Darlington Pair

A single BC547 transistor gives you modest current gain, but a Darlington pair chains two transistors to multiply that gain into the hundreds of thousands. Motor driver circuits and relay switches use this trick to let a microcontroller pin, carrying just a few microamps, switch several amps of load current.

Core Concept

In a Darlington pair, the emitter of Q1 connects directly to the base of Q2. The collector of Q1 connects to the collector of Q2. Any current entering the base of Q1 gets amplified by Q1, and then that amplified current gets amplified again by Q2.

The overall current gain is approximately the product of the two individual gains. Two BC547 transistors each with β = 200 produce a combined gain of around 40,000. The TIP120 is a popular monolithic Darlington device rated at 60V, 5A, with a guaranteed minimum gain of 1000.

The price of this gain is a higher saturation voltage. Because two base-emitter junctions sit in series, VBE_total is about 1.4V instead of 0.7V. This increases power loss when the device is used as a switch.

Key Equations

Overall current gain: β_D ≈ β1 × β2 + β1 + β2

Simplified for matched transistors: β_D ≈ β²

Combined VBE: VBE_D = VBE1 + VBE2 ≈ 1.4V at 300K

Collector current: IC = β_D × IB where IB is the input base current in amperes.

Saturation voltage: VCE_sat ≈ 0.9 to 1.1V (higher than a single BJT due to the stacked junctions).

Given:
  β1 = β2 = 150 (two BC547 transistors)
  IB1 = 10 µA = 10 × 10⁻⁶ A
  VCC = 12V
  RC = 470 Ω

Why this formula:
  The emitter current of Q1 drives the base of Q2, so gains multiply.

Formula:
  β_D = β1 × β2 + β1 + β2

Substitution:
  β_D = 150 × 150 + 150 + 150

Calculation:
  β_D = 22500 + 300 = 22800

  IC = β_D × IB1
     = 22800 × 10 × 10⁻⁶
     = 0.228 A

  VC = VCC - IC × RC
     = 12 - 0.228 × 470
     = 12 - 107.2 V  → transistor saturates

  At saturation, IC_max = (VCC - VCE_sat) / RC
                        = (12 - 1.0) / 470
                        = 23.4 mA

Final Answer:
  β_D = 22800
  The pair saturates; maximum switched current = 23.4 mA

Exam Tip: GATE often asks you to calculate β_D and then find collector current. Students forget the +β1+β2 term and use just β1×β2. For β=100 each, the error is only 2%, but the exact formula is β_D = β1×β2 + β1 + β2. Also remember VBE_total = 1.4V, not 0.7V. Using 0.7V in a bias calculation for a Darlington will give wrong base voltage and wrong operating point.

Key Properties

• Current gain is approximately β1 × β2. Two transistors with β = 150 each give β_D ≈ 22800.
• VBE_total ≈ 1.4V because two p-n junctions are in series. This must be accounted for in any biasing calculation.
• VCE_sat is typically 0.9V to 1.1V, much higher than the 0.2V of a single BJT. This limits efficiency in switching applications.
• Switching speed is slower than a single transistor because Q2 cannot turn off until Q1 has discharged its stored charge first.
• The TIP120 / TIP122 series are ready-made NPN Darlington transistors rated at 5A continuous, 60V, in a TO-220 package.
• A freewheeling diode is always added across an inductive load (relay coil, motor) when a Darlington is used as a switch.
• The input impedance is very high, approximately β_D × RE. This makes the pair ideal for interfacing high-impedance sensors.

Quick Revision

• β_D ≈ β1 × β2 + β1 + β2 for two cascaded BJTs.
• VBE_total ≈ 1.4V (two junctions in series).
• VCE_sat ≈ 0.9–1.1V, not 0.2V like a single BJT.
• TIP120 is a popular monolithic NPN Darlington: 5A, 60V, β ≥ 1000.
• Both collectors are tied together; only the outer base and outer emitter are used as terminals.
• Slower turn-off than a single BJT due to charge storage in Q2.
• Used in relay drivers, motor controllers, and Darlington arrays like the ULN2003 (7 NPN pairs, 500mA each).
• Exam trap: students use VBE = 0.7V instead of 1.4V when calculating the base bias resistor for a Darlington switch. This gives a base current that is too small to saturate Q2.