1 / 26

VCE Physics

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.

amaya-nguyen
Download Presentation

VCE Physics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. VCE Physics Unit 3 Electronics & Photonics Revision Questions

  2. Electronics & Photonics RevisionQuestion Type: Ohm’s Law 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

  3. Electronics & Photonics RevisionQuestion Type: Potential Difference 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

  4. Electronics & Photonics RevisionQuestion Type: Electrical Energy Q: Determine the electrical energy dissipated in the 100 Ω resistor of Figure 1 in 1 second. In your answer provide the unit. A: Electrical energy W = VQ = VIt = (4.0)(40 x 10-3)(1) = 0.16 Joule

  5. A B 20 VRMS C VOUT D E Electronics & Photonics RevisionQuestion Type: Voltage Divider Network Q: What is the RMS output voltage, VOUT? 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. 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. You can now redraw the circuit.

  6. 50 Ω A B 100 Ω 20 VRMS C A B 50 Ω VOUT 50 Ω 20 VRMS D E (R1) VOUT Figure 3 C 150 Ω (R2) VIN R2 D E 50 Ω VOUT = (R1 + R2) 20(150) Figure 4 = (50 + 150) Electronics & Photonics RevisionQuestion Type: Voltage Divider Network The series circuit in Figure 3 can be further simplified as shown in Figure 4 The original question (value of VOUT) can now be answered, using the Voltage Divider formula: = 15 V

  7. A B 20 VRMS C VOUT D E Electronics & Photonics RevisionQuestion Type: Current Flows 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. A: Alternative D

  8. Electronics & Photonics RevisionQuestion Type: Ohm’s Law 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

  9. VOUT 0.5 Voltage Gain = = (10 x 10-3) VIN Electronics & Photonics RevisionQuestion Type: Transistor Characteristics Q: Calculate the change in collector voltage, and determine the small-signal voltage gain of the amplifier circuit. Show your working. 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. 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. 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 = 50

  10. Electronics & Photonics RevisionQuestion Type: Transducer Properties 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.

  11. 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 a photodiode 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.

  12. Electronics & Photonics RevisionQuestion Type: Ohm’s Law 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. 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? 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

  13. Electronics & Photonics RevisionQuestion Type: Linear and Non Linear Devices 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(mA) I(Amp) V(volts) V(volts) a non linear device – an LED a linear device – a resistor

  14. R2 VIN R1 VOUT 6 5000 30 VIN VOUT = = A: (5000 + R2) (R1 + R2) R1 Electronics & Photonics RevisionQuestion Type: Voltage Divider Network 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. 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. R2 = 20,000 Ω = 20kΩ

  15. R2 VIN R1 VOUT R2 VIN R1 VOUT Electronics & Photonics RevisionQuestion Type: Parallel Resistors 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. A: Connect two 10k resistors together in parallel

  16. 15k R2 VIN 100 mV 5k R1 VOUT 100 x 10-3 20 x 103 Electronics & Photonics RevisionQuestion Type: Ohm’s Law 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. A: Series circuit – add resistances so RT = 20kΩ Use Ohm’s Law to find current V = IR so I = = 5μA

  17. Electronics & Photonics RevisionQuestion Type: Capacitor Properties 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 resistor At high frequency the capacitor acts like a short circuit (zero resistance) allowing most current to flow through it and very little through the resistor.

  18. Vcc = +20V ic R2 RC = 2kΩ C CIN iB B E output input CE RE R1 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. 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).

  19. Vcc = +20V ic R2 RC = 2kΩ C CIN iB B E output input CE RE R1 Electronics & Photonics RevisionQuestion Type: Transistor Characteristics 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

  20. Electronics & Photonics RevisionQuestion Type: Sketch Graphs 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

  21. Electronics & Photonics RevisionQuestion Type: Ohm’s Law 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

  22. Electronics & Photonics RevisionQuestion Type: Ohm’s Law 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. A: Increased output 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

  23. 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 converted A. 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

  24. Electronics & Photonics RevisionQuestion Type: Ohm’s Law 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

  25. 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.

  26. Electronics & Photonics RevisionQuestion Type: Sketch Graphs 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. A: As shown the slope on the fall and rise represents the 8 μs delay in the response of the phototransistor

More Related