Darlington Transistor Calculations

Darlington transistor is a well known and popular connection using a pair of bipolar transistor junction transistor (BJT), designed for operating like an unified"superbeta"transistor. The following diagram shows the details of the connection.

Definition

A Darlington transistor can be defined as a connection between two BJTs that allows them to form a single composite BJT acquiring a substantial amount of current gain, which may range beyond thousand typically.

The main advantage of this configuration is that the composite transistor behaves like a single device having an enhanced当前增益equivalent to the product of the current gains of the each transistor.

If the Darlington connection comprises of two individual BJTs with current gains β₁and β₂可以使用公式来计算合并的电流增益：

β_D= β₁β₂-------- (12.7)

When matched transistors are used in a Darlington connection such that β₁= β₂= β , the above formula for the current gain gets simplified as:

β_D= β²-------- (12.8)

包装的达灵顿晶体管

Due to its immense popularity, Darlington transistors are also manufactured and available ready made in a single package which have two BJTs internally wired up as one unit.

The following table provides the datasheet of an example Darlington pair within a single package.

The indicated current gain, is the net gain from the two BJTs. The unit comes with 3 standard terminals externally, namely base, emitter, collector.

这种包装的达林顿晶体管external features similar to a normal transistor but have very high and enhanced current gain output, compared to the normal single transistors.

How to DC Bias a Darlington Transistor Circuit

The following figure shows a common Darlington circuit using transistors with a very high current gain β_D.

Here the base current can be calculated using the formula:

I_B= V_CC- V_BE/ R_B+ β_DR_E-------------- (12.9)

Although this may look similar to theequation which is normally applied for any regular BJT, the value β_Din the above equation will be substantially higher, and the V_BEwill be comparatively larger. This has been also proven in the sample datasheet presented in the previous paragraph.

Therefore, the emitter current can be calculated as:

I_E= (β_D+ 1)I_B≈ β_DI_B-------------- (12.10)

DC voltage will be:

V_E= I_ER_E-------------- (12.11)

V_B= V_E+ V_BE-------------- (12.12)

Solved Example 1

From the data given in the following figure, calculate the bias currents and voltages of the Darlington circuit.

Solution: Applying Eq.12.9 the base current is determined as:

I_B= 18 V - 1.6 V / 3.3 MΩ + 8000(390Ω) ≈ 2.56 μA

Applying Eq.12.10, the emitter current may be evaluated as:

I_E≈8000(2.56μ)≈20.28 mA≈我_C

Emitter DC voltage can be calculated using equation 12.11, as:

V_E= 20.48 mA(390Ω) ≈ 8 V,

Finally collector voltage can be assessed by applying Eq. 12.12 as given below:

V_B= 8 V + 1.6 V = 9.6 V

In this example the supply voltage at the collector of the Darlington will be:
V_C= 18 V

AC Equivalent Darlington Circuit

In the figure shown below, we can see aBJT发射机追随者circuit connected in Darlington mode. The base terminal of the pair is connected to an ac input signal through capacitor C1.

The output ac signal obtained through capacitor C2 is associated with the emitter terminal of the device.

上图显示上述配置的仿真结果。在这里，可以看到达灵顿晶体管被具有输入电阻的交流当量电路取代r_iand an output source of current represented asβ_DI_b

The AC Input Impedance can be calculated as explained below:

交流基础电流通过r_iis:

I_b= V_i- V_o/ r_i---------- (12.13)

自从
V_o=（i_b+ β_DI_b)R_E----------（12.14）

If we apply Eq 12.13 in Eq. 12.14 we get:

I_br_i= V_i- V_o= V_i- I_b(1 + β_D)R_E

解决上述V_i:

V_i= I_b[r_i+ (1 + β_D)R_E]

V_i/ I_b= r_i+ β_DR_E

Now, examining the transistor base, its ac input impedance can be evaluated as:

Z_i= R_B॥ r_i+ β_DR_E---------- (12.15)

Solved Example 2

Now let's solve a practical example for the above AC equivalent emitter follower design:

Determine the input impedance of the circuit, given r_i= 5 kΩ

Applying Eq.12.15 we solve the equation as given below:

Z_i= 3.3 MΩ ॥ [5 kΩ + (8000)390 Ω)] = 1.6 MΩ

Practical Design

Here's a practical Darlington design by connecting a2N3055 power transistorwith a small signal BC547 transistor.

A 100K resistor is used at the signal input side to reduce the current to a few millamps.

Normally with such low current at the base, the 2N3055 alone can never illuminate a high current load such as a 12V 2 amp bulb. This is because the current gain of 2N3055 is very low to process the low base current into high collector current.

However as soon as another BJT which is a BC547 here is connected with 2N3055 in a Darlington pair, the unified current gain jumps up into a very high value, and allows the lamp to glow at full brightness.

2N3055的平均电流增益（HFE）约为40，而对于BC547来说，这是400。当两者合并为达灵顿对时，增益大大增加到40 x 400 = 16000，太棒了。这就是我们能够从达灵顿晶体管配置中获得的功能，并且只需简单的修改即可将普通外观的晶体管变成一个额定评分的设备。