How To Analyze This Transistor Stage (NPN + PNP) At 5 V Supply And Show That The PNP Is Off?

Introduction

In this article, we will delve into the analysis of a transistor stage that consists of one NPN (Q1) and one PNP (Q2) transistor. The circuit is powered by a 5 V supply, and we will determine the conditions under which the PNP transistor (Q2) is off. We will use the given circuit specifics to analyze the behavior of the transistors and show that the PNP transistor is indeed off under certain conditions.

Circuit Specifics

The circuit specifics are as follows:

• Supply Voltage (𝑉𝐶𝐶): 5 V
• Load Resistance (𝑅𝐿): 100 Ω
• Lower Rail (𝑉𝐸𝐸​): Can be either 0 V (GND) or 5 V

Understanding the Transistor Stage

The transistor stage consists of two transistors: an NPN transistor (Q1) and a PNP transistor (Q2). The NPN transistor is typically used as a common emitter amplifier, while the PNP transistor is used as a common base amplifier. In this circuit, we will assume that the NPN transistor (Q1) is in the active region, while the PNP transistor (Q2) is in the cutoff region.

Analyzing the Circuit

To analyze the circuit, we need to determine the voltage and current relationships between the transistors. We will use the following assumptions:

• The NPN transistor (Q1) is in the active region, with a collector-emitter voltage (𝑉𝐶𝐸) of 5 V.
• The PNP transistor (Q2) is in the cutoff region, with a collector-emitter voltage (𝑉𝐶𝐸) of 0 V.

Calculating the Base Current of the NPN Transistor

To calculate the base current of the NPN transistor (Q1), we need to use the following equation:

𝐼𝐵 = 𝐼𝐶 / 𝐵𝐸

where 𝐼𝐵 is the base current, 𝐼𝐶 is the collector current, and 𝐵𝐸 is the current gain of the NPN transistor.

Assuming a current gain (𝐵𝐸) of 100 for the NPN transistor, we can calculate the base current as follows:

𝐼𝐵 = 𝐼𝐶 / 100

To determine the collector current (𝐼𝐶), we need to use the following equation:

𝐼𝐶 = (𝑉𝐶𝐶 - 𝑉𝐶𝐸) / 𝑅𝐿

where 𝑉𝐶𝐶 is the supply voltage, 𝑉𝐶𝐸 is the collector-emitter voltage, and 𝑅𝐿 is the load resistance.

Substituting the given values, we get:

𝐼𝐶 = (5 V - 0 V) / 100 Ω = 50 mA

Now, we can calculate the base current as follows:

𝐼𝐵 = 50 mA / 100 = 0.5 mA

Calculating the Collector Current of the PNP Transistor

To calculate the collector current of the PNP transistor (Q2), we need to use the following equation:

𝐼𝐶 = (𝑉𝐶𝐸 - 𝑉𝐸𝐸) / 𝑅𝐿

where 𝑉𝐶𝐸 is the collector-emitter voltage, 𝑉𝐸𝐸 is the emitter-base voltage, and 𝑅𝐿 is the load resistance.

Since the PNP transistor is in the cutoff region, the collector-emitter voltage (𝑉𝐶𝐸) is 0 V. Therefore, the collector current (𝐼𝐶) is also 0 A.

Conclusion

In this article, we analyzed a transistor stage that consists of one NPN (Q1) and one PNP (Q2) transistor. We determined the conditions under which the PNP transistor (Q2) is off and showed that the PNP transistor is indeed off when the collector-emitter voltage (𝑉𝐶𝐸) is 0 V. We also calculated the base current of the NPN transistor (Q1) and the collector current of the PNP transistor (Q2) using the given circuit specifics.

References

• "Transistor Amplifiers" by Robert F. Coughlin
• "The Art of Electronics" by Paul Horowitz and Winfield Hill

Appendix

The following appendix provides additional information and derivations used in this article.

Derivation of the Base Current of the NPN Transistor

The base current of the NPN transistor (Q1) can be calculated using the following equation:

𝐼𝐵 = 𝐼𝐶 / 𝐵𝐸

where 𝐼𝐵 is the base current, 𝐼𝐶 is the collector current, and 𝐵𝐸 is the current gain of the NPN transistor.

Assuming a current gain (𝐵𝐸) of 100 for the NPN transistor, we can calculate the base current as follows:

𝐼𝐵 = 𝐼𝐶 / 100

To determine the collector current (𝐼𝐶), we need to use the following equation:

𝐼𝐶 = (𝑉𝐶𝐶 - 𝑉𝐶𝐸) / 𝑅𝐿

where 𝑉𝐶𝐶 is the supply voltage, 𝑉𝐶𝐸 is the collector-emitter voltage, and 𝑅𝐿 is the load resistance.

Substituting the given values, we get:

𝐼𝐶 = (5 V - 0 V) / 100 Ω = 50 mA

Now, we can calculate the base current as follows:

𝐼𝐵 = 50 mA / 100 = 0.5 mA

Derivation of the Collector Current of the PNP Transistor

The collector current of the PNP transistor (Q2) can be calculated using the following equation:

𝐼𝐶 = (𝑉𝐶𝐸 - 𝑉𝐸𝐸) / 𝑅𝐿

where 𝑉𝐶𝐸 is the collector-emitter voltage, 𝑉𝐸𝐸 is the emitter-base voltage, and 𝑅𝐿 is the load resistance.

Frequently Asked Questions

In this Q&A article, we will address some of the most common questions related to analyzing a transistor stage that consists of one NPN (Q1) and one PNP (Q2) transistor. We will provide detailed answers to help you understand the behavior of the transistors and show that the PNP transistor is indeed off under certain conditions.

Q: What is the purpose of the NPN transistor (Q1) in the circuit?

A: The NPN transistor (Q1) is used as a common emitter amplifier in the circuit. Its purpose is to amplify the input signal and provide a high current gain.

Q: What is the purpose of the PNP transistor (Q2) in the circuit?

A: The PNP transistor (Q2) is used as a common base amplifier in the circuit. Its purpose is to provide a high voltage gain and a low current gain.

Q: Why is the PNP transistor (Q2) in the cutoff region?

A: The PNP transistor (Q2) is in the cutoff region because the collector-emitter voltage (𝑉𝐶𝐸) is 0 V. This means that the transistor is not conducting any current.

Q: How do you calculate the base current of the NPN transistor (Q1)?

A: To calculate the base current of the NPN transistor (Q1), you need to use the following equation:

𝐼𝐵 = 𝐼𝐶 / 𝐵𝐸

where 𝐼𝐵 is the base current, 𝐼𝐶 is the collector current, and 𝐵𝐸 is the current gain of the NPN transistor.

Q: How do you calculate the collector current of the PNP transistor (Q2)?

A: To calculate the collector current of the PNP transistor (Q2), you need to use the following equation:

𝐼𝐶 = (𝑉𝐶𝐸 - 𝑉𝐸𝐸) / 𝑅𝐿

where 𝑉𝐶𝐸 is the collector-emitter voltage, 𝑉𝐸𝐸 is the emitter-base voltage, and 𝑅𝐿 is the load resistance.

Q: What is the significance of the collector-emitter voltage (𝑉𝐶𝐸) in the circuit?

A: The collector-emitter voltage (𝑉𝐶𝐸) is the voltage between the collector and emitter terminals of the transistor. In the circuit, the collector-emitter voltage (𝑉𝐶𝐸) is 0 V for the PNP transistor (Q2), which means that the transistor is in the cutoff region.

Q: How do you determine the conditions under which the PNP transistor (Q2) is off?

A: To determine the conditions under which the PNP transistor (Q2) is off, you need to analyze the circuit and determine the values of the collector-emitter voltage (𝑉𝐶𝐸) and the emitter-base voltage (𝑉𝐸𝐸). If the collector-emitter voltage (𝑉𝐶𝐸) is 0 V, then the PNP transistor (Q2) is in the cutoff region and is off.

Q: What are the implications of the PNP transistor (Q2) being off in the circuit?

A: The implications of the PNP transistor (Q2) being off in the circuit are that the transistor is not conducting any current and is not amplifying any signal. This means that the circuit is not functioning as intended, and the PNP transistor (Q2) needs to be replaced or repaired.

Conclusion

In this Q&A article, we addressed some of the most common questions related to analyzing a transistor stage that consists of one NPN (Q1) and one PNP (Q2) transistor. We provided detailed answers to help you understand the behavior of the transistors and show that the PNP transistor is indeed off under certain conditions.