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Darlington pair Amplifier
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(A) Darlington connection

  • It is a popular connection of two BJT's to obtain one 'super beta' transistor.

  • It is shown in following figure:

enter image description here

  • Let, $\beta_{1}$ is gain of transistor Q$_{1}$ & $\beta_{2}$ is gain of transistor Q$_{2}$

Then, due to darlington connection, overall gain will be,

$\beta$ = $\beta_{1}.\beta_{2}$

  • This connection was first introduced by Dr. Sidney Darlington in 1953, hence the name.

(B) Darlington pair amplifier

  • Consider following circuit diagram, it shows darlington pair amplifier with emitter-follower configuration.

enter image description here

  • By applying KVL to input side of above circuit, we get,

V$_{CC}$ = I$_{B_{1}}R_{B}$ + V$_{BE_{1}}$ + V$_{BE_{2}}$ + I$_{E_{2}}R_{E}$

$\therefore$ By simplifying above eqn,

I$_{B_{1}}$ = $\frac{V_{CC} - V_{BE_{1}} - V_{BE_{2}}}{R_{B} + \beta_{D}.R_{E}}$ ....(I$_{E_{2}}$ = $\beta_{D}$) ....(I)

Also, the emitter current of Q$_{1}$ is equal to base current of Q$_{2}$

$\therefore$ I$_{E_{2}}$ = $\beta_{2}I_{B_{2}}$

= $\beta_{2}I_{E_{1}}$

= $\beta_{2}.(\beta_{1}.I_{B_{1}})$

$\therefore$ I$_{C_{2}}$ $\approx$ I$_{E_{2}}$ = $\beta_{1}.\beta_{2}$.I$_{B_{1}}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;$ = $\beta_{D} I_{B_{1}}$ .....(II)

  • Now, the collector voltage of both transistors,

$V_{C_{1}}$ = $V_{C_{2}}$ = $V_{CC}$

  • The emitter voltage of Q$_{2}$ is,

$V_{E_{2}}$ = I$_{E_{2}}R_{E}$

  • The base voltage of Q$_{1}$ is,

$V_{B_{1}}$ = $V_{CC}$ - I$_{B_{1}}R_{B}$

= $V_{E_{2}}$ + $V_{BE_{1}}$ + $V_{BE_{2}}$

$\therefore$ The collector emitter voltage of Q is,

$V_{CE_{2}}$ = $V_{C_{2}}$ - $V_{E_{2}}$

$V_{CE_{2}}$ = $V_{CC}$ - $V_{E_{2}}$ .............(III)

Eg. Find the value of I$_{E}$ and V$_{CE}$ for given darlington configuration

enter image description here

Given: $\beta_{1}$ = 80, $\beta_{2}$ = 100, V$_{BE}$ = 1.6V

Given that:

$\beta_{1}$ = 80

$\beta_{2}$ = 100

V$_{BE}$ = 1.6V

+V$_{CC}$ = 18V

To find:-

(i) I$_{E}$

(ii) V$_{CE}$

Solution:-

We know that,

I$_{E}$ = $\beta_{1}$.$\beta_{2}$.I$_{B}$ ....................................(I)

& I$_{B}$ = $\frac{V_{CC} - (V_{BE})}{R_{B} + \beta_{D}.R_{E}}$

= $\frac{V_{CC} - V_{BE}}{R_{B} + \beta_{1}.\beta_{2}.R_{E}}$

= $\frac{18 - 1.6}{3.3 \times 10^6 + (80 \times 100 \times 390)}$

I$_{B}$ = 2.55 $\mu$A.

$\therefore$ Substituting in above eqn (I),

we get,

I$_{E}$ = $80 \times 100 \times 2.55 \times 10^-6$

I$_{E}$ =20.4 mA .......................................(A)

Now,

V$_{CE}$ = V$_{CC}$ - V$_{E2}$

= V$_{CC}$ - I$_{E}$R$_{E}$

= 18 - 20.4 $\times 10^-3 \times390$

V$_{CE}$ = 10.044V ..................................(B)

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