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From Wall Outlet 120Vrms (340Vpp) Transformer DO NOT Connect these two wires! BR1 하 Q1 www R1 하 +++ N GND Figure 15: BJT Series Linear Voltage Regulator Probe Probe GND BJT Series Linear Voltage Regulator As you observed in the 2nd diode laboratory, ripple is a source of noise that cannot be completely eliminated using simple filtering alone or combined with a Zener diode shunt regulator. One way to remove ripple in a power supply design is to use a linear voltage regulator. One of the simplest implementations (short of using an actual linear voltage regulator - that is a lot more expensive) is a no hum (no ripple) BJT based series linear voltage regulator circuit as seen in Figure 15. For the 5.0V no hum circuit in Figure 15, we are going to solve for Rz. Z₁=5.6V Zener Diode (@20mA), Ri=100kQ, RL=1kQ, Q1=2N3904 (ẞ=150), VBE = 0.6V, VRL = 5.0V, Maximum voltage at the Collector = 10V. Assume Forward Active Region. Hint: We assume the circuit has reached steady state and treat all caps as open circuits. To solve for Rz, we need to determine the voltage drop across it and the current flowing through it. We are given a collector voltage of 10V, which is also the voltage applied to the top of Rz. The bottom or R= (remember the caps are all open circuits - you may want to redraw the circuit that way) is the Zener voltage, 5.6V. The current through the Zener is given in the problem description as 20 mA. The only other current flowing through Rz is the base current of the BJT. To solve for the base current, you need to solve for the emitter current and then divide that by (ẞ+1). We know the output voltage is 5V and the total load is Ri||RL, but R₁ is 100 times larger than R₁ so the parallel combination is ~1k. By applying simple ohms law, we can solve for the emitter current. Now we have both the Zener current and the base current and can now solve for Rz.

From Wall Outlet 120Vrms (340Vpp) Transformer DO NOT Connect these two wires! BR1 하 Q1 www R1 하 +++ N GND Figure 15: BJT Series Linear Voltage Regulator Probe Probe GND BJT Series Linear Voltage Regulator As you observed in the 2nd diode laboratory, ripple is a source of noise that cannot be completely eliminated using simple filtering alone or combined with a Zener diode shunt regulator. One way to remove ripple in a power supply design is to use a linear voltage regulator. One of the simplest implementations (short of using an actual linear voltage regulator - that is a lot more expensive) is a no hum (no ripple) BJT based series linear voltage regulator circuit as seen in Figure 15. For the 5.0V no hum circuit in Figure 15, we are going to solve for Rz. Z₁=5.6V Zener Diode (@20mA), Ri=100kQ, RL=1kQ, Q1=2N3904 (ẞ=150), VBE = 0.6V, VRL = 5.0V, Maximum voltage at the Collector = 10V. Assume Forward Active Region. Hint: We assume the circuit has reached steady state and treat all caps as open circuits. To solve for Rz, we need to determine the voltage drop across it and the current flowing through it. We are given a collector voltage of 10V, which is also the voltage applied to the top of Rz. The bottom or R= (remember the caps are all open circuits - you may want to redraw the circuit that way) is the Zener voltage, 5.6V. The current through the Zener is given in the problem description as 20 mA. The only other current flowing through Rz is the base current of the BJT. To solve for the base current, you need to solve for the emitter current and then divide that by (ẞ+1). We know the output voltage is 5V and the total load is Ri||RL, but R₁ is 100 times larger than R₁ so the parallel combination is ~1k. By applying simple ohms law, we can solve for the emitter current. Now we have both the Zener current and the base current and can now solve for Rz.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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From Wall Outlet 120Vrms (340Vpp) Transformer DO NOT Connect these two wires! BR1 하 Q1 www R1 하 +++ N GND Figure 15: BJT Series Linear Voltage Regulator Probe Probe GND
Transcribed Image Text:From Wall Outlet 120Vrms (340Vpp) Transformer DO NOT Connect these two wires! BR1 하 Q1 www R1 하 +++ N GND Figure 15: BJT Series Linear Voltage Regulator Probe Probe GND
BJT Series Linear Voltage Regulator As you observed in the 2nd diode laboratory, ripple is a source of noise that cannot be completely eliminated using simple filtering alone or combined with a Zener diode shunt regulator. One way to remove ripple in a power supply design is to use a linear voltage regulator. One of the simplest implementations (short of using an actual linear voltage regulator - that is a lot more expensive) is a no hum (no ripple) BJT based series linear voltage regulator circuit as seen in Figure 15. For the 5.0V no hum circuit in Figure 15, we are going to solve for Rz. Z₁=5.6V Zener Diode (@20mA), Ri=100kQ, RL=1kQ, Q1=2N3904 (ẞ=150), VBE = 0.6V, VRL = 5.0V, Maximum voltage at the Collector = 10V. Assume Forward Active Region. Hint: We assume the circuit has reached steady state and treat all caps as open circuits. To solve for Rz, we need to determine the voltage drop across it and the current flowing through it. We are given a collector voltage of 10V, which is also the voltage applied to the top of Rz. The bottom or R= (remember the caps are all open circuits - you may want to redraw the circuit that way) is the Zener voltage, 5.6V. The current through the Zener is given in the problem description as 20 mA. The only other current flowing through Rz is the base current of the BJT. To solve for the base current, you need to solve for the emitter current and then divide that by (ẞ+1). We know the output voltage is 5V and the total load is Ri||RL, but R₁ is 100 times larger than R₁ so the parallel combination is ~1k. By applying simple ohms law, we can solve for the emitter current. Now we have both the Zener current and the base current and can now solve for Rz.
Transcribed Image Text:BJT Series Linear Voltage Regulator As you observed in the 2nd diode laboratory, ripple is a source of noise that cannot be completely eliminated using simple filtering alone or combined with a Zener diode shunt regulator. One way to remove ripple in a power supply design is to use a linear voltage regulator. One of the simplest implementations (short of using an actual linear voltage regulator - that is a lot more expensive) is a no hum (no ripple) BJT based series linear voltage regulator circuit as seen in Figure 15. For the 5.0V no hum circuit in Figure 15, we are going to solve for Rz. Z₁=5.6V Zener Diode (@20mA), Ri=100kQ, RL=1kQ, Q1=2N3904 (ẞ=150), VBE = 0.6V, VRL = 5.0V, Maximum voltage at the Collector = 10V. Assume Forward Active Region. Hint: We assume the circuit has reached steady state and treat all caps as open circuits. To solve for Rz, we need to determine the voltage drop across it and the current flowing through it. We are given a collector voltage of 10V, which is also the voltage applied to the top of Rz. The bottom or R= (remember the caps are all open circuits - you may want to redraw the circuit that way) is the Zener voltage, 5.6V. The current through the Zener is given in the problem description as 20 mA. The only other current flowing through Rz is the base current of the BJT. To solve for the base current, you need to solve for the emitter current and then divide that by (ẞ+1). We know the output voltage is 5V and the total load is Ri||RL, but R₁ is 100 times larger than R₁ so the parallel combination is ~1k. By applying simple ohms law, we can solve for the emitter current. Now we have both the Zener current and the base current and can now solve for Rz.
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