# JFET with Two-Supply Source Bias

JFET(Junction Field Effect Transistor) are biased so that they can be operated with stable operating point in the ohmic or active region. When operated in stable operating point the transistor operates with pre determined circuit voltages and currents so that any environmental variation such as temperature or using same transistor with different electrical parameters does not vary the output. There are couple of JFET biasing methods. The fixed gate bias, self bias and voltage divider bias method were explained in the tutorial How to bias JFET transistor? and JFET Biasing Worked Out Example Calculation. Here another JFET biasing method known as two supply source bias is explained with calculation involved. The JET two supply source bias method is used to bias the JFET transistor in the active region.

The circuit diagram of a JFET biased with two supply source is shown below.

Here we have used 2N5459 N-channel JFET with two supply source $$V_{DD}=+5V$$ and $$V_{SS}=-5V$$.  A gate resistor $$R_G=1M\Omega$$ is used which has no effect on the biasing the transistor. That we can simply remove the gate resistor shown in the above circuit. Because with or without it the current into the gate of the transistor will be zero.

Now we will show how to calculate the value for the drain resistor $$R_D$$ and source resistor $$R_S$$ so that the JFET is biased in the active region.

By applying KVL(Kirchhoff's voltage law) around the input circuit we can write,

$$V_G = V_{GS}+I_D R_S - V_{SS}$$

Since $$V_G = 0$$ we have,

$$I_D R_S = V_{SS} - V_{GS}$$   --------------->(1)

and therefore drain current can be determined by,

$$I_D = \frac{V_{SS} - V_{GS}}{R_S}$$    --------------->(2)

and the source resistor can be determined by,

$$R_S = \frac{V_{SS} - V_{GS}}{I_D}$$     --------------->(3)

The usual approach is to find drain current$$I_D$$ using either the transconductance equation or using graphical method. And then calculate the drain resistor value($$R_D$$). The transconductance equation or the Shockley equation is as follows,

$$I_D=I_{DSS}(1-\frac{V_{GS}}{V_P})^2$$  --------------->(4)

So to find $$I_D$$ using this equation requires the knowledge of $$I_{DSS}$$, $$V_GS$$ and $$V_P$$. The value of $$I_DSS$$ and $$V_P$$ are obtained from the JFET datasheet. The $$V_GS$$ value is a user choice.

Here we will be using Proteus software to find the solution. When using software for circuit prototype, the solution relies upon the spice model of the JFET transistor. This can often be different from the datasheet. Nevertheless, the steps involved in solving the problem is similar.

To solve the above circuit problem, we need the drain characteristics curve for the JFET used in the circuit simulation. In the tutorial import spice model in Proteus and draw JFET drain curve, we have shown how to draw the drain curve. The following graph shows the drain curve for the 2N5459 JFET in Proteus software.

From the drain curve graph we can see that the drain saturation current $$I_{DSS}$$ is 10mA and pinch off voltage $$V_P$$ is 1.2V.

Having found $$I_{DSS}$$ and $$V_p=V_{GS(off)}$$ from graph as illustrated above or referring to JFET datasheet, the next thing we need is the gate to source voltage, $$V_{GS}$$. This is user choice and by choosing appropriately $$V_{GS}$$ we can control the drain current as dictated by the Shockley equation. That is why JFET is called voltage controlled current device. We don't want to set $$V_{GS}$$ arbitrarily but instead use a value such that the output drain current $$I_D$$ has maximum possible swing between 0 and $$I_{DSS}$$. A usual approach is to set the $$V_{GS}$$ half way between 0V and $$-(V_{GS(off)}$$($$=V_P$$). This is illustrated below.

Thus,

$$V_{GS} = \frac{V_{GS(off)}}{2} = \frac{V_P}{2} = \frac{1.2V}{2}$$

therefore, $$V_{GS}=0.6V$$

And from the Shockley equation(4),

$$I_D=10mA (1-\frac{0.6V}{1.2V})^2$$

gives, $$I_D=2.5mA$$

With $$I_D=2.5mA$$, the source resistor value using equation(3) turns out to be,

$$R_S = \frac{5 - (-0.6)}{2.5mA}=2.24K/Omega$$

Next we have to solve for the drain resistor $$R_D$$. Applying Kirchhoff's voltage law in the output circuit we can write,

$$V_{DD}=V_D+I_D R_D$$   ----------------------->(5)

Rearranging for $$R_D$$,

$$R_D = \frac{V_{DD}-V_D}{I_D}$$  ---------------->(6)

Here we need the drain voltage which is user choice. Let the drain voltage(V_D=2.5V\) then,

$$R_D = \frac{5V-2.5V}{2.5mA}$$

this gives, $$R_D = 1k/Omega$$

With these calculated resistor values the circuit and the simulation result is shown below.

The above circuit simulation picture shows that the voltage and current calculated above are as expected.

In the above circuit calculation we have calculated the drain current $$I_D=2.5mA$$ but using the drain graph above we find that for gate to source voltage $$V_{GS}=-600mV$$, the drain current is $$I_D=1.6mA$$. See graphed picture below.

Using $$I_D=1.6mA$$ in finding source resistor $$R_S$$ with equation(3) and drain resistor $$R_D$$ with equation(6), we get

$$R_S=3.5 k\Omega$$ and $$R_D=1.56 k\Omega$$

With this calculated resistor value the following shows the simulation result.

Thus we showed two ways of solving JFET with Two-Supply Source Bias with worked out calculation. Such two supply source bias method is used in application like creating JFET amplifier and modulators. See some JFET applications:

Many times the gate voltage may not be high enough stabilize gate to source voltage. In this situation we can use current source bias which is illustrated in the next tutorial JFET Current Source Bias Worked Out Example.