Two stage amplifier question
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Two stage amplifier question

[From: ] [author: ] [Date: 13-11-05] [Hit: ]
5V + 708mV = 8.208VI know now that the current through the collector resistor of Q₁ is:I( 10kΩ ) = (15V - 8.208V) ⁄ (10kΩ) = 679.2μAHowever, the collector current of Q₁ (Ic₁) is that value, less the base current Ib₂ required by Q₂,......

        Vbe₂ ≈ 656mV + 25.9mV ⋅ ln( (7.463mA) ⁄ 1mA ) ≈ 708mV

Adding that to Ve₂, I get:

        Vb₂ = 7.5V + 708mV = 8.208V

I know now that the current through the collector resistor of Q₁ is:

        I( 10kΩ ) = (15V - 8.208V) ⁄ (10kΩ) = 679.2μA

However, the collector current of Q₁ (Ic₁) is that value, less the base current Ib₂ required by Q₂, so:

        Ic₁ = 679.2μA - 37.3μA = 641.9μA

If Q₁ is in the normal active mode, as well, then it's base and emitter currents will be:

        Ib₁ = Ic₁ ⁄ β = (641.9μA) ⁄ 200 ≈ 3.21μA
        Ie₁ = ((β+1)⋅Ic₁) ⁄ β ≈ 645.1μA

Now that we know Ie₁, we can compute Ve₁:

        Ve₁ = Ie₁ ⋅ 1800Ω  ≈ 1.161V

Again, we can estimate Vbe₁ now as:

        Vbe₁ ≈ 656mV + 25.9mV ⋅ ln( (641.9μA) ⁄ 1mA ) ≈ 645mV

Adding that to Ve₁, we get the quiescent base voltage for Q₁ as:

        Vb₁ = Ve₁ + Vbe₁ = 1.161V + 645mV = 1.806V

We know now the current through the 220kΩ resistor as:

        I( 220kΩ ) = (15V - 1.806V) ⁄ (220kΩ) ≈ 60μA

(Hmm. Nice number.)

But we know that Ib₁ must be subtracted from that to get the current in R₁. And since we know the voltage across R₁, it's easy to now compute R₁ as:

        R₁ = (1.806V) ⁄ (60μA - 3.21μA) ≈ 31.8kΩ

The above analysis answers (b) and (c).

The little-re of the 1st stage is about 40Ω, so the input resistance of the amplifier is:

        R( input ) = 220kΩ || 31.8kΩ || (β+1)⋅(1.8kΩ+40Ω) ≈ 25.8kΩ

This answers (d).

The voltage gain of the emitter follower, Q₂, is only slightly less than 1:

        Av₂ ≈ α ≈ β ⁄ (β+1) ≈ .995

But the voltage gain of the common emitter, Q₁, is about:

        Av₁ ≈ -(10kΩ) ⁄ (1800Ω) ⋅ α

Combining these two, you get:

        Av ≈ -(10kΩ) ⁄ (1800Ω) ⋅ α² ≈ -5.5

This answers (e).

If Ce were connected, Av would be quite different and would depend upon the little re value of 40Ω (because the 1800Ω value would be bypassed.)
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