Electronics I

1. Electronics I Tutorial 2 Tutorial 2

2. Question 1 For the circuit shown below: • If Vs = 0.8sin(ωt), VBB = 2V, RB = 2K, and Vcc = 15V, determine the maximum value of Rc that ensures that the transistor is not saturated. • If Vs = 0.5sin(ωt), VBB = 2V, RB = 10K, and Rc = 1K, determine the minimum value of Vcc that ensures that the transistor is not saturated. • If Vs = 0.5sin(ωt), RB = 5K, Vcc = 20V and Rc = 1K, determine the range of VBB that ensures that the transistor operates in the active region. Tutorial 2

3. Question 1 (Solution 1a) • If Vs = 0.8sin(ωt), VBB = 2V, RB = 2K, and Vcc = 15V, determine the maximum value of Rc that ensures that the transistor is not saturated. • At Saturation VCE = 0 • Vcc = IcxRc + VCE = Ic(sat) x Rc  Ic(sat) = Vcc/Rc Also: Ic(sat) = β x IB(sat) Worse case when IB is maximum, or when Vs is maximum IB = (VBB + Vs – VBE)/RB IB(max) = (2 + 0.8 – 0.7)/2K = 2.1/2K = 1.05mA IC(max) = β x IB(max) = 100 x 1.05 = 105mA  Rc(min) = Vcc/Ic(max) = 15V/105mA = 0.143K Tutorial 2

4. Question 1 (Solution 1b) • If Vs = 0.5sin(ωt), VBB = 2V, RB = 10K, and Rc = 1K, determine the minimum value of Vcc that ensures that the transistor is not saturated. • At Saturation VCE = 0 • Vcc = IcxRc + VCE = Ic(sat) x Rc  Ic(sat) = Vcc/Rc Also: Ic(sat) = β x IB(sat) Worse case when IB is maximum, or when Vs is maximum IB = (VBB + Vs – VBE)/RB IB(max) = (2 + 0.5 – 0.7)/10K = 1.8/10K = 0.18mA IC(max) = β x IB(max) = 100 x 0.18 = 18mA  Vcc(min) = Ic(max) x Rc = 18mA x 1K = 18V Tutorial 2

5. Question 1 (Solution 1c) • If Vs = 0.5sin(ωt), RB = 5K, Vcc = 20V and Rc = 1K, determine the range of VBB that ensures that the transistor operates in the active region. • We must ensure that when VBB is max the transistor is not saturated.  Ic(sat) = Vcc/Rc = β x IB(sat)  IB(sat) = Vcc/(β x Rc) = 20/(100 x 1) =0.2mA VBB(max) = IB(sat) x RB + 0.7 – Vs(max) = 0.2 x 5 + 0.7 -0.5 = 1.2V (b) To ensure that the transistor is not cut-off, VBE must not drop below 0.7V. At the limit point IB = 0. • VBB(min) = IB x RB + 0.7 – Vs(min) • = 0 + 0.7 –(-0.5) = 1.2V • If VBB is not exactly 1.2V then the transistor will either be saturated or cut-off. With VBB = 1.2V and Vs = 0.5sin(ωt) the transistor will operate between cut-off and saturation Tutorial 2

6. Question 2 For the circuit shown below assume that Vs = 0.3sin(ωt), VBB = 1.5V, RB = 10K, Rc = 1K, and Vcc = 15V. • Determine the Q point (IB(Q), IC(Q) and VCE(Q)) as well as the saturation collector current IC(sat) and the saturation base current (IB(sat)), and draw the DC load line. • Draw the waveform of the input (Vs) and the output voltage (Vo). Show all necessary values on the waveforms (VCE(Q), VCE(min), and VCE(max)). Calculate also the voltage gain of the amplifier Tutorial 2

7. Question 2 (Solution 2a) For the circuit shown below assume that Vs = 0.3sin(ωt), VBB = 1.5V, RB = 10K, Rc = 1K, and Vcc = 15V. Determine the Q point (IB(Q), IC(Q) and VCE(Q)) as well as the saturation collector current IC(sat) and the saturation base current (IB(sat)), and draw the DC load line. • At the Q-Point Vs = 0  IB(Q) = (VBB – 0.7)/RB = (1.5 – 0.7)/10K = 0.08mA  IC(Q) = β x IB(Q) = 100 x 0.08 = 8mA  VCE(Q) = Vcc – IC(Q) x Rc = 15 – 8 x 1 = 7V At saturation VCE = 0 • IC(sat) = Vcc / Rc = 15/1K = 15mA • IB(sat) = IC(sat) / β = 15/100 0,15mA Tutorial 2

8. Question 2 (Solution 2b) For the circuit shown below assume that Vs = 0.3sin(ωt), VBB = 1.5V, RB = 10K, Rc = 1K, and Vcc = 15V. Draw the waveform of the input (Vs) and the output voltage (Vo). Show all necessary values on the waveforms (VCE(Q), VCE(min), and VCE(max)). Calculate also the voltage gain of the amplifier • When Vs = +0.3V  IB = (VBB + 0.3 – 0.7)/RB = (1.5 + 0.3 – 0.7)/10K = 0.11mA  IC = β x IB = 100 x 0.11 = 11mA  VCE = Vcc – IC x Rc = 15 – 11 x 1 = 4V (VCE(min)) When Vs = -0.3V •  IB = (VBB - 0.3 – 0.7)/RB = (1.5 - 0.3 – 0.7)/10K = 0.0.5mA •  IC = β x IB = 100 x 0.05 = 5mA • VCE = Vcc – IC x Rc = 15 – 5 x 1 = 10V (VCE(max)) Voltage Gain = ΔVo/ΔVs = (10 – 4)/(0.3 – (-0.3)) = 6 / 0.6 = 10 Tutorial 2

9. Question 3 Repeat question 2b assuming that Vs = 1sin(ωt) When Vs = +1V  IB = (VBB + 1 – 0.7)/RB = (1.5 + 1 – 0.7)/10K = 0.15mA>IB(sat)  Vs(sat) = IB x RB + 0.7 - VBB = 0.15 x 10 +0.7 – 1.5 = 0.7V  VCE = VCE(sat) = 0 (VCE(min)) When Vs = -1V will be cut-off since (VBB + Vs) <0.7 Vs(min) = 0.7 -VBB = 0.7 – 1.5 = -0.8V Tutorial 2

10. Question 4 For the common emitter amplifier shown below: • Determine the Q-point and draw the DC load line. • Draw the small signal model equivalent circuit. • Determine the input impedance of the circuit. • Calculate the voltage gain of the amplifier Tutorial 2

11. Question 4 (Solution 4a) • Determine the Q-point and draw the DC load line. • Under DC conditions all capacitors are assumed to be open circuit. • Use the Thevenis Theorem for the input stage  RTH = R1//R2 = (R1 x R2)/(R1 + R2) • RTH = (10K x 40K)/(10K + 40K) = 8K • VTH = (Vcc x R2)/(R1 + R2) • VTH = (15 x 10K)/(10K + 40K) = 3V • Input Loop: VTH = IBxRTH +VBE + IE x RE β>>1 (IE = IC) and (RTH << βxRE)  VTH = VB = VBE + IE x RE • IE(Q) = (VTH – VBE)/RE = (3 – 0.7)/2K = 1.15mA • IB(Q) = IE / β = 1.15mA)/100 = 11.5uA Tutorial 2

12. Question 4 (Solution 4a – cont.) • Determine the Q-point and draw the DC load line. • Output Loop: IC(Q) = IE(Q) = 1.15mA Vcc = IC(Q) x RC + VCE(Q) + IE(Q) x RE  VCE(Q) = VCC – (RC + RE) x IC(Q) = 15 – (1K+2K)x1.15mA = 11.55V Vcc = IC(sat) x RC + 0 + IE(sat) x RE  IC(sat) = VCC / (RC + RE) = 15/(1K+2K) = 5mA 5mA Q-point 1.15mA 11.55 15V Tutorial 2

13. Question 4 (Solution 4b) • Draw the small signal model equivalent circuit. Simplified AC circuit Small signal model equivalent circuit: (Replace transistor with the simplified h-model) Tutorial 2

14. Question 4 (Solution 4c) • Determine the input impedance of the circuit. • Calculate the voltage gain of the amplifier Input impedance: Zin = R1//R2//(β x re) = RTH // (β x re) re = 25mV/IE(Q) = 25mV/1.15mA = 22Ω • Zin = 8K // (22x100) = (8K x 2.2K)/(8K + 2.2K) = 176K/30 = 1.7K Voltage Gain = -Rc/re = 1K/22Ω = 45 Tutorial 2