MOSFET characteristics
Time：2020.08.27
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1. The field effect tube is a voltage control device, the gate basically does not take current, and the transistor is a current control device, and the base must take a certain current. For this reason, the field effect tube is often used when the rated current of the signal source is extremely small. .
2. The field effect transistor is multi-carrier conduction, and the transistor is the conduction of two kinds of carriers. But the low birthrate is obviously affected by the environment.
3. The field effect tube FET has the same amplifying effect as the transistor BJT, and there are electrode correspondences G-b, S-e, D-c between these two amplifying elements. Therefore, according to the BJT circuit, the corresponding FET amplifier circuit can be obtained. But it cannot be simply replaced, otherwise the circuit may not work normally. In addition to being used as amplifiers and controllable switches, field effects can also be used as voltage-controlled variable linear resistors.
4. The S pole and D pole of the field effect tube are symmetrical and can be used interchangeably. The gate-source voltage of the depletion-type MOS tube can be used positive or negative, which is more flexible than the transistor.
5. When FETs compose an amplifier circuit, like transistors, a suitable static operating point must be selected. The grid must have a suitable bias voltage, but no bias current will appear. Different types of FETs have different requirements for the polarity of the bias voltage. The special list is as follows:
Note: JFET stands for junction field effect tube, DMOS stands for depletion mode field effect tube; EMOS stands for enhanced field effect tube.
self-supplied bias circuit
is a self-supplied bias circuit as shown in the figure. The capacitor acts as a bypass to Rs-the source bypass capacitor.
Self-supplied bias circuit
Note: The enhancement mode MOSFET only has ID when the voltage between G and S reaches a certain turn-on voltage VT, so self-bias is not applicable to enhancement mode MOSFET.
Analysis: When Is flows through Rs, the voltage drop of Vs is negative to ground. So, VGS=VG-VS=-ISRS=-IDRS
The question is: VGS is determined by ID and ID changes with VGS. How can the values of ID and VGS be determined? The following content introduces two solutions.
Graphical method
Steps:
(1) As a DC load line (return from output)
The intersection of the DC load line at the different VGS of the output characteristic curve, that is, the simultaneous solution of the tube internal equation and the output loop equation, indicates that the values of ID, VDS, and VGS in the circuit must be at these intersections.
(2) Use the points on the DC load line as the transfer characteristic curve.
(3) Fixed static operating point Q (starting from the G-pole circuit) Since Q should also satisfy the input circuit, the DC load line VGS=-IDRS of the input circuit is used as shown in the figure.
(4) Substitute the data, get Q: VGS, VDS, ID
Calculation method
calculation method uses the following formula to solve:
The IDSS of the formula is called the saturation current, that is, the value of ID when VGS=0; VP is called the pinch-off voltage, when VGS=VP, the drain current ID is 0.
Let us look at an example, the circuit is shown in the figure.
Known: VP=-4V, IDSS=2mA. Substituting the above two formulas into:
Solving these two equations, the solution is: ID1=0.82mA, ID2=0.30mA
Because -ID1RS< VP, ID1=0.82mA is unsuitable for the title, so discard it, so the static drain current IDQ is
IDQ=0.30mA;
The static tube pressure drop VGSQ and VDSQ are respectively
