1 / 32

Microelectronics 2

Electrical Engineering 2. Lecture 19. Microelectronics 2. Dr. Peter Ewen. (Room G08, SMC; email - pjse). I E. I C. e c. Fig. 117: Output Characteristics – CB Configuration. V BE. b. I C (V CB ). V CB. I B. I C / mA. Active region I C ≈ α I E , α ≈ 1. I E = 1.0 mA. 1

cyrus-rosa
Download Presentation

Microelectronics 2

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. Electrical Engineering 2 Lecture 19 Microelectronics 2 Dr. Peter Ewen (Room G08, SMC; email - pjse)

  2. IE IC e c Fig. 117: Output Characteristics – CB Configuration VBE b IC(VCB) VCB IB IC / mA Active region IC≈ αIE, α≈ 1 IE = 1.0 mA 1 0.75 0.5 0.25 IE = 0.75 mA Breakdown region IE = 0.50 mA Saturation region IE = 0.25 mA ICBO Cutoff region IE = 0 -1 0 1 2 3 4 5 6 7 8 VCB / V

  3. Fig. 118: Input Characteristic – CE Configuration IB / A VCE = 20V VCE = 5V 10 8 6 4 2 0 Increasing VCE 0 0.2 0.4 0.6 0.8 VBE / V

  4. IC Fig. 119: Output Characteristics – CE Configuration c IB b VCE e VBE IE IC / mA Active region IB = 40 A 4 3 2 1 IB = 30 A Breakdown region Saturation region IB = 20 A IB = 10A ICEO Cutoff region IB = 0 0 5 10 15 20 VCE / V

  5. CCCE IE(VCE)  IC(VCE) Fig. 121: Output Characteristics – CC Configuration IE ≈ IC since α ≈ 1 IE / mA Active region IB = 40 A 4 3 2 1 IB = 30 A Breakdown region Saturation region IB = 20 A IB = 10A ICEO Cutoff region IB = 0 0 5 10 15 20 VCE / V

  6. IE Fig. 120: Input Characteristics – CC Configuration 0.7V e IB b VCE c VCB IB vs. VCB for different values of VCE IC IB / A Transistor on  VBE ≈ 0.7V VCE = 15V VCE = 10V VCE = 5V VCB≈ VCE – 0.7V 80 60 40 20 As VCB VCE, VBE  0, transistor turns off 4.3 0 5 10 15 20 VCB / V

  7. THE EARLY EFFECT, OR BASE-WIDTH MODULATION depletion regions base emitter collector effective width of base n p n e-b junction c-b junction VCB=1V VCB=2V VCB=3V VCB=4V VCB=6V VCB=5V IB VCB VBE + +

  8. If the effective width of the base decreases: – 1. There will be less recombination in the base, so α (and hence β) will increase. • 2. The minority carrier concentration gradient (Δn/Δx) will increase: • so |IE| will increase. |IE|  Δn/Δx (Δx is the basewidth) 3. The c-b depletion region may extend all the way over to the e-b junction – PUNCH-THROUGH

  9. LECTURE 19  Heat transfer in semiconductor devices • Thermal resistance • Derating

  10. 10. Bandgap voltage reference (a) Show that for the circuit below, I2 increases as temperature, T, increases. (b) Given that VBE for a BJT decreases as temperature increases, design a circuit, based on that in (a), which will produce an output voltage that is independent of temperature. Ip I2 (constant) Q1 Q2 VBE2 R1 VBE1

  11. Ip I2 (constant) Q1 Q2 VBE2 R1 VBE1 I2 I2 = Ipex x1 x2 x (a) If the transistors are matched then IS1 = IS2, so dividing the second equation by the first and re-arranging for I2: Since VBE1 must be greater than VBE2 (as part of VBE1 is dropped across R1), I2 must increase as T increases (as T increases, the magnitude of the negative quantity in square brackets decreases).

  12. (b) Given that VBE for a BJT decreases as temperature increases, design a circuit, based on that in (a), which will produce an output voltage that is independent of temperature. ● Add resistor R2 to convert I2 to a voltage. ● Add transistor Q3 to provide a voltage with a negative temperature coefficient (VBE3) Vo = VBE3 + I2R2 Vo R3 R2 ● As T↑, VBE3↓ but from part (a) I2↑ (and hence I2R2↑). By choosing R2 appropriately, Vo will be independent of T. Ip I2 (constant) Q3 VBE3 Q1 Q2 VBE2 ● Note that the constant current, Ip, is provided by R3 and the constant output voltage Vo. R1 VBE1

  13. VCC ● Add power supply Vo ● Vo ≈ 1.2V (1.2eV is the bandgap energy for Si at 0K, hence the circuit is known as a “bandgap voltage reference” – it provides a low value reference voltage that is independent of temperature.) R3 R2 Ip I2 Q3 (constant) VBE3 Q1 Q2 VBE2 R1 VBE1

  14. VCC IC Power, P, dissipated in a BJT is: P ≈ VCE x IC VCE Vo Vi In power devices temperatures can be ~ 150° C α increases as T increases β increases as T increases VBE decreases by 2mV/°C ICBO doubles for every 10°C rise in temperature. ICEO = (1+β)ICBO

  15. moving medium, e.g. air HEAT TRANSFER heat source conduction radiation convection Fig. 123

  16. Fig. 124: Heat sinks

  17. THERMAL RESISTANCE Heat Flow  Charge Flow THERMAL RESISTANCE – a measure of the opposition to the flow of heat. Thermal Resistance, θ (°CW-1)  Electrical Resistance Temperature Difference, ΔT  Voltage Difference Rate of Heat Flow, P  Current cf. Ohm’s Law:

  18. Fig. 125: Thermal resistances between the device and ambient atmosphere

  19. “Thermal” circuit diagram for transistor on heat sink.

  20. 11. Thermal resistance (Bogart, 4th Ed., Example 16-2, p.733) • The collector-base junction of a BJT dissipates 2W. The thermal resistance from junction to case is 8°C/W and that from case to air is 20°C/W. The ambient temperature is 25°C. • What is the junction temperature? • What is the case temperature?

  21. TJ (junction) TC (case) TA (ambient) 25C 2W θJC (8°C/W) θCA (20°C/W) • The collector-base junction of a BJT dissipates 2W. The thermal resistance from junction to case (θJC) is 8°C/W and that from case to air (θCA) is 20°C/W. The ambient temperature is 25°C. • What is the junction temperature? Solution: Start by drawing the “thermal circuit diagram” θT = θJC + θCA = 8 + 20 = 28°C/W TJ – 25oC 2W 28°C/W TJ – TA = θTP TJ = TA + θTP = 25 + 28 x 2 = 81°C

  22. TJ (junction) TC (case) TA (ambient) 25C 2W θJC (8°C/W) θCA (20°C/W) The collector-base junction of a BJT dissipates 2W. The thermal resistance from junction to case (θJC) is 8°C/W and that from case to air (θCA) is 20°C/W. The ambient temperature is 25°C. (b) What is the case temperature? Solution: TJ – TC = θJCP TC = TJ - θJCP = 81 - 8 x 2 = 65°C

  23. DERATING

  24. DERATING 25 15 5 0 Gradient of line gives the DERATING FACTOR, D (in W/°C): D = 25/150 = 0.17W/°C Pd , Maximum allowed power dissipation / W 0 25 75 125 175 TC, case temperature / °C derating factor = [Pd(T) - Pd(T+T)] / T

  25. 12. Derating (Bogart, 4th Ed., p.735) A device has a maximum rated dissipation, Pd, of 20W at 25°C ambient and a derating factor of 100mW/°C for temperatures above this. What is the maximum permissible dissipation at 100°C? Solution: derating factor = [Pd(T) - Pd(T+T)] / T Pd(T+T) = Pd(T) - T x derating factor Pd(100) = Pd(25) – (100 – 25) x 100 x 10-3 = 20 – 75 x 0.1 = 12.5W

  26. 13. Thermal resistance (Bogart, 4th Ed., Example 16-4, p.735) • A semiconductor device has Pd(max) = 5W at 50°C (case). The thermal resistance from case to ambient is 5°C/W. • If the ambient temperature is 40°C, can the device be operated at maximum power without auxiliary cooling (heat sink or fan)? • If not, what is the maximum permissible power dissipation, with no auxiliary cooling, at 40°C ambient. • (c) What is the derating factor (for ambient temperature) for the device?

  27. A semiconductor device has Pd(max) = 5W at50°C (case). The thermal resistance from case to ambient (θCA) is 5°C/W. • If the ambient temperature is 40°C, can the device be operated at maximum power without auxiliary cooling (heat sink or fan)? TC 5W θCA TA Solution: We need to find the case temperature, TC , and check that it is not greater than 50°C. TC – TA = θCAP TC – 40 = 5 x 5 TC= 65C Hence device cannot be operated without auxiliary cooling.

  28. A semiconductor device has Pd(max) = 5W at 50°C(case). The thermal resistance from case to ambient (θCA) is 5°C/W. (b) If not, what is the maximum permissible power dissipation, with no auxiliary cooling, at 40°C ambient. TC P θCA TA Solution: TC – TA 50 - 40 = 2W = = 5 θCA

  29. 5 2 Pd , Maximum allowed power dissipation / W 0 25 30 35 40 TA, ambient temperature / °C A semiconductor device has Pd(max) = 5W at 50°C(case). The thermal resistance from case to ambient (θCA) is 5°C/W. (c) What is the derating factor (for ambient temperature) for the device? Solution: Find ambient temperature for 5W dissipation at 50C (case): TC – TA = θCA  5 = (50 – TA)/5  TA = 25C From part (b) Gradient of line gives the DERATING FACTOR, D D = (5 - 2)/(40 - 25): = 0.2W/C

  30. HEAT TRANSFER IN SEMICONDUCTOR DEVICES • Because of power dissipation the temperature inside a device can be much higher than ambient, hence the heat generated must be removed. • There are three mechanisms of heat transfer: • CONDUCTION – energy transferred between atoms • RADIATION – heat transferred by IR radiation • CONVECTION – heat transferred to a moving medium • Heat sinks conduct heat away from device package and get rid of it by radiation and convection.

  31. Heat flow is analogous to flow of electrical charge: • P – rate of heat flow (= power) • ΔT – temperature difference • θ – thermal resistance • DERATING FACTOR indicates how the maximum allowable power dissipation decreases with increasing case or ambient temperature.

  32. Next week both lecture slots will be run as examples classes looking at previous exam questions. Dec 2010

More Related