Draw and explain JFET characteristics and prove an equation.

Show that for a JFET

$$g_m = \dfrac{2}{\vert V_p \vert}\sqrt{l_{DSS}.l_{DS}} $$

1 Answer

Characteristics of JFETS

There are two types of static characteristics viz

  1. Output or drain characteristics and

  2. Transfer characteristic.

1) Output or Drain Characteristic.

  • The curve drawn bertween drain current $I_p$ and drain-source voltage VDS with gate-to source voltage VGS as the parameter is called the drain or output characteristic. This characteristic is analogous to collector characteristic of a BJT:

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2) Transfer Characteristics of JFET

  • The transfer characteristic for a JFET can be determined experimentally, keeping drain-source voltage, $V_{DS}$ constant and determining drain current, $I_D$ for various values of gate-source voltage, $V_{GS}$. The circuit diagram is shown in fig. The curve is plotted between gate-source voltage, $V_{GS}$ and drain current, $I_D$ , as illustrated in fig. It is similar to the transconductance characteristics of a vaccum tube or a transistor. It is observed that

(i) Drain current decreases with the increase in negative gate-source bias (ii) Drain current, ID = IDSS when VGS = 0 (iii) Drain current, ID = 0 when VGS = VD

The transfer characteristic can also be derived from the drain characteristic by nothing values of drain current, $I_D$ corresponding top various values of gate-source voltage, $V_{GS}$ for a constant drain-source voltage and plotting them.

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The pinch off voltage is the value of $V_{DS}$. at which the drain current reaches its constant saturation value. Any further increase in $V_DS$ does not have any effect on the value of $I_D$. Do not make a confusion between the cut off and pinch off voltage is denoted by $V_P$.

$g_m = \frac{2}{|V_P|} \sqrt{I_{DSS}. I_{DS}}$

we know that,

$g_m = g_mo \bigg[1 - \frac{V_GS}{V_P}\bigg]$

where, g_m0 = $\frac{2I_{DSS}}{|V_P|}$

$\therefore g_m = \frac{2I_{DSS}}{|V_P|} \bigg[1 - \frac{V_GS}{V_P}\bigg]$

Also $I_{DS} = I_{DSS}[1-\frac{V_GS}{V_P}] \hspace{2cm}....(1)$

Also $I_{DS} = I_{DSS}\bigg[1 - \frac{V_GS}{V_P}\bigg]^2 \hspace{2cm} ... (2)$

$\hat{a}\epsilon \times \hat{a}\epsilon \times \bigg[1 - \frac{V_GS}{V_P}\bigg]^2 = \frac{I_{DS}}{I_{DSS}}$

$\therefore \bigg[1 - \frac{V_GS}{V_P}\bigg] = \sqrt{\frac{I_{DS}}{I_{DSS}}} \hspace{2cm}....(3)$

substituting equation (3) into equation (1) we get,

$g_m = \bigg[\frac{2I_{DSS}}{|V_P|}\bigg] \times \sqrt{\frac{I_{DS}}{I_{DSS}}}$

$\therefore g_m = \frac{2}{|V_P|} \sqrt{I_{DSS}.I_{DS}} \hspace{2cm}... (proved)$

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