A single transistor can switch a heavy load, but it asks for a heavy base current in return: a β of 100 means 2 A of collector current needs 20 mA at the base — right at the limit of a microcontroller pin. The Darlington pair solves this by stacking two transistors so the first one drives the second: the current gains multiply, and a whisper of base current switches amps. This post designs one: a 5 V logic signal switching a 12 V, 2 A lamp.

Figure 1: The Darlington pair (a) and the base current it saves (b). Q1's emitter feeds Q2's base, so the two current gains multiply: β_total ≈ 100 × 100 = 10,000. (Schematic, not to scale.)

How the gain multiplies

In a Darlington pair the emitter of the first transistor (Q1, the driver) connects straight into the base of the second (Q2, the output), and the two collectors join together. Q1 amplifies the base current by β₁, and Q2 amplifies that result again by β₂:

β_total ≈ β₁ × β₂

With two ordinary transistors of β = 100 each, β_total ≈ 100 × 100 = 10,000 (the exact figure is β₁β₂ + β₁ + β₂ = 10,200, so 10,000 is a safe round number to design with). The pair behaves like one super-transistor with a single base, a single collector and a single emitter — with two small differences: the base-emitter voltage is doubled, V_BE(total) = 0.7 + 0.7 = 1.4 V, and the saturation voltage is higher, about 0.9 V instead of 0.2 V, because Q1 cannot saturate while Q2 is on. You can buy the pair ready-made in one package — the TIP120 is the classic example, with a datasheet minimum β of 1000.

Worked example — 5 V logic switches a 12 V lamp

Take a 12 V lamp with a resistance of 6 Ω (about 2 A at 12 V) and switch it with a 5 V microcontroller pin through a Darlington pair, using V_CE(sat) ≈ 0.9 V as the example value (the TIP120 datasheet allows up to 2 V at 3 A, so 0.9 V at 2 A is a fair working figure):

I_C = (12 − 0.9) / 6 = 1.85 A  ·  P_lamp = 1.85² × 6 = 20.5 W

The lamp sees 12 − 0.9 = 11.1 V instead of the full 12 V, so it draws 1.85 A and gives 20.5 W rather than its nominal 24 W — the price of the Darlington's higher saturation voltage. The power lost in the pair itself is:

P = V_CE(sat) × I_C = 0.9 × 1.85 = 1.67 W

Now the base current. The pair needs only I_C / β_total = 1.85 A / 10,000 = 185 µA. Give it a margin and drive it with 240 µA from the 5 V pin. The base resistor drops 5 − 1.4 = 3.6 V (the 1.4 V is the pair's double base-emitter voltage):

R_B = 3.6 V / 0.24 mA = 15 kΩ  ·  I_B = 3.6 / 15,000 = 0.24 mA

15 kΩ is a standard E24 value. Check saturation: 0.24 mA × 10,000 = 2.4 A of available collector current, more than the 1.85 A the lamp takes — the forced β is 1.85 / 0.00024 = 7,708, below the pair's 10,000, so the pair is properly saturated and fully on. The microcontroller pin supplies only 0.24 mA, far inside its ~20 mA limit.

How much base current does a single transistor need?

The same 1.85 A through one transistor with β = 100 would need I_B = 1.85 / 100 = 18.5 mA — almost the whole 20 mA a microcontroller pin can safely give, with no margin at all. A TIP120 Darlington (β minimum 1000 from its datasheet) needs 1.85 / 1000 = 1.85 mA. Two discrete β = 100 transistors need only 0.185 mA. That is the whole point of the pair: it turns a load the logic pin cannot drive into a load it barely notices.

The price — heat, and a slower turn-off

Nothing is free. The 1.67 W lost in the pair becomes heat, and a TO-220 package without a heatsink has a junction-to-ambient thermal resistance of about 62.5 °C/W (TIP120 datasheet figure):

ΔT = 1.67 × 62.5 = 104 °C  ·  T_j = 25 + 104 = 129 °C

That is below the 150 °C maximum junction temperature, but far too close for comfort — a hot day or a slightly higher load would push it over. Fit a small heatsink (about 25 °C/W total) and the rise is only 1.67 × 25 = 42 °C, giving T_j = 67 °C — safe with plenty of margin. The second price is speed: Q1's stored charge has nowhere to go when the base drive is removed, so a Darlington turns off more slowly than a single transistor. A resistor of about 1 kΩ from Q2's base to ground gives that charge an escape path and sharpens the turn-off — worth adding when the pair switches fast or often.

Key points to remember

• Darlington gain: β_total ≈ β₁ × β₂ — two β = 100 transistors give about 10,000 (exactly 10,200), so the pair acts like one super-transistor.
• Worked example: a 12 V, 6 Ω lamp with V_CE(sat) ≈ 0.9 V draws 1.85 A and 20.5 W; the pair loses 0.9 × 1.85 = 1.67 W.
• Base drive: the pair needs only 185 µA; a 15 kΩ resistor from 5 V logic gives 0.24 mA (after the 1.4 V double V_BE), with 2.4 A available — safely saturated, forced β 7,708.
• Comparison for the same 1.85 A: a single BJT (β = 100) needs 18.5 mA — at the ~20 mA microcontroller pin limit; a TIP120 (β min 1000) needs 1.85 mA; two discrete transistors need 0.185 mA.
• Heat: 1.67 W in a TO-220 with no heatsink (62.5 °C/W) reaches 129 °C junction — too hot; a small heatsink (25 °C/W) brings it to 67 °C.
• The trade-offs: V_CE(sat) ≈ 0.9 V instead of 0.2 V (the lamp gets 11.1 V, not 12 V), V_BE total is 1.4 V, and turn-off is slower — add a ~1 kΩ resistor from Q2's base to ground if it switches fast.