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3.1 Name TWO types of junction field effect transistors (JFETs) - NSC Electrical Technology Electronics - Question 3 - 2018 - Paper 1

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3.1 Name TWO types of junction field effect transistors (JFETs). 3.2 Explain how the construction of the JFET was modified to overcome the leakage current between t... show full transcript

Worked Solution & Example Answer:3.1 Name TWO types of junction field effect transistors (JFETs) - NSC Electrical Technology Electronics - Question 3 - 2018 - Paper 1

Step 1

3.1 Name TWO types of junction field effect transistors (JFETs).

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Answer

Two types of junction field effect transistors (JFETs):

  1. N-channel JFET (NFET)
  2. P-channel JFET (PFET)

Step 2

3.2 Explain how the construction of the JFET was modified to overcome the leakage current between the gate terminal and drain-source channel.

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Answer

In order to overcome leakage currents in a JFET, the gate terminal is electrically insulated from the drain-source channel. This is achieved by using an extremely narrow layer of metal-oxide-silicon (MOS) which effectively reduces the leakage path between the gate and the channel, thus enhancing the device's performance.

Step 3

3.3.1 Identify the type of MOSFET used in this circuit.

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Answer

The type of MOSFET used in this circuit is an Enhancement-mode N-channel MOSFET.

Step 4

3.3.2 Explain when the lamp in this circuit will switch ON.

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Answer

The lamp will switch ON as soon as the gate voltage VGSV_{GS} is raised to a sufficient level. This voltage enables current to flow through the internal channel of the MOSFET, thereby activating the lamp.

Step 5

3.3.3 Describe what will happen if Rgs is short-circuited.

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Answer

If RGSR_{GS} is short-circuited, the internal conductive channel of the MOSFET will disperse, cutting the current flow and switching the lamp OFF. Additionally, this could lead to the gate of the MOSFET being tied directly to ground (0 V), which might further affect its operation.

Step 6

3.4.1 Identify region E.

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Answer

Region E is identified as the cut off region in the UJT characteristic curve.

Step 7

3.4.2 Explain what happens in the UJT between points C and D of the characteristic curve.

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Answer

Between points C and D, the UJT transitions from a non-conducting state to a conducting state. As the voltage approaches point C, the device will start to turn ON, and current flow will increase substantially, resulting in the characteristic negative resistance observed in this segment of the curve.

Step 8

3.5.1 Identify the configuration in which the transistors are connected.

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Answer

The transistors are connected in a Darlington pair configuration.

Step 9

3.6.1 Label pin 3.

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Answer

Pin 3 is the Inverting input of the 741 operational amplifier.

Step 10

3.6.2 Name the type of package in which the integrated circuit above is constructed.

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Answer

The integrated circuit is typically constructed in a Dual-in-line package outline (DIP) or Surface mount package (SMP).

Step 11

3.6.3 Draw the output signal on ANSWER SHEET 3.6.3 if the signals in FIGURE 3.6 are applied to the inputs of an op amp.

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Answer

The output signal from the op amp will have a typical form, such as a sine wave, depending on the input signals and the characteristics of the op amp used. The output can vary in amplitude and phase relative to the input.

Step 12

3.7 Explain the difference between open-loop gain and closed-loop gain with reference to op amps.

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Answer

Open-loop gain is the gain of an op-amp without any feedback from the output to the input, meaning the op-amp amplifies the input voltage to its maximum level. Closed-loop gain, on the other hand, is the gain of an op-amp that includes feedback, effectively stabilizing the output and reducing the overall gain of the circuit.

Step 13

3.8 Calculate the output voltage of the op amp in FIGURE 3.8 below.

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Answer

Given that:

  • VIN=5V_{IN} = 5 mV
  • RF=220extkriangleextR_{F} = 220 ext{ k} riangle ext{ }
  • RIN=440extriangleR_{IN} = 440 ext{ } riangle

The output voltage can be calculated using the formula: VOUT=VIN(1+RFRIN)V_{OUT} = V_{IN} \left(1 + \frac{R_{F}}{R_{IN}}\right) Substituting the values: VOUT=5×103(1+220000440)V_{OUT} = 5 \times 10^{-3} \left(1 + \frac{220000}{440}\right) =5×103(1+500)= 5 \times 10^{-3} \left(1 + 500\right) =5×103×501=2.51extV= 5 \times 10^{-3} \times 501 = 2.51 ext{ V}

Step 14

3.9 Refer to FIGURE 3.9 below and explain the operation of the 555 timer when the circuit is running.

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Answer

The 555 timer operates in either astable or monostable modes. When in astable mode, it continually oscillates between high and low states, creating a square wave output. In monostable mode, when triggered, it produces a single output pulse of a specific duration. The timing for both modes is determined by external resistors and capacitors.

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