Darlington Pair

Related Concepts: lab 7 darlington

Based on Laboratory Exercise 15: Darlington Pair from your manual (note: this is a second experiment numbered 15 in the manual, appearing after Exercise 23), here is the revision note.

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1) Experiment Snapshot

Experiment Title: Darlington Pair (High Current Gain Amplifier)

Objective(s):

• Verify that the current gain of a Darlington pair is much greater than that of a single transistor emitter follower.
• Calculate and compare the input impedance of a Darlington emitter follower vs. a single transistor emitter follower.

Core Principle / Theory:

• Darlington Configuration: Two transistors connected in cascade (collectors tied together, emitter of Q1 feeds base of Q2).
• Super Beta: The total current gain (${\beta }_D$) is the product of the individual gains: ${\beta }_D\approx {\beta }_1\times {\beta }_2$.
• Input Impedance: Extremely high, calculated as $Z_i\approx {\beta }_D\times R_E$.
• Output Impedance: Very low.

Key Formula(s):

• Emitter Current: $I_E=V_E/R_E$
• Current Gain ($\beta$): $Gain=\frac{\Delta I_E}{\Delta I_B}$
• Input Impedance: $Z_i=\beta \times R_L$ (where $R_L$ is the emitter load resistor)

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2) Apparatus Setup

Connection / Block Diagram:

• Part A (Single Transistor): Common Collector setup. Potentiometer $R1$ controls Base Current ($I_B$). Output taken from Emitter across Load $R3$ ($100\Omega$).
• Part B (Darlington): Two transistors (Q1, Q2) connected as a Darlington pair. Emitter of Q1 $\to$ Base of Q2. Collectors tied to $V_{CC}$. Output taken from Q2 Emitter across Load $R4$ ($47\Omega$).
• Supply: $V_{CC}=9V$ DC.

Precautions:

• Meter Placement: Ensure the microammeter is in series with the base to measure $I_B$ correctly.
• Range: Use a low voltage range (e.g., 2.5V or 10V) for measuring $V_E$ as it might be small.
• Wiring: Verify the Darlington connections; the Base of Q2 must only connect to the Emitter of Q1 (no external bias resistors at that junction).

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3) Procedure

Part A: Single Transistor Gain (Reference)

1. Connect the single emitter follower (Fig 15-2). $V_{CC}=9V$.
2. Adjust Potentiometer $R1$ to set Base Current ($I_B$) to 4 $\mu$A.
3. Measure Voltage across Emitter Resistor $R3$ ($V_E$). Record it.
4. Adjust $R1$ to increase Base Current ($I_B$) to 8 $\mu$A.
5. Measure and record new $V_E$.
6. Calculate $I_E$ for both steps and determine Current Gain ($\Delta I_E/\Delta I_B$).

Part B: Darlington Pair Gain 7. Connect the Darlington circuit (Fig 15-3). 8. Adjust $R1$ to set Base Current ($I_B$) to 2 $\mu$A. 9. Measure Voltage across Emitter Resistor $R4$ ($V_E$). Record it. 10. Adjust $R1$ to increase $I_B$ to 4 $\mu$A. 11. Measure and record new $V_E$. 12. Calculate $I_E$ and Darlington Current Gain.

Part C: Input Impedance 13. Calculate theoretical Input Impedance for both circuits using $Z_i=\beta \times R_{Load}$.

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4) Data Taken / Measurements

Table Template:

Parameter	Single Transistor (Part A)	Darlington Pair (Part B)	Unit
Step 1
Input $I_{B1}$	4	2	$\mu$A
Load Resistor	$R3=100$	$R4=47$	$\Omega$
Output $V_{E1}$	________	________	Vdc
Calc $I_{E1}$	$V_{E1}/100$	$V_{E1}/47$	A
Step 2
Input $I_{B2}$	8	4	$\mu$A
Output $V_{E2}$	________	________	Vdc
Calc $I_{E2}$	$V_{E2}/100$	$V_{E2}/47$	A
Results
$\Delta I_B$	$4\mu A$	$2\mu A$	A
$\Delta I_E$	________	________	A
Current Gain	________	________	-

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5) Calculations

1. Emitter Current: $I_E=\frac{V_E}{R_{load}}$
2. Current Gain (Experimental): $Gain=\frac{I_{E2}-I_{E1}}{I_{B2}-I_{B1}}$ (Manual Example: Single $\approx 150$; Darlington $\approx 20,000$)
3. Input Impedance: $Z_i=\text{Gain}\times R_{load}$ (Manual Example: Single $\approx 15k\Omega$; Darlington $\approx 940k\Omega$)

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6) Results & Outcome

• Single Transistor: Showed moderate current gain ($\approx 100-200$).
• Darlington Pair: Showed massive current gain ($\approx 10,000+$). The gain is roughly the square of the single stage gain (if ${\beta }_1\approx {\beta }_2$).
• Input Impedance: The Darlington pair presented a much higher input impedance ($\approx 1M\Omega$) compared to the single stage ($\approx 15k\Omega$).

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7) Why This Outcome Happens (Reasoning)

• Multiplication of Beta: In a Darlington pair, the emitter current of the first transistor becomes the base current of the second ($I_{E1}=I_{B2}$). Since $I_{C2}={\beta }_2{I}_{B2}={\beta }_2({\beta }_1{I}_{B1})$, the total gain is ${\beta }_{total}\approx {\beta }_1{\beta }_2$.
• High Input Impedance: Resistance at the emitter is “reflected” back to the base multiplied by the current gain. Since the gain is huge, the reflected resistance ($Z_{in}$) is huge.
• Trade-off: While gain is high, the voltage drop from base to output is higher ($V_{BE1}+V_{BE2}\approx 1.2V$ to $1.4V$) compared to a single transistor ($0.7V$).

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8) Viva / Lab Test Quick Prep

Q1: What is the main advantage of a Darlington pair? A: Extremely high current gain and very high input impedance.

Q2: What is the approximate total beta (${\beta }_D$) if two identical transistors with $\beta =100$ are used? A: $100\times 100=10,000$.

Q3: Why is the input impedance of a Darlington pair so high? A: Because the load resistance is multiplied by the total beta (${\beta }_D$) when viewed from the base.

Q4: What is a disadvantage regarding the Base-Emitter voltage in a Darlington pair? A: It has double the voltage drop ($\approx 1.2V-1.4V$) because there are two junctions in series.

Q5: How are the collectors connected in a standard Darlington pair? A: The collectors are tied together and connected to the supply.

Common Mistakes:

• Calculation Error: Forgetting to convert units (e.g., subtracting $\mu$A from mA without conversion) when calculating gain.
• Resistor ID: Confusing the load resistor $R3$ ($100\Omega$) used in Part A with $R4$ ($47\Omega$) used in Part B.