Darlington Transistor: What is it? (Darlington Pair)

Contents

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Key learnings:

Darlington Transistor Definition: A Darlington transistor is a semiconductor device that pairs two BJTs to achieve a very high current gain, functioning as a unified component.
Current Amplification: The current gain of a Darlington pair is significantly higher than that of standard transistors, making it ideal for applications needing strong amplification.
Circuit Integration: In a Darlington pair, the emitter of the first transistor connects to the base of the second, allowing for compounded current amplification in a compact form.
Switching Capability: Darlington transistors are particularly effective in switching applications where high currents are controlled by low input currents.
Performance Characteristics: While they offer high current gains and compact design, Darlington transistors have slower switching speeds and higher base-emitter voltages, which can be a limitation in rapid-switching or low-voltage applications.

What is a Darlington Transistor?

A Darlington transistor, or Darlington pair, combines two BJTs (Bipolar Junction Transistor) to provide high current gain. The emitter current of the first transistor drives the base of the second, so the pair needs much less input current than a single power BJT at the same collector current.

The two-BJT connection is called a Darlington Pair. It may use separate transistors or a three-terminal integrated package with one external base, collector and emitter. Sidney Darlington patented the configuration in 1953.

If the current gain values are β1 and β2, the exact common-emitter current gain is β1β2 + β1 + β2. The product β1β2 is a useful approximation only when both gains are large. “Super Beta Transistor” is another name for this high-gain connection.

Darlington Transistor Circuit

A Darlington pair uses two PNP transistors or two NPN transistors of the same polarity. Their collectors are connected together, and the first emitter drives the second base. “Back to back” does not describe this connection.

The input is applied to the first base and the external emitter is the second emitter. At the common collector terminal, current is the sum of the collector currents of both devices. An integrated Darlington exposes these nodes as three terminals, as shown in the figure.

The circuit contains transistors Q1 and Q2.

I_b1 = Base current of transistor Q1
I_e1 = Emitter current of transistor Q1
I_b2 = Base current of transistor Q2
I_e2 = Emitter current of transistor Q2

The package input current is the base current of Q1:

$I_{b1} = I_b$

The external emitter current is the emitter current of Q2. However, the protected equation below labels Q2 collector current as total collector current; for a common-collector Darlington, total collector-terminal current is the sum of Q1 and Q2 collector currents.

$I_{c2} = I_c$

V_be1 = base-emitter voltage of transistor Q1
V_be2 = base-emitter voltage of transistor Q2

The input-to-output base-emitter voltage is the sum of the two junction voltages:

$V_{be} = V_{be1} + V_{be2}$

β₁ = Current gain of transistor Q1
β₂ = Current gain of transistor Q2

The symbol β_D normally denotes common-emitter collector current gain. The protected derivation below instead uses external emitter current divided by base current, so its exact result includes one more than the collector-current gain.

$Current \, gain \, \beta_D = \frac{Output \, Current}{Input \, Current}$

(1) $\begin{equation*} \beta_D = \frac{i_e}{i_b} = \frac{i_{e2}}{i_{b1}} \end{equation*}$

For each BJT, emitter current is the sum of base and collector current, and β relates collector current to base current:

$i_e = i_c + i_b$

$i_e = \beta i_b + i_b$

(2) $\begin{equation*} i_e = (\beta + 1) i_b \end{equation*}$

For transistor Q2:

$i_{e2} = (\beta_2 + 1) i_{b2}$

Using equation 1:

$\beta_D = \frac{( \beta_2 + 1) i_{b2}}{i_{b1}}$

From the circuit, Q1 emitter current becomes Q2 base current:

$i_{e1} = i_{b2}$

$\beta_D = \frac{( \beta_2 + 1) i_{e1}}{i_{b1}}$

For transistor Q1:

$i_{e1} = (\beta_1 + 1) i_{b1}$

$\beta_D = \frac{( \beta_2 + 1) (\beta_1 + 1) i_{b1} }{i_{b1}}$

$\beta_D = ( \beta_2 + 1) (\beta_1 + 1)$

$\beta_D = \beta_1 \beta_2 + \beta_1 + \beta_2$

The product β₁β₂ dominates the sum β₁ + β₂ when both gains are large. For example, let β₁=100 and β₂=100.

Then β₁β₂ = 10000 and β₁+β₂ = 200. The β₁+β₂ correction makes the exact collector-current gain 10,200. The earlier protected emitter-current derivation should have an additional +1 term before this approximation.

$\beta_D = \beta_1 \beta_2$

PNP and NPN Darlington Transistor

Two PNP devices form a PNP Darlington, while two NPN devices form an NPN Darlington. The figure shows both connections.

Both types connect the collectors together and connect the first emitter to the second base. Current directions and voltage polarities reverse between the NPN and PNP versions.

An integrated Darlington can reduce component count and board area. Separate devices offer more choice in ratings and thermal layout.

Darlington Transistor Switch

The following example compares a single BJT and a Darlington as a microcontroller-driven low-side switch. It is illustrative only: a real design must use the transistor datasheet, the controller’s guaranteed output limits, a base resistor, load transients and thermal calculations.

The example assumes a 5 A load and a microcontroller output limited to 20 mA. That limit is device-specific, and a microcontroller sources current into the base rather than drawing it.

The assumed saturated collector current is:

$I_c (sat) = 5 A$

The example assumes an active-region current gain β of 100:

$\beta = 100$

The following division estimates active-region base current. It does not prove saturation because forced beta, temperature and device spread must be considered:

$I_b = \frac{I_c(sat)}{\beta}$

$I_b = \frac{5}{100}$

$I_b = 50mA$

The claim that deep saturation always needs five times this current is not a general rule. Use a datasheet VCE(sat) test condition or a documented forced-beta target; 250 mA would exceed ordinary microcontroller output ratings.

Under the stated 20 mA output limit, the microcontroller cannot directly provide the example’s assumed single-BJT base drive.

The next calculation assumes βd = 10000. Do not use nominal active-region gain alone to size switch drive at saturation; verify the guaranteed base-current and saturation-voltage conditions in the selected device’s datasheet.

$I_{bd} = \frac{I_c (sat)}{\beta_d}$

$I_{bd} = \frac{5}{10000}$

$I_{bd} = 0.5mA$

The same unsupported factor of five gives 2.5 mA in this worked arithmetic. A practical Darlington may need more base current and can drop several volts at high collector current, so check controller voltage, base resistance, load voltage and heat dissipation.

A Darlington can reduce required drive current, but it is not automatically suitable for the same load. For high-current, low-voltage switching, a logic-level MOSFET may waste less power because it can have a much lower on-state voltage.

TIP120 Darlington Transistor

TIP120 is an NPN power Darlington. The ST datasheet guarantees DC gain of at least 1000 only at specified active-region test points, not under every current, voltage or temperature. Suitability for an Arduino or other microcontroller depends on base drive, load transients, voltage drop and thermal design.

For the cited TO-220 package, pin 1 is base, pin 2 and the tab are collector, and pin 3 is emitter. Confirm the pinout in the datasheet for the exact manufacturer and package.

In a common NPN low-side switch, pin 1 receives current through a base resistor, pin 2 connects to the load and pin 3 connects to the return. Other circuit configurations use the terminals differently.

The ST datasheet lists 5 A collector current and 8 A peak collector current as absolute maximum ratings, not guaranteed operating points. Actual current is limited by the safe-operating-area curve, pulse duration, junction temperature, saturation loss and heatsinking. At 5 A and 20 mA base current, specified VCE(sat) can be as high as 4 V.

TIP120 has a 60 V collector-emitter rating. TIP121 and TIP122 are rated at 80 V and 100 V respectively, but voltage rating alone does not establish suitability. Check current, power, safe operating area, switching and transient requirements.

The equivalent circuit of TIP120 is as shown in the below figure.

How to Test a Darlington Transistor?

A diode-mode check can find shorts or open junctions, but it cannot confirm full gain, leakage, switching or safe operation. Disconnect power and isolate the device from paths that can affect the reading before testing.

Step-1: Identify the base, collector and emitter from the exact datasheet and determine whether the device is NPN or PNP. The steps below describe an NPN Darlington; reverse meter polarity for PNP.

Step-2: Set a digital multimeter (DMM) to diode mode. Low resistance mode is less informative and may not supply enough voltage to forward-bias two junctions.

Step-3: Place the positive lead on base and the negative lead on emitter. Then repeat from base to collector, accounting for any internal resistors shown in the datasheet.
A base-to-emitter path normally shows about two forward junction drops; diode mode displays voltage, not hFE.

Step-4: Reverse each lead pair and check collector to emitter in both directions with no base drive.
Unexpected near-zero readings suggest a short. Internal resistors, protection components and in-circuit paths can prevent a simple open-circuit result, so compare readings with the device schematic.

Advantages of Darlington Transistor

A Darlington pair offers these advantages over a single BJT in suitable circuits:

High current gain allows a small base current to control a larger collector current.
In an emitter-follower circuit, high input impedance and local feedback can provide a lower output impedance than one transistor.
An integrated package reduces part count and simplifies PCB placement.

Disadvantages of Darlington Transistor

The two-transistor path also introduces these trade-offs:

Stored charge can make turn-off slower than for a suitable single transistor or MOSFET.
The input-to-emitter voltage includes two base-emitter junction drops.
Higher saturation voltage can cause substantial conduction loss and heat at high current.
The additional transistor and capacitance limit high-frequency performance.
Its extra high-frequency phase shift can reduce stability margin at a critical frequency in a feedback circuit.