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Question

When VDS is the drain voltage and VDS(max) is the maximum drain voltage, the JFET will breakdown if:

The correct answer is

VDS > VDS(max)

Understanding JFET Breakdown Voltage

A Junction Field-Effect Transistor (JFET), like other semiconductor devices, has limits on the voltages it can withstand. One important limit is the maximum drain voltage, often denoted as \(V_{DS(max)}\).

The drain voltage, \(V_{DS}\), is the voltage difference between the drain and source terminals of the JFET. As \(V_{DS}\) is increased, the current through the JFET changes depending on the operating region (ohmic, saturation). However, if \(V_{DS}\) becomes too high, a phenomenon called breakdown occurs.

Breakdown happens when the electric field across a region within the semiconductor device becomes so strong that it causes a rapid increase in current, often due to avalanche multiplication of charge carriers. This sudden increase in current can lead to permanent damage to the JFET if the power dissipation is not limited.

The maximum drain voltage, \(V_{DS(max)}\), is specified by the manufacturer as the voltage limit that the device can handle without entering the breakdown region under normal operating conditions. If the applied drain voltage \(V_{DS}\) exceeds this maximum limit, the device will enter breakdown.

Analyzing the Condition for JFET Breakdown

Let's look at the given options in relation to the concept of breakdown voltage:

  • Option 1: \(V_{DS} < V_{DS(max)}\) - This means the drain voltage is less than the maximum rated voltage. In this condition, the JFET is operating within its safe voltage limits and will not experience breakdown due to excessive \(V_{DS}\).
  • Option 2: \(V_{DS} = V_{DS(max)}\) - This means the drain voltage is equal to the maximum rated voltage. The device is at the edge of its voltage limit, but breakdown typically occurs when the voltage *exceeds* the limit.
  • Option 3: \(V_{DS} > V_{DS(max)}\) - This means the drain voltage is greater than the maximum rated voltage. When the applied voltage surpasses the device's voltage rating (\(V_{DS(max)}\)), the electric field becomes high enough to cause breakdown.
  • Option 4: \(V_{DS} = 0.25 V_{DS(max)}\) - This means the drain voltage is only a quarter of the maximum rated voltage. The JFET is operating well within its safe voltage limits.

Therefore, the JFET will enter breakdown when the drain voltage \(V_{DS}\) exceeds the maximum drain voltage \(V_{DS(max)}\).

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Important Questions from Field Effect Transistors

  1. FET is like a switched on condition when it operates in ______ mode.

  2. Which of the following is the characteristic of Field-effect transistor?

  3. Which of the following is true about the transconductance of a MOSFET in saturation (I Dis the Drain Current)?

  4. Transconductance in an FET indicates how effectively the input voltage controls the

  5. A FET has

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