There are many types of MOSFET biasing methods such as zero bias, fixed bias, self bias, voltage divider bias, two supply bias and current source bias. FET transistors can either be biased in ohmic or active region(linear region). If the purpose of the FET is to use it as an amplifier then it is biased in the active region. To bias depletion MOSFET in either depletion mode or enhancement mode in the active region we can use either self bias, voltage divider bias or the two supply bias. Here voltage divider biasing of depletion MOSFET is illustrated with worked out example calculation.
Circuit & Operation
The following shows the circuit diagram of depletion MOSFET biased using voltage divider biasing.
In this example the LND150 depletion MOSFET is used. Also 5V power supply is used. The biased circuit is applied with input signal Vin of 100mV amplitude and frequency of 1kHz. The output signal appears at the 10kOhm load resistor. The biasing resistor R1 and R2 provides a gate bias voltage which fixes the gate to source voltage. This circuit produces stable operating point and also uses only one power supply.Choose bias point
To bias the circuit we have to select the operating point in the active region. The following shows the location of the bias point Q in the active region.
The selected Q-point or the bias point is at,VDSQ = 4V, IDQ = 3.62mA
Calculate gate to source voltage
After setting the drain current we can know the gate to source voltage required so that IDQ = 3.62mA. The gate to source voltage relation to drain current is given by the Shockley equation.
ID=IDSS(1−VGSVGS(off))2
which can be resolved to,
VGS=VGS(off)(1−√IDQIDSS)
From the graph above or the datasheet we have,
VGS(off)=2V and IDSS=2.33mA
Therefore,
VGS=2V(1−√3.62mA2.33mA)
that is, VGS=0.49V
Choose gate voltage and source voltage
The gate voltage and the source voltage can be chosen equal to the gate to source voltage.
VG=VD=VGS=0.49V
Calculate voltage divider resistors
Knowing the gate voltage we can then determine the voltage divider resistor R1 and R2.
We have,
VG=R2R1+R2VDD
Rearranging,
R2=VGVDD−VGR1
or, R2=VGVDD−VGR1
Let R1=10kΩ
then, R2=0.49V5V−0.49V10kΩ
that is, R2=1kΩ
Calculate source resistor
The source voltage is selected equal to gate to source voltage VGS,
VS=VGS=0.49V
Then, the drain resistor is,
RS=VSID
or, RS=0.49V3.62mA
that is, RS≃135ΩCalculate drain resistor
The drain resistor is,
RD=VDD−VDID
or, RD=5V−0.49V3.62mA=4.51V3.62mA
that is, RD≃1.2kΩ
Calculate input/output impedance
The input impedance is due to the voltage divider circuit is,
Zi≃R1||R2=R1R2R1+R2
or, Zi≃R1||R2=1kΩ×10kΩ1kΩ+10kΩ
that is, Zi≃909Ω
The output impedance is,
Zo≃RD
that is, Zo≃1.2kΩ
Calculate coupling capacitor
The input coupling capacitor is,
C1=12πf(0.1)Zi
or, C1=12π(1kHz)(0.1)(909Ω)
that is, C1=1.75μF
The output coupling capacitor is,
C2=12πf(0.1)Zo
or, C2=12π(1kHz)(0.1)(1.2kΩ)
that is, C2=1.32μF
Result
The following is the completed voltage divider biased depletion MOSFET amplifier.
When the input signal is 100mV amplitdue and frequency is 1kHz then the input and output waveform as shown in oscilloscope is shown below.
Recommendation
The following are recommended tutorials:
- How to design JFET source follower?
- How to design Common Gate JFET Amplifier
- How to design common source JFET Amplifier
- Comparison of BJT, JFET, D-MOSFET, E-MOSFET construction and operation