Common Circuit Application(Current source)

The drain current is set by R_S such that V_GS = I_DR_S. Any value of current can be chosen between zero and I_DSS.

Regions of JFET operation

Cut-off Region: -

The transistor is off. There is no conduction between the drain and the source when the gate-source voltage is greater than the cut-off voltage (I_D = 0 for V_GS > V_GSoff)

Active Region (also called the Saturation Region): -

The transistor is on. The drain current is controlled by the gate-source voltage (V_GS) and relatively insensitive to V_DS. In this region the transistor can be an amplifier.

In the active Region I_D = I_DSS (1 – V_GS / V_GSoff) ²

Ohmic Region: -

The transistor is on, but behaves as a voltage proportional to the source-drain voltage and is controlled by the gate voltage.

I_D = I_DSS [ 2 (1 – V_GS / V_GSoff) V_DS / - V_GSoff - (V_DS / V_GSoff) ² ]

In the Ohmic Region: R_DS ≈ V_GSoff / 2I_DSS (V_GS - V_GSoff) = 1 / g_m

Input Impedance of a Transistor

Impedance is defined as Z = V/I. In linear circuits (with resistors, capacitors, inductors, batteries, etc.) this ratio is the reciprocal of the slope of the I versus V graph. In circuits with nonlinear elements such as a transistor, the input impedance of the resistor is defined as the reciprocal of the slope of the I versus V graph. This is simply the derivative of V_in with respect to I_in-

Z_in = dV_in / dI_in

We can easily find Z_in from what we know already of the behavior of the transistor. We know that the sum of V_BE and the IR drop across R_E must equal V_in.

V_in = V_B = V_BE +V_E = V_BE + I_ER_E [I_R = I_C + I_B = βI_B + I_B = (β + 1)I_B]

V_in =V_BE + I_ER_E = I_B(β + 1)R_E [I_B = I_in]

V_in = V_BE + I_in (β + 1) R_E

Taking the derivative of V_in with respect to I_in, remembering that V_BE is a constant, we get the result:

Z_in = dV_in/dI_in = d/dI_in(V_BE + I_in (β + 1)R_E) = (β + 1)R_E

Zin = (β + 1)RE ≈ βRE

Because I_E = I_B (β + 1). The IR drop across R_E is greater then it would be for I_B alone. The amplification of the base current causes R_E to appear larger to a source looking into the input by (β + 1).

Definition of Gain

Gain is defined as the ratio of the output signal to the input signal. Because transistor amplifiers often have a quiescent output (a non zero output when the input is zero) we define gain as the derivative of the output with respect to the input. Thus gain is defined as the ratio of the change in output to the change in input.

So far we have not specified the output quantity, the reason is that we can define the gain with respect to any given output and input quantity.

General definition: A =d(Output) / d(Input) if (Output) = 0 when (Input) = 0, then A = (Output) / (Input)

Voltage Gain: A_v = dV_out/ dV_in if V_out = 0 when V_in = 0, then A = V_out / V_in

Current Gain: A_I = dI_out / dI_in if I_out = 0 when I_in = 0, then A = I_out / I_in

Power Gain: A_p = dP_out / dP_in if P_out = 0 when P_in = 0, then A = P_out / P_in

Note that a negative gain means that the sign of the signal is inverted. Negative gain is not possible for Power Gain. |A| less than unity indicate that the output is smaller than the input.

The quantities need not be the same. If the input and output quantities are different, the gain is no longer unitless. The most common examples are transimpedancc gain and transadmittancc gain.

Transmpedancc Gain: A_Z = dV_out / dI_in if V_out = 0 when I_in = 0, then I = I_out / I_in

Transadmittancc Gain: A_Y = dI_out / dV_in if I_out = 0 when V_in = 0, then A = I_out / I_in