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VCE Physics Unit 3 Electronics & Photonics Revision Questions

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VCE Physics. Unit 3 Electronics & Photonics Revision Questions. Electronics & Photonics Revision Question Type:. Ohm’s Law. Figure 1 shows a resistor, a linear circuit component, with resistance R = 100 Ω. - PowerPoint PPT Presentation

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VCE Physics

Unit 3

Electronics & Photonics

Revision Questions

Electronics & Photonics RevisionQuestion Type:

Figure 1 shows a resistor, a linear circuit component, with resistance R = 100 Ω.A DC current, I = 40 mA, passes through this resistor in the direction shown by the arrows.

Q: What is the voltage drop across this resistor? Express your answer in volts. A: V = IR

= (40 x 10-3)(100) = 4.0 V

Ohm’s Law

3

Electronics & Photonics RevisionQuestion Type:

Q: Which one of the following statements (A to D) concerning the voltage across the resistor in Figure 1 is true?

A. The potential at point A is higher than at point B.B. The potential at point A is the same as at point B.C. The potential at point A is lower than at point B.D. The potential at point A varies in sign with time compared to that at point B.

A: Alternative A

Potential Difference

Electronics & Photonics RevisionQuestion Type:

Q: Determine the electrical energy dissipated in the 100 Ω resistor of Figure 1 in 1 second. In your answer provide the unit.

Electrical Energy

A: Electrical energy W = VQ = VIt = (4.0)(40 x 10-3)(1)

= 0.16 Joule

Electronics & Photonics RevisionQuestion Type:

20 VRMS

VOUT

A B

C

D E

In Figure 2, five identical 100 Ω resistors are used to construct a voltage divider. The voltage source across this voltage divider is an AC supply with an RMS voltage of 20 V. The resistors are labelled by the letters A to E as shown.

Q: What is the RMS output voltage, VOUT?

Voltage Divider Network

A: Step 1 Determine the equivalent resistance for A and B and D and E

For A and B: 1/RE = 1/RA + 1/RB = 1/100 + 1/100 = 2/100 So RE = 100/2 = 50 Ω

Replace A and B with one 50 Ω resistor, same with D and E.

You can now redraw the circuit.

Electronics & Photonics RevisionQuestion Type:

The series circuit in Figure 3 can be further simplified as shown in Figure 4

20 VRMS

VOUT

A B

C

D E

50 Ω

50 Ω

100 Ω

Figure 3

20 VRMSVOUT

A B

C

D E

50 Ω

50 Ω

150 Ω

Figure 4

(R1)

(R2)

VOUT =VIN R2

(R1 + R2)

The original question (value of VOUT) can now be answered, using the Voltage Divider formula:

= 20(150)

(50 + 150) = 15 V

Voltage Divider Network

Electronics & Photonics RevisionQuestion Type:

Q: Which one of the following statements (A to D) concerning the RMS currents in the circuit of Figure 2 is true?A. The current in resistor A is identical to the current in resistor C.B. The current in resistor D is twice the current in resistor C.C. The current in resistor B is twice the current in resistor E.D. The current in resistor A is identical to the current in resistor D.

20 VRMS

VOUT

A B

C

D E

A: Alternative D

Current Flows

Electronics & Photonics RevisionQuestion Type:Figure 3 shows a typical n-p-n transistor voltage amplifier circuit. You may assume that the transistor is correctly biased and the circuit is operating in the linear-amplification region. The DC collector current is 20 mA.

Q: Show that the voltage between point C and Earth is 20 V.

A: A Collector current of 20 mA means RC has 20 mA flowing through it.V across Rc = IR = (20 x 10-3)(500) = 10 V

So, VC = Vcc – VRc

= 30 – 10 = 20 V

Ohm’s Law

Electronics & Photonics RevisionQuestion Type:

The AC current gain of the transistor is 200.A small time-varying AC voltage, VIN, is applied at the input. As VIN rises from 0 to 10 mV this causes the base current to increase from 0 to 5 µA.

Q: Calculate the change in collector voltage, and determine the small-signal voltage gain of the amplifier circuit. Show your working.

IC = βIB = (200)(5 x 10-6) = 0.001 A

Thus the Voltage drop across RC is VC = IRC = (0.001)(500) = 0.5 V

Voltage Gain =VOUT

VIN

A: The rise in base current switches the transistor “on” causing a current to flow in the collector circuit. The base and collector currents are related through the current gain.

= 0.5

(10 x 10-3)= 50

Transistor Characteristics

Electronics & Photonics RevisionQuestion Type:

Q: Describe the basic purpose of each of the following electronic transducers.i. Light-Emitting Diode (LED)ii. Photodiode

A: (i) Emits visible light when a current flows through it. Converts electrical energy to light.

(ii) Switches on (allows a current to flow through it) when exposed to light. Converts light to electrical energy.

Transducer Properties

Electronics & Photonics RevisionQuestion Type:

The information on an audio CD is represented by a series of pits (small depressions) in the surface that are scanned by laser light. When there is no pit the reflected light gives a maximum light intensity, I1, detected by aphotodiode circuit. When the laser light strikes a pit, the light intensity is reduced to I0. A plot of a typical light intensity incident on the photodiode is shown in Figure 4.

Electronics & Photonics RevisionQuestion Type:

Q: With no light incident upon the photodiode, the current in the photodiode circuit, the “dark current”, is 5 µA.What is the output voltage, VOUT, across the 100 Ω resistor in the circuit of Figure 5b?

Ohm’s Law

A: The photodiode and resistor are in series.The same current flows through each. VOUT = V100Ω = IR = (5 x 10-6)(100)

= 5 x 10-4 V

The variation in current as a function of light intensity for the photodiode is shown in Figure 5a, together with the circuit used to determine this, which is shown in Figure 5b.

Electronics & Photonics RevisionQuestion Type:

A resistor is a linear device. An example of a non-linear device is a light-emitting diode (LED).

Q: On the axes provided, sketch a typical current-voltage characteristic curve for each of the devices mentioned.In both cases label the axes and indicate appropriate units.

I(Amp)

V(volts)

a linear device – a resistor a non linear device – an LED

V(volts)

I(mA)

Linear and Non Linear Devices

Electronics & Photonics RevisionQuestion Type:

An essential component in some of the practical circuits covered in this exam paper is the voltage divider. A DC voltage divider circuit is shown in Figure 1.

VIN

VOUT

R2

R1

For the circuit of Figure 1, VIN = 30 V, R1 = 5 kΩ and the output voltage VOUT = 6 V.

Q: What is the value of the resistance R2? Show your working.

Voltage Divider Network

A: VOUTVIN

(R1 + R2)R1

= 65000

=30

(5000 + R2)R2 = 20,000 Ω = 20kΩ

Electronics & Photonics RevisionQuestion Type:

You wire up the circuit shown in Figure 1 but only have 10 kΩ resistors to work with. Q: Explain how you would construct the R1 = 5 kΩ resistor using only 10 kΩ resistors. Include a sketch to show the connections between the appropriate number of 10 kΩ resistors.

VIN

VOUT

R2

R1

A: Connect two 10k resistors together in parallel

VIN

VOUT

R2

R1

Parallel Resistors

Electronics & Photonics RevisionQuestion Type:

In Figure 1 the 30 V DC input to the voltage divider is replaced by a 100 mV (peak-to-peak) sinusoidal AC input voltage. The resistance values are now R1 = 5 kΩ and R2 = 15 kΩ.

Q: What is the current through resistor R2? Show your working, and express your answer as a peak-to-peak current in μA.

VIN

VOUT

R2

R1100 mV

5k

15k

A: Series circuit – add resistances so RT = 20kΩ

Use Ohm’s Law to find current V = IR so I = 100 x 10-3

20 x 103= 5μA

Ohm’s Law

Electronics & Photonics RevisionQuestion Type:

Figure 1 is modified so that R1 has a capacitor, C, placed in parallel with it. The new circuit is shown in Figure 2, and can now have an AC or DC input voltage, VIN.

Q: It is observed that if VIN is a DC voltage, then in the R1–C parallel combination, a DC current passes through R1, but not through C. However, if VIN is a high frequency AC voltage, then an AC current passes mainly through C with very little through R1. Explain this observation.

A: At low frequency the Capacitor acts like an open circuit (infinite resistance) forcing all current to flow through the resistorAt high frequency the capacitor acts like a short circuit (zero resistance) allowing most current to flow through it and very little through the resistor.

Capacitor Properties

Electronics & Photonics RevisionQuestion Type:

These ideas are incorporated into the design of an n-p-n transistor amplifier circuit, shown in Figure 3, to amplify the small, high frequency AC voltage from a microphone. You may assume that the transistor is correctly biased.The collector and base currents in Figure 3 are denoted respectively using lower case iC and iB to emphasise their AC time variation and small magnitude.

Vcc = +20V

ic

RC = 2kΩ

R1

R2

iB

RE CE

E

C

B

input

CIN

output

In this circuit the AC collector current, iC, and the AC base current, iB, are related by iC = 100 iB.The 10 mV (peak-to-peak) input voltage from the microphone gives rise to a time-varying base current of iB = 10 μA (peak-to-peak).

Electronics & Photonics RevisionQuestion Type:

Vcc = +20V

ic

RC = 2kΩ

R1

R2

iB

RE CE

E

C

B

input

CIN

output

Q: Calculate the AC voltage across the collector resistor, RC, and show that the magnitude of the voltage gain of this transistor amplifier circuit is 200.

A: iC = 100iB and iB = 10 μA So iC = (100)(10 x 10-6) = 0.001 A

VRc = iCRC

= (0.001)(2000) = 2.0 V

Gain = VOUT/VIN = 2.0/(10 x10-3) = 200

Transistor Characteristics

Electronics & Photonics RevisionQuestion Type:

Q: Which one of the following cathode ray oscilloscope (CRO) traces of the output voltage (A to D) correctly identifies the distortion arising from transistor cut-off for the circuit of Figure 3? The input to this transistor amplifier circuit is a sinusoidal voltage.

A: Alternative B

Sketch Graphs

Electronics & Photonics RevisionQuestion Type:

You are asked to investigate the properties of an optical coupler, sometimes called an opto-isolator. This comprises a light-emitting diode (LED) that converts an electrical signal into light output, and a phototransistor (PT) that converts incident light into an electrical output. Before using an opto-isolator chip you consider typical LED and PT circuits separately.A simple LED circuit is shown in Figure 4 along with the LED current-voltage characteristics. The light output increases as the forward current, IF , through the LED increases.

Q: Using the information in Figure 4, what is the value of the resistance, RD, in series with the LED that will ensure the forward current through the LED is IF = 10 mA?

A: For 10 mA to flow through LED requires a voltage of 1.5 V (read from graph)Because LED and R are in series VRD = 10 – 1.5 = 8.5 V

VRD = IRD so RD = VRD/I = 8.5/(10 x 10-3) = 850 Ω

Ohm’s Law

Electronics & Photonics RevisionQuestion Type:

Q: Will the light output of the LED increase or decrease if the value of RD is a little lower than the value you have calculated in the last question? Justify your answer.

Justification: Reducing the value of RD will not affect the voltage drop across it.The Voltage across RD is controlled by the LED which will remain at 1.5 V thus VRD will still equal 8.5 V. So if V remains the same and R goes

down I must go up. So a larger current flows through the LED meaning an increased light output

A: Increased output

Ohm’s Law

Electronics & Photonics RevisionQuestion Type:

The LED in Figure 4 is an electro-optical converter. Which one of the following statements (A to D) regarding energy conversion for the LED is correct?

All the electrical energy supplied from the DC power supply is convertedA. only to heat energy in both the resistor, RD, and the LED.B. partly to heat energy in the resistor, RD, the remainder to light-energy output from the LED.C. partly to heat energy in both the resistor, RD, and the LED, with the remainder to light-energy output from the LED.D. to heat energy in the LED, with the remainder to light-energy output from the LED.

A: Alternative C

Electronics & Photonics RevisionQuestion Type:

You now consider the phototransistor (PT) circuit of Figure 5 with RC = 2.2 kΩ. The light is incident upon the base region of the PT and produces a collector current, IC.

Q: As the light intensity incident on the PT increases, which one of the following statements concerning the PT-circuit of Figure 5 is correct?A. The collector current remains constant, but VOUT increases.B. The collector current remains constant, but VOUT decreases.C. The collector current increases, but VOUT decreases.D. The collector current decreases and VOUT decreases.

A: Alternative C

Ohm’s Law

Electronics & Photonics RevisionQuestion Type:

Rather than use a separate LED and PT, you choose an opto-isolator chip (the region within the dotted lines) shown in the circuit of Figure 6. The opto-isolator is a linear device, that means that the PT collector current, IC, and the LED forward current, IF , are related by IC = const × IF.In addition you are also told that the opto-isolator has a switching time of 8 μs. This represents the time it takes for the PT collector current, IC, to respond to a sudden change in light output from the LED.

Electronics & Photonics RevisionQuestion Type:

Q: Figure 7a shows the LED forward current, IF , as a function of time. On Figure 7b sketch the opto-isolator output collector current, IC, as a function of time, given the initial condition (at t = 0) IC = 4 mA when IF = 20 mA.

Sketch Graphs

A: As shown the slope on the fall and rise represents the 8 μs delay in the response of the phototransistor