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5.6 Ideal current-voltage relationship:
Consider the structure of JFET as shown in figure 3. As the ideal solution is much complicated, therefore, we will consider a differential volume for the ease of calculations. In the figure dx is the differential distance across the channel considering a small increment. If we move gradually from source to drain we will cover the total channel from source to drain.
Let us consider the differential volume of the channel= V[h-w(x)]dx
Resistance of the differential volume = (10)
Then the conductivity can be
There is no effect of incremental distance on the channel, but if we consider the differential voltage (dV) across the channel , we can express the current as:
ID= dV (11)
In this way*…show more content…*

As an example, if a change in gate voltage of 0.1 V causes a change in drain current of 0.3 mA, then, Transconductance, g_m = 0.3 mA 0.1 V = 3 mA/V = 3 × 10−3 A/V or mho or S (siemens) = 3 × 10−3 × 106 µ mho = 3000 µ mho (or μS) From Schokley equation we have the expression of drain current I_(DS )=I_(DSS ) (1-V_GS/V_P )^2 Differentiating both sides we have, (dI_D)/(dV_GS )=2I_(DSS ) ⌊1-V_GS/V_P ⌋(-1/V_P ) g_m=-(2I_DSS)/V_P ⌊1-V_GS/V_P ⌋ When, V_GS=0, g_m=g_mo g_mo=-(2I_DSS)/V_P Then we can write, g_m=g_mo ⌊1-V_GS/V_P ⌋ iii) Amplification factor ( µ ): It is the ratio of change in drain-source voltage (ΔVDS) to the change in gate-source voltage (ΔVGS) at constant drain current i.e. Amplification factor, µ = ΔVDS/ΔVGS at constant ID Amplification factor of a JFET indicates how much more control the gate voltage has over drain current than has the drain voltage. For instance, if the amplification factor of a JFET is 50, it means that gate voltage is 50 times as effective as the drain voltage in controlling the drain current. The relationship among JFET parameters can be established as under: We know µ = ΔVDS/ΔVGS Multiplying the numerator and denominator on R.H.S. by ΔID, we get, µ = rd × gm i.e. amplification factor = a.c. drain resistance ×

As an example, if a change in gate voltage of 0.1 V causes a change in drain current of 0.3 mA, then, Transconductance, g_m = 0.3 mA 0.1 V = 3 mA/V = 3 × 10−3 A/V or mho or S (siemens) = 3 × 10−3 × 106 µ mho = 3000 µ mho (or μS) From Schokley equation we have the expression of drain current I_(DS )=I_(DSS ) (1-V_GS/V_P )^2 Differentiating both sides we have, (dI_D)/(dV_GS )=2I_(DSS ) ⌊1-V_GS/V_P ⌋(-1/V_P ) g_m=-(2I_DSS)/V_P ⌊1-V_GS/V_P ⌋ When, V_GS=0, g_m=g_mo g_mo=-(2I_DSS)/V_P Then we can write, g_m=g_mo ⌊1-V_GS/V_P ⌋ iii) Amplification factor ( µ ): It is the ratio of change in drain-source voltage (ΔVDS) to the change in gate-source voltage (ΔVGS) at constant drain current i.e. Amplification factor, µ = ΔVDS/ΔVGS at constant ID Amplification factor of a JFET indicates how much more control the gate voltage has over drain current than has the drain voltage. For instance, if the amplification factor of a JFET is 50, it means that gate voltage is 50 times as effective as the drain voltage in controlling the drain current. The relationship among JFET parameters can be established as under: We know µ = ΔVDS/ΔVGS Multiplying the numerator and denominator on R.H.S. by ΔID, we get, µ = rd × gm i.e. amplification factor = a.c. drain resistance ×

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