voltage variation by an external resistor connected between the drain and the supply voltage. That gives you voltage amplification from the input gate to the output. So the transistor does not give voltage amplification unless there is an external load impedance.
     
  5. MrChips

    Moderator

    Oct 2, 2009
    12,449
    3,365
    We had a related question recently about solving for a diode operating point using the resistor load line.

    [​IMG]

    In the chart above, assume the supply voltage Vdd is 8V and the series load resistor is 1kΩ.

    The maximum Vds voltage at 0 Id will be 8V on the X-axis.
    The maximum Id = Vdd/R = 8V/1k = 8mA as shown on the Y-axis.
    Draw a straight line connecting these two extreme points. This is the resistor load line.

    Now at Vgs = 1.75V, Vds is approximately 2.8V
    At Vgs = 1.25V, Vds is approx. 4.7V

    Hence a change of Vgs of 0.5V produces a change of -1.9V at the drain.
    This is a gain of -1.9/0.5 = -3.8
     
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