A \(p\) -channel JFET has device parameters of \(I_{D S S}=7.5 \mathrm{~mA}\) and \(V_{P}=4 \mathrm{~V}\). Sketch the transfer characteristics.

Short Answer

Expert verified
To sketch the transfer characteristics for the given p-channel JFET, utilise the given parameters and the relationship between the drain current and gate-source voltage. As this is a p-channel JFET, the graph would be in the third quadrant with the curve starting from \(I_D = I_{DSS}\) at \(V_{GS} = 0\) and ending at \(I_D = 0\) at \(V_{GS} = V_P\). Remember to calculate individual points for varying \(V_{GS}\) and plot them correlatively.

Step by step solution

01

Determine the Relationship

In this step, we'll use the relationship for the drain current in a p-channel JFET which is given as \(I_D = I_{DSS} (1 - V_{GS}/V_{P})^2\). The parameters given are \(I_{DSS}\) as 7.5 mA and \(V_{P}\) as 4 V.
02

Construct the Transfer Characteristics

The relationship we found above is applied to create the transfer characteristic graph between \(I_D\) and \(V_{GS}\). For this, vary the gate-source voltage (\(V_{GS}\)) from 0 to the value of \(V_P\) which is 4V. For each \(V_{GS}\) compute the corresponding \(I_D\) using the relationship. Plot these points on the graph.
03

Sketch the Curve

Now, these points are connected to give the transfer characteristic. Note that the curve starts from \(I_D = I_{DSS}\) at \(V_{GS} = 0\) and reaches \(I_D = 0\) at \(V_{GS} = V_P\). Remember that as this is a p-channel JFET, the curve should be in third quadrant of the graph because both \(I_D\) and \(V_{GS}\) will be negative.

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