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Electrical Engineering - Circuits - Darlington Configuration

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Homework Statement



Determine [itex]A_{V}[/itex] and [itex]A_{i}[/itex] given [itex]β_{D} = 8000[/itex] and [itex]V_{BE} = 1.6 V[/itex].

http://img801.imageshack.us/img801/6272/44tv.png [Broken]

Homework Equations





The Attempt at a Solution



I'm not exactly sure how to solve this problem. This is a problem that my professor solved in class. The only problem is that I'm having a hard time understanding how he solved. I believe he made the approximation of assuming that the two transistors are one transistor. I don't see how else one is supposed to solve this problem. Under this assumption I drew the small signal equivalent.

http://img43.imageshack.us/img43/4683/y7v6.png [Broken]

I start off by calculating [itex]I_{B}[/itex].

[itex]I_{B} = \frac{V_{CC} - V_{BE}}{R_{B} + β_{D}R_{E}} = \frac{18 V - 1.6 V}{3.3 MΩ \frac{10^{6} Ω}{MΩ} + (8000)390 Ω} ≈ 2.555x10^{-6} A[/itex]

Next I calculate [itex]I_{O}[/itex].

[itex]I_{C} = β_{D}I_{B} = 8000(2.555x10^{-6} A) = 2.04x10^{-2} A ≈ I_{E} = I_{O}[/itex]

Also [itex]V_{CC} = V_{C} = 18 V[/itex].

Next I calculate [itex]V_{O}[/itex].

[itex]V_{O} = V_{E} = V_{R_{E}} = I_{E}R_{E} = (2.04x10^{-2} A)(390 Ω) = 7.956 V[/itex]

Next I calculate [itex]V_{CE}[/itex].

[itex]V_{CE} = V_{C} - V{E} = 18 V - 7.956 V = 10.044 V[/itex]

Next I calculate [itex]r_{e}[/itex].

[itex]r_{e} = \frac{26 mV \frac{V}{10^{3} mV}}{2.04x10^{-2} A} ≈ 1.275 Ω[/itex]

Next I calculate [itex]β_{D}r_{e}[/itex].

[itex]β_{D}r_{e} = 8000(1.275 Ω) = 10200 Ω[/itex]

Next I calculate [itex]V_{βr_{e}}[/itex].

[itex]V_{βr_{e}} = I_{B}βr_{e} = (2.555x10^{-6} A)(10200 Ω) ≈ 2.606x10^{-2} V[/itex]

Next I calculate [itex]V_{I}[/itex].

[itex]V_{I} = V_{B} = V_{E} + V_{βr_{e}} = 7.956 V + 2.606x10^{-2} V ≈ 7.982 V[/itex]

Now I'm able to calculate [itex]A_{V}[/itex].

[itex]A_{V} = \frac{V_{O}}{V_{I}} = \frac{7.956 V}{7.982 V} ≈ 0.997[/itex]

Next I calculate [itex]I_{R_{B}}[/itex]

[itex]I_{R_{B}} = \frac{V_{B}}{R_{B}} = \frac{7.982 V}{3.3 MΩ \frac{10^{6} Ω}{MΩ}} ≈ 2.419x10^{-6} A[/itex]

Next I calculate [itex]I_{I}[/itex] using Kirchhoff's Current Law.

[itex]I_{I} = I_{R_{B}} + I_{B} = 2.419x10^{-6} A + 2.555x10^{-6} A = 4.974x10^{-6} A[/itex]

Finally I'm able to calculate [itex]A_{I}[/itex].

[itex]A_{I} = \frac{I_{O}}{I_{I}} = \frac{2.04x10^{-2} A}{4.974x10^{-6} A} ≈ 4104.628[/itex]

I don't know if these approximations are accurate and I'm not exactly sure how to properly draw the small signal analysis and how to solve this problem another way without being given β for each transistor. I'm concerned that my [itex]A_{V} < 1[/itex] and that my [itex]A_{I}[/itex] is so big. I would appreciate any help with this problem.

Thanks.
 
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Answers and Replies

  • #2
rude man
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Your numbers are certainly approximately correct.

For an emitter follower the voltage gain is always < 1.

Your current gain seems a bit low, but it might be spot-on. I personally don't use models and assume the base voltage does not change with input voltage, in which case the current gain = beta = 8000. So your current gain is certainly not high. (You on the other hand are not allowed to make that assumption! :frown: )

In real lfe this circuit would never work. Beta is a very unstable (with temperature) and non-repeatable parameter for any transistor from unit to unit, so the emitter voltage would vary all over the place. Typically, circuits are designed assuming beta = infinity

Anyway, you were right to assume the Darlington is a single transistor with Vbe ~ 1.6V and beta = 8000.
 

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