Circuit Diagram for Common Collector Amplifier with 2.5V Output and 700mA

rushikeshkumawat93 wrote:

(...) ill be 2.5V and current upto 700mA will be OK. We have tried it but the 0.6v is consumed by transistor as a biasing hence our output is stuck at 1.9V. hence, please provide how to increase the amplitude of that 1.9v and make it to 2.5V. Also give all components values which will be used in the circuit.

Hello! It sounds like you’re working on a buffer amplifier using the common collector configuration (also known as an emitter follower), and you’re facing an issue with the voltage drop across the transistor. Let's tackle this together.

### Step 1: Understanding the Common Collector Configuration
In a common collector configuration:
- The output voltage is taken across the emitter resistor.
- The transistor’s base-emitter junction inherently has a voltage drop (about 0.6V to 0.7V), which means the emitter voltage will always be slightly less than the base voltage.

### Step 2: Circuit Design to Meet Your Specifications
Since you want an output voltage of 2.5V and a current capability of up to 700mA, you need to ensure that the base voltage is higher than 2.5V + 0.7V (Vbe drop) = 3.2V approximately.

### Step 3: Increasing the Input Voltage
To achieve 2.5V at the emitter with up to 700mA current, you must increase your input voltage to at least 3.2V. This is to compensate for the inherent Vbe drop of the transistor.

### Step 4: Component Selection and Circuit Diagram
1. Transistor: Since your current requirement is up to 700mA, a suitable transistor could be the 2N4401, which can handle up to 600mA. If you expect to push to 700mA regularly, consider a beefier transistor like the BD139, which can handle currents up to 1.5A.
2. Base Drive Resistor: To ensure proper biasing and safe operation, we need to calculate the base resistor. Assuming a supply voltage of 5V:
- We need to set the base at about 3.2V to ensure the emitter is at 2.5V.
- If using a BD139, with a base-emitter voltage of about 0.7V, the voltage across the base resistor will be 5V - 3.2V = 1.8V.
- Choosing a base current of about 1mA (a conservative choice for stability), the base resistor $ R_b $ would be $ R_b = \frac{1.8V}{1mA} = 1.8k\Omega $.
3. Emitter Resistor: To set the output voltage, an emitter resistor is added, but since you want the full 2.5V with reference to ground (and assuming the load will handle current regulation), you might not need an additional emitter resistor. If fine voltage adjustment is required, a small value can be added.

### Schematic:


### Step 5: Verification and Testing
Once the circuit is built, it’s crucial to test it under actual load conditions to ensure it meets the required 2.5V output at up to 700mA. Measure the output voltage without the load and with the load to check for any significant drop.

### Follow-Up Questions
- Is the power supply capable of providing higher voltages if needed?
- What type of load will be connected to this amplifier? This might affect stability and component choice.

Let's work through each step, and feel free to ask questions or request further details on each part!

rushikeshkumawat93 wrote:

Does the Darligton Pair will fullfill my needs ? If yes then what extra components I needed ? Please give us values also.

Yes, using a Darlington pair can be a good solution to increase the current gain, which is beneficial in driving high current loads while maintaining a relatively stable output voltage. A Darlington pair essentially combines two bipolar junction transistors (BJTs) in a configuration that results in a high current gain, approximately the product of the gains of the two individual transistors.

### Benefits of Using a Darlington Pair:
1. High Current Gain: This makes it suitable for applications where a small base current needs to control a large load current.
2. Better Drive Capability: Can handle more current without needing a large base current, which can be beneficial if your control signal is weak.

### Considerations for Using a Darlington Pair:
1. Increased Voltage Drop: A Darlington pair will have a double Vbe drop (about 1.2V to 1.4V), which is higher than a single transistor. This needs to be compensated for in the input drive voltage.

### Circuit Design with Darlington Pair:
To design a Darlington pair circuit that can provide an output of 2.5V at 700mA, and considering the higher Vbe drop, you’ll need to adjust the input (base) voltage accordingly.

#### Components Needed:
1. Transistors: You can choose two medium power transistors or a ready-made Darlington pair. For instance, combining two BD139 transistors or using an integrated Darlington like the TIP122 which can handle up to 5A.
2. Base Resistor (Rb): This resistor is crucial for setting the correct base current considering the increased base-emitter voltage.
3. Input Voltage: You need to provide an input voltage that is higher than what you used in the single transistor setup due to the double Vbe drop.

#### Example Calculation for TIP122:
- Base-Emitter Voltage: Approximately 1.4V
- Desired Emitter Voltage: 2.5V
- Required Base Voltage: 2.5V + 1.4V = 3.9V

Assuming a 5V supply:
- Voltage across Rb: 5V - 3.9V = 1.1V
- Assuming a base current of 1mA (since Darlington pairs have high current gain, a smaller base current is usually sufficient):
- $ Rb = \frac{1.1V}{1mA} = 1.1k\Omega $

#### Schematic:


### Follow-Up Questions:
- Do you have a specific voltage level for your input, or is it adjustable?
- What sort of thermal management do you have in place? Darlington transistors can generate more heat due to their higher saturation voltage.

Using a Darlington pair will certainly meet your current amplification needs, but ensure your power supply can handle the slightly higher voltage requirement and that your design includes considerations for the additional voltage drop and heat generation. Would you like to explore more about how to handle these additional considerations?