Differential Amplifier Design
In this assignment, you will design the microphone pre-amplifier highlighted with red in the block diagram, which is a differential amplifier as shown in the schematics below.
Your design should satisfy the required differential gain, input impedance, single-ended common-mode gain, and the power dissipation specifications provided in Section A below. Then in Sections B and C, you will simulate your circuit on LTSpice to compare the simulation results with hand calculations.
Hand design: Design the bipolar differential amplifier and the current source and bias network (π 1, π3, πππ π4) above such that:
(i) Differential gain: π΄π β₯ 200 π π,
(ii) Input differential resistance: π ππ β₯ 50 πΞ©,
(iii) π΄ππ < 0.1 where π΄ππ is the single-ended common-mode gain (the gain to a common-mode input signal when the output is measured not differentially but from one of the outputs with respect to ground).
(iv) To keep the battery discharge lifetime long, the power consumption of the differential amplifier, πππππ΄πΏ, must be less than 2 milliwatts.
Design the circuit with BJTs having π½ = 200, ππ΄ = 100 π, πππ ππ΅πΈ = ~0.7 π ππ πΉπππ€πππ π΄ππ‘ππ£π. Use +ππ·π· = 9 π and βπππ = β9 π. Clearly show your steps.
Design Suggestion for Part A.
Derive the expression for π ππ (ignore ππ of Q1 and Q2.). Replace the small signal parameter(s) with βdc currentsβ, βresistorsβ, βππ‘β”, βπ½β, βππ΄”, etc. and simplify the expressions to the extent possible (e.g., manipulate the expressions and then replace ππ΄ = 100 π, π½ = 200, etc.). The π ππ design specification will let you find an upper limit for Q1 (and Q2) dc collector current, πΌπΆ1(= πΌπΆ2).
The total power dissipation is πππππ΄πΏ = πππ·π· + ππππ = 9π β (πΌππ·π· + πΌπππ). Here, πΌππ·π· is the sum of all dc currents leaving the +ππ·π· and πΌπππ is the sum of all dc currents entering the βπππ. In πππππ΄πΏ write πΌππ·π· and πΌπππ in terms of Q1 (or Q2) collector currents (ignore the current on the reference current generation branch formed by π 1 and π4). The πππππ΄πΏ design specification will let you find another upper limit for πΌπΆ1(= πΌπΆ2).
Derive the expressions for π΄π and π΄ππ. (When deriving π΄π, include ππ of Q1 or Q2. When deriving π΄ππ ignore ππ of Q1 and Q2 but include ππ of Q3.). Replace the small signal parameter(s) with βdc currentsβ, βresistorsβ, βππ‘β”, βπ½β, βππ΄”, etc. and simplify the expressions to the extent possible (e.g., manipulate the expressions and then replace ππ΄ = 100 π, π½ = 200, etc.). The π΄π and π΄ππ design specifications will let you respectively find lower and upper limits for the product π πΆ β πΌπΆ1.
Consider the forward-active region requirement of Q1 (or Q2). For ππΆπ = 0 π, find another upper limit for the product π πΆ β πΌπΆ1.
Consider the upper limits for πΌπΆ1 to pick an πΌπΆ1. Find π 1 based on the value you pick for πΌπΆ1. Then use the upper and lower limits for π πΆ β πΌπΆ1 to pick π πΆ.
In your simulations, use the BJT model 2N2222 of NXP, which has a SPICE model as below with ππ΄ and π½ highlighted:
DC Analysis: In LTSpice do a DC operating point simulation (.op) with both inputs connected to ground(i.e., ππΆπ = 0 π).
Find the simulated DC values for πΌπ 1, πΌπΆ3, πΌπΆ4, πΌπΆ1, πΌπΆ2, ππ΅3, ππΈ1,2, ππ1, ππ2. Compare them with your hand calculations. Additionally, comment on the matching between πΌπ 1 πππ πΌπΆ3 and comment on the theoretical vs. simulated match between πΌπ 1 πππ πΌπΆ3.
Transient Analysis: In LTSpice do a transient simulation (.tran) for 100 ms.
For differential small-signal input simulations:
Apply π£ππ = 1 πππ π πππ’π πππππ π πππππ ππ‘ 100 π»π§. [i.e., π£ππ1 = + π£ππ 2 = 0.5 ππ sin (2 β π β 100π»π§ β π‘) and π£ππ2 = β π£ππ 2 = 0.5 ππ sin ((2 β π β 100π»π§ β π‘) + π) with DC offset = 0V. ]
For common-mode small-signal input simulations:
Apply π£ππ = 1 πππ π πππ’π πππππ π πππππ ππ‘ 100 π»π§. [i.e., π£πππ1 = π£πππ2 = π£ππ = 1 ππ sin (2 β π β 100π»π§ β π‘) with DC offset = 0V.]
For the differential small-signal input, what is the expected emitter voltage of Q1 and Q2, π£π1(= π£π2)? Plot the simulated waveform. What is the simulated value of π£π1(= π£π2)?
Plot π£ππ, π£ππ(π£ππ = π£π2 β π£π1), πππ πππ. Note that πππ is the base current of Q1 (πππ = ππ1).
Calculate the simulated π΄π = π£ππ/π£ππ and π ππ = π£ππ/πππ. Compare the values with your design targets.
If the simulation results do not match the design constraints, tune your circuit to achieve the goals.
For the common-mode small-signal input, plot π£ππ and π£πππ. (π£πππ = π£πππ2 = π£πππ1 π€βππ π‘βπ ππππ’π‘ ππ π ππππππ β ππππ π πππππ)