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# ECE 252:Analogue Electronics

**Lecture: DC and AC Analysis of PNP BJT Circuits**

### Voltage Gain Calculation

### **Introduction to Bipolar Junction Transistors (BJTs)**

A **Bipolar Junction Transistor (BJT)** is a three-terminal semiconductor device used for amplification and switching applications. It has two main types: **NPN** and **PNP**. In this lecture, we will focus on **PNP transistors** in two common configurations:

* **Common Base (CB) Configuration**
* **Common Emitter (CE) Configuration**

These configurations are used in analog circuits such as amplifiers and signal processing systems.

***

#### **2. Prerequisite Knowledge**

Before diving into the DC and AC analysis, you should be familiar with:

**a) Semiconductor Basics**

* **PN Junction**: Understanding how current flows through a P-N junction is essential.
* **Doping**: How semiconductors are modified with impurities to create P-type and N-type materials.

**b) Transistor Operation**

* **Active Region**: The region where a BJT can amplify signals.
* **Cutoff and Saturation Regions**: Used for switching applications.

**c) Circuit Analysis Basics**

* **Kirchhoff’s Voltage and Current Laws (KVL & KCL)**: Essential for analyzing circuits.
* **Ohm’s Law**: Used for calculating voltages, currents, and resistances in a circuit.

***

#### **3. Importance of DC and AC Analysis in BJTs**

Analyzing BJTs under **DC and AC conditions** is crucial for designing stable and efficient circuits.

**Why DC Analysis?**

* Determines the **biasing conditions** of the transistor.
* Ensures the transistor operates in the **correct region** (Active Region for amplification).
* Helps in designing circuits that work under different temperature and voltage variations.

**Why AC Analysis?**

* Evaluates how the transistor responds to **small input signals**.
* Determines **gain, input impedance, and output impedance**, which are key for amplifier performance.
* Helps in designing **high-frequency circuits** for communication systems.

***

#### **4. Common Base (CB) and Common Emitter (CE) Configurations**

**a) Common Base (CB) Configuration**

* The **Base** is the common terminal for both input and output.
* Input is applied to the **Emitter** and output is taken from the **Collector**.
* Provides **low input impedance** and **high output impedance**.
* Mainly used in **high-frequency applications**.

**DC Equivalent Circuit of CB Configuration**

* Determines the **operating point** of the transistor.
* Capacitors are treated as **open circuits**.
* Resistors and DC sources define the **biasing conditions**.

**AC Equivalent Circuit of CB Configuration**

* Used to analyze the **small-signal response** of the transistor.
* Capacitors are treated as **short circuits**.
* DC voltage sources are replaced with **ground**.

**b) Common Emitter (CE) Configuration**

* The **Emitter** is the common terminal for both input and output.
* Input is applied to the **Base**, and output is taken from the **Collector**.
* Provides **moderate input impedance** and **high gain**.
* Widely used in **audio and RF amplifiers**.

**DC Equivalent Circuit of CE Configuration**

* Establishes the **DC operating point**.
* Shows how the transistor is **biased** to function in the active region.

**AC Equivalent Circuit of CE Configuration**

* Helps in understanding how the circuit amplifies signals.
* Shows the transistor’s **gain, impedance, and response to AC signals**.

***

#### **5. Biasing in DC Analysis**

For a PNP transistor, we must ensure proper **biasing** to keep the transistor in the **active region**:

* **Emitter-Base Junction**: Forward biased (Emitter more negative than Base).
* **Collector-Base Junction**: Reverse biased (Collector more positive than Base).

Biasing methods include:

* **Fixed Bias**: Simple but lacks stability.
* **Voltage Divider Bias**: Provides better stability.
* **Emitter Bias**: Helps in **temperature stability**.

***

#### **6. Small-Signal Model for AC Analysis**

For AC analysis, the transistor is replaced with a **small-signal model**:

* **h-parameter model**: Uses **h\_FE (current gain)** to describe behavior.
* **Hybrid-Pi Model**: More accurate for **high-frequency applications**.

Important parameters:

* **Voltage Gain (Av)** = VoutVin\frac{V\_{\text{out}}}{V\_{\text{in}}}Vin​Vout​​
* **Input Impedance (Z\_in)**: Resistance seen by the input source.
* **Output Impedance (Z\_out)**: Resistance seen by the load.

***

#### **7. Applications of Common Base and Common Emitter Circuits**

| Configuration           | Applications                                                      |
| ----------------------- | ----------------------------------------------------------------- |
| **Common Base (CB)**    | Used in **high-frequency amplifiers** and RF circuits.            |
| **Common Emitter (CE)** | Used in **audio amplifiers, oscillators, and signal processing**. |

***

#### **8. Conclusion**

* **DC Analysis** is used to **set up the transistor’s operating point**.
* **AC Analysis** helps in **understanding amplification and signal response**.
* **Common Base (CB)** is used for **high-frequency applications**.
* **Common Emitter (CE)** is widely used in **amplifiers and signal processing**.

***

#### **9. Further Reading**

* **Sedra & Smith – Microelectronic Circuits**
* **Millman & Grabel – Microelectronics**
* **Analysis and Design of Analog Integrated Circuits by Gray, Hurst, Lewis & Meyer**

####  **DC Equivalent of a Common Base PNP BJT Circuit**

A **DC equivalent circuit** is obtained by:

* Replacing capacitors with open circuits (since they block DC).
* Replacing AC sources with short circuits.
* Keeping DC voltage sources and biasing resistors as they are.

For a **common base PNP transistor**, the **base** is the common terminal for both input and output signals. The **DC equivalent circuit** mainly consists of:

* A PNP transistor with proper DC biasing.
* A current source (representing collector current).
* Voltage sources to establish proper emitter-base and collector-base junction biases.

**Diagram (DC Equivalent – Common Base PNP)**

```mathematica
     Vcc (+)
         │
         R_C
         │
         C (Collector)
         |
         |    
         B (Base)
         |
         R_B
         |
         GND (0V)
         |
         E (Emitter)
         |
         V_EE (-)
```

***

#### **2. AC Equivalent of a Common Base PNP BJT Circuit**

The **AC equivalent circuit** is obtained by:

* Replacing capacitors with short circuits (since they act as short circuits for AC signals).
* Replacing DC voltage sources with ground.
* Replacing the transistor with its small-signal equivalent model (h-parameter or hybrid-pi model).

In the **AC equivalent circuit**, the behavior of the transistor is analyzed in terms of small-signal parameters such as **input resistance, output resistance, and gain**.

**Diagram (AC Equivalent – Common Base PNP)**

```mathematica
     AC output
         │
         R_C
         │
         │
         C
         |
         │    
         B ─────── (AC ground)
         |
         R_B
         |
         E
         |
      AC input
```

***

#### **3. DC Equivalent of a Common Emitter PNP BJT Circuit**

In a **common emitter** configuration, the **emitter** is the common terminal for both input and output.

For the **DC equivalent**, the capacitors are replaced with open circuits, and the transistor is analyzed in its DC biasing state.

**Diagram (DC Equivalent – Common Emitter PNP)**

```mathematica
      Vcc (+)
         │
         R_C
         │
         C (Collector)
         |
         |
         B (Base)
         |
         R_B
         |
         GND (0V)
         |
         E (Emitter)
         |
         V_EE (-)
```

***

#### **4. AC Equivalent of a Common Emitter PNP BJT Circuit**

The **AC equivalent** circuit replaces the transistor with its small-signal equivalent, and capacitors are treated as short circuits.

**Diagram (AC Equivalent – Common Emitter PNP)**

```mathematica
     AC output
         │
         R_C
         │
         │
         C
         |
         │    
         B ─────── (AC ground)
         |
         R_B
         |
         E
         |
      AC input
```

***

#### **Explanation of Terms**

1. **DC Equivalent Circuit**:
   * Used to analyze the transistor’s operating point (biasing).
   * Shows how the transistor behaves with **DC voltage and currents**.
   * Ignores capacitors because they block DC.
2. **AC Equivalent Circuit**:
   * Used to analyze **signal amplification and response**.
   * Shows how the circuit behaves with **AC signals**.
   * Capacitors act as **short circuits** since they allow AC to pass.
   * Transistor is replaced with **small-signal models**.
3. **Common Base Configuration**:
   * The base is the **common** terminal for input and output.
   * Has **low input impedance** and **high output impedance**.
   * Used in **high-frequency applications**.
4. **Common Emitter Configuration**:
   * The emitter is the **common** terminal.
   * Provides **high gain** and is widely used in amplifiers.
   * Has **moderate input impedance** and **high output impedance**.

#### **Additional Explanations and Problem-Solving Examples for PNP BJT Circuits**

Now that we have covered the theoretical aspects, let’s apply the knowledge to some **practical problems** and **examples** related to DC and AC analysis of PNP BJTs in Common Base (CB) and Common Emitter (CE) configurations.

***

### **1. Example: DC Biasing Analysis of a Common Base PNP BJT Circuit**

#### **Problem Statement:**

Given the following **Common Base PNP BJT circuit**, determine the **DC operating point** 

```mathml
1. **Voltage gain** \( A_V \): How much amplification is needed?  
2. **Power supply** \( V_{CC} \): The available DC voltage.  
3. **Collector current** \( I_C \): Set based on the desired power level.  
4. **Impedances**: Choose suitable values for input and output matching.  

### **Example Specifications:**  
- **Voltage gain** \( A_V = -10 \) (10× amplification, inverted signal).  
- **Power supply** \( V_{CC} = 12V \).  
- **Collector current** \( I_C = 2mA \).  
- **Input impedance** \( Z_{in} > 10k\Omega \).  
- **Load resistor** \( R_L = 2k\Omega \).  

```

1. **Voltage gain** ( A\_V ): How much amplification is needed?
2. **Power supply** ( V\_{CC} ): The available DC voltage.
3. **Collector current** ( I\_C ): Set based on the desired power level.
4. **Impedances**: Choose suitable values for input and output matching.

#### **Example Specifications:**

* **Voltage gain** ( A\_V = -10 ) (10× amplification, inverted signal).
* **Power supply** ( V\_{CC} = 12V ).
* **Collector current** ( I\_C = 2mA ).
* **Input impedance** ( Z\_{in} > 10k\Omega ).
* **Load resistor** ( R\_L = 2k\Omega ).

***

(IE,IC,IB,VBE,VCBI\_E, I\_C, I\_B, V\_{BE}, V\_{CB}IE​,IC​,IB​,VBE​,VCB​).

**Circuit Parameters:**

* **V\_CC** = 12V
* **V\_EE** = -5V
* **R\_B** = 100kΩ
* **R\_C** = 2kΩ
* **R\_E** = 1kΩ
* **β\betaβ (current gain)** = 50
* **V\_{BE}** = 0.7V (PNP transistor assumption)

***

#### **Step 1: Calculate the Emitter Current IEI\_EIE​**

Using Kirchhoff’s Voltage Law (KVL) around the **Emitter-Base Loop**:

VEE+IERE+VBE=0V\_EE + I\_E R\_E + V\_{BE} = 0VE​E+IE​RE​+VBE​=0

Substituting values:

−5V+IE(1kΩ)+0.7V=0-5V + I\_E (1kΩ) + 0.7V = 0−5V+IE​(1kΩ)+0.7V=0 IE=5V−0.7V1kΩ=4.3V1kΩ=4.3mAI\_E = \frac{5V - 0.7V}{1kΩ} = \frac{4.3V}{1kΩ} = 4.3mAIE​=1kΩ5V−0.7V​=1kΩ4.3V​=4.3mA

***

#### **Step 2: Calculate the Collector Current ICI\_CIC​**

Since IC≈IEI\_C \approx I\_EIC​≈IE​ for large β\betaβ:

IC≈IE=4.3mAI\_C \approx I\_E = 4.3mAIC​≈IE​=4.3mA

***

#### **Step 3: Calculate the Base Current IBI\_BIB​**

IB=ICβ=4.3mA50=0.086mAI\_B = \frac{I\_C}{\beta} = \frac{4.3mA}{50} = 0.086mAIB​=βIC​​=504.3mA​=0.086mA

***

#### **Step 4: Calculate VCBV\_{CB}VCB​**

Applying Kirchhoff’s Voltage Law in the **Collector Circuit**:

VCB=VC−VBV\_{CB} = V\_C - V\_BVCB​=VC​−VB​

First, calculate VCV\_CVC​:

VC=VCC−ICRCV\_C = V\_{CC} - I\_C R\_CVC​=VCC​−IC​RC​ VC=12V−(4.3mA×2kΩ)V\_C = 12V - (4.3mA \times 2kΩ)VC​=12V−(4.3mA×2kΩ) VC=12V−8.6V=3.4VV\_C = 12V - 8.6V = 3.4VVC​=12V−8.6V=3.4V

For VBV\_BVB​, using VB=VE+VBEV\_B = V\_E + V\_{BE}VB​=VE​+VBE​:

VB=(−5V)+(0.7V)=−4.3VV\_B = (-5V) + (0.7V) = -4.3VVB​=(−5V)+(0.7V)=−4.3V

Thus:

VCB=VC−VB=3.4V−(−4.3V)=7.7VV\_{CB} = V\_C - V\_B = 3.4V - (-4.3V) = 7.7VVCB​=VC​−VB​=3.4V−(−4.3V)=7.7V

Since VCBV\_{CB}VCB​ is **positive**, the transistor is operating in the **active region**, confirming the design is correct.

***

### **2. Example: AC Gain Calculation of a Common Emitter PNP Amplifier**

#### **Problem Statement:**

Given a **Common Emitter PNP amplifier**, determine:

1. Voltage gain (AVA\_VAV​)
2. Input impedance (ZinZ\_{in}Zin​)
3. Output impedance (ZoutZ\_{out}Zout​)

**Circuit Parameters:**

* **R\_C** = 2kΩ
* **R\_E** = 500Ω
* **h\_FE (β\betaβ)** = 100
* **Internal resistance of transistor** re=25mVIEr\_e = \frac{25mV}{I\_E}re​=IE​25mV​ (thermal voltage approximation)

***

#### **Step 1: Calculate rer\_ere​ (Small-Signal Emitter Resistance)**

Using the thermal voltage approximation:

re=25mVIEr\_e = \frac{25mV}{I\_E}re​=IE​25mV​

Assume **I\_E = 1mA**:

re=25mV1mA=25Ωr\_e = \frac{25mV}{1mA} = 25Ωre​=1mA25mV​=25Ω

***

#### **Step 2: Calculate Voltage Gain AVA\_VAV​**

For a **Common Emitter Amplifier**:

AV=−RCre+REA\_V = -\frac{R\_C}{r\_e + R\_E}AV​=−re​+RE​RC​​ AV=−2kΩ25Ω+500ΩA\_V = -\frac{2kΩ}{25Ω + 500Ω}AV​=−25Ω+500Ω2kΩ​ AV=−2000525=−3.8A\_V = -\frac{2000}{525} = -3.8AV​=−5252000​=−3.8

So, the amplifier has a **gain of -3.8**, meaning it **inverts the signal** and amplifies it **3.8 times**.

***

#### **Step 3: Calculate Input Impedance ZinZ\_{in}Zin​**

Zin=(β+1)×(re+RE)Z\_{in} = (\beta + 1) \times (r\_e + R\_E)Zin​=(β+1)×(re​+RE​) Zin=(100+1)×(25Ω+500Ω)Z\_{in} = (100 + 1) \times (25Ω + 500Ω)Zin​=(100+1)×(25Ω+500Ω) Zin=101×525Ω=53.03kΩZ\_{in} = 101 \times 525Ω = 53.03kΩZin​=101×525Ω=53.03kΩ

***

#### **Step 4: Calculate Output Impedance ZoutZ\_{out}Zout​**

For a **Common Emitter Amplifier**, ZoutZ\_{out}Zout​ is approximately **R\_C**:

Zout≈RC=2kΩZ\_{out} \approx R\_C = 2kΩZout​≈RC​=2kΩ

***

### **3. Summary of Key Takeaways**

#### **Common Base (CB) Circuit**

* **Low input impedance, high output impedance**.
* **Used in high-frequency amplifiers**.
* **Voltage gain is greater than 1**.

#### **Common Emitter (CE) Circuit**

* **Moderate input impedance, high gain**.
* **Used in most amplifier circuits**.
* **Inverts the signal (negative gain)**.

***

### **4. Further Exercises**

Try solving these problems to strengthen your understanding:

1. **Design a Common Emitter PNP Amplifier** with **a gain of -10**. Choose suitable resistor values.
2. **Modify the first example**: What happens to IE,IC,IBI\_E, I\_C, I\_BIE​,IC​,IB​ if RER\_ERE​ increases?
3. **For an amplifier with AV=−5A\_V = -5AV​=−5 and Zin=40kΩZ\_{in} = 40kΩZin​=40kΩ**, determine the values of RCR\_CRC​ and RER\_ERE​.

***

### **5. Conclusion**

* **DC Analysis** determines the **operating point** (Q-point).
* **AC Analysis** helps in **gain calculation and impedance matching**.
* **Understanding these concepts** is crucial for designing stable **transistor amplifiers**.