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Diseño de un amplificador con BJT:

• a gain of 10 a gain of 10
• 50 K Ω or higher input impedance 50 K Ω or higher input impedance
• 50 Ω or lower output impedance 50 Ω or lower output impedance
• Low freq response: 20 Hz Low Frequency response: 20 Hz
• High freq response: 100kHz. High Frequency Response: 100kHz.

Required: Required:

Design an amplifier using BJTs Design a BJT amplifier

The amplifier must-have: The amplifier must have:

• a gain of 10 a gain of 10
• 50 K Ω or higher input impedance 50 K Ω or higher input impedance
• 50 Ω or lower output impedance 50 Ω or less output impedance
• Low freq response: 20 Hz Low Frequency response: 20 Hz
• High freq response: 100kHz High Frequency Response: 100kHz

1) Use Standard Resistor Values \u200b\u200b 1) Use standard values

resistance

We are going to use two stages for the amplifier , A common emitter stage Followed by a common collector stage. We will use two-stage amplifier, a common-emitter stage followed by a common collector stage.

Assuming β = 300, Vbe = 0.7 V, Vcc = 15 V Assuming that β = 300, VBE = 0.7 V, Vcc = 15 V

Common Collector Stage common collector stage

Let Rc2 = 2.2K Ω, Re2 = 2.2K Ω, VCEQ = 7.5 V That Tc2 = 2.2 K Ω, K = 2.2 Ω RE2, = 7.5 V VCEQ

Let

R3 / / R4 = Rth = 60 Ω That R3 / / R4 = 60 Ω Rth =

Solving for R3 & R4 Solving for R3 and R4

R3 = 188K Ω, R4 = 88K = 188K Ω Ω R3, R4 = 88K Ω

Using standard resistor values \u200b\u200busing the standard values \u200b\u200bof resistance

R3 = 220K Ω, R4 = R3 = 100K 220K Ω Ω, R4 = 100K Ω

Rth (new) = 68.75K Ω Rth (new) = 68.75K Ω

The Equivalent Circuit Ac ac equivalent circuit

Rout: omitting the output resistance of the first stage Path: The omission of the output resistance of the first stage

Which is Less Than 50 Ω What is less than 50 Ω

2) Common Emitter Stage 2) common-emitter stage

This stage must-have to gain = 10, and input impedance 50KΩ ≥ This stage should have a gain = 10, and ≥ 50KΩ input impedance

A rough Approximation for the gain ≈ Rc1/Re1 An approach to gain ≈ Rc1/Re1

Let Rc1 = 10k, RE1 = 820Ω, V = 7.5 VCEQ That Rc1 = 10k, Re1 = 820Ω, V = 7.5 VCEQ

The Equivalent Circuit Ac ac equivalent circuit

We want Rin Rin ≥ ≥ 50KΩ 50KΩ We

Rin = Rth / / Rib Rin = Rth / / Rib

50K = Rth / / 258K 50K = Rth / / 258k

Solving for Solving Rth Rth

Rth = Rth = 62KΩ 62KΩ

Solving R3 & R4 for Solving for R3 and R4

R1 = 658K Ω, R2 = 68K Ω Ω R1 = 658K, R2 = 68 K Ω

Using standard resistor values \u200b\u200busing the standard values \u200b\u200bof resistance

R1 = 470K + 220K Ω Ω Ω = 690K, R2 = 68K Ω R1 = 470K + 220K = 690K Ω Ω Ω, R2 = 68 K Ω

Rout = 10k Ω Rc1 = Path = = 10K Ω Rc1

3) Frequency Response 3) Frequency response

Fl = 20 Hz, Fh = 100KHz Fl = 20 Hz, Fh = 100KHz

Using standard capacitor values \u200b\u200busing the standard values \u200b\u200bof capacitors

C1 = 0.16 uF , C2 = 0.12 uF, C3 = 0.033 uF = 0.16 uF C1, C2 = 0.12 mF, C3 = 0.033 mF

4) Simulation 4) Simulation

That Which shows VCEQ 1 = 8125-0566 = 7559 V VCEQ This shows that 1 = 8.125 to 0.566 = 7.559 V

VCEQ 2 = 11354-3658 = 7696 V VCEQ 2 = 11.354-3.658 = 7.696 V

PREPARED BY:

NERWIN ANTONIO MORA REINOSO

CI: 17,557,095

ESS SECTION 1