Electronic – Biasing of CE amplifier with emitter resistor for specific gain

I am trying to design a Common-Emitter amplifier to the following specifications:

1. Midband voltage gain of 50
2. Frequency range 100Hz to 20kHz
3. Load \$5k\Omega\$ with a coupling capacitor (not shown below)
4. 12V supply lines
5. Input source resistance \$100 \Omega\$

Using a 2N2222 BJT transistor and the following CE configuration:

I've drawn the small-signal model as follows, assuming that in the midband coupling capacitors are treated as shorts and bypass and load capacitors are treated as open circuits:

I began my design by picking a maximum current. I want to keep that low so I choose \$2mA\$ and plot the IC vs VCE curves for this specific transistor in a spice simulator:

Choosing a point half way on the load line for symmetrical swing, I obtain:

• \$V_{CE} = 6V\$
• \$I_C = 1mA\$
• \$I_B = 5.5\mu A\$

I calculate \$ \beta = \frac{I_C}{I_B} = \frac{1mA}{5.5\mu A} = 182\$

\$R_c = \frac{V_{cc}}{I_c}=\frac{12}{2mA}=6000 \Omega\$

I pick \$R_E = 0.1R_c = 600 \Omega\$ based on a rule of thumb for beta stability.

Now I would like to design my bias network \$R_1, R_2\$ to enforce the bias conditions above and also give a gain of 50. The voltage gain expression is given as:

\$A_v = – \frac{\beta R_c || R_l}{r_\pi + (1+\beta)R_E}(\frac{R_i}{R_i+R_s})\$

I first calculate

\$r_\pi = \frac{V_T}{I_B} = \frac{0.026V}{5.5\mu A} = 4727 \Omega\$

\$R_{ib} = r_\pi + (1+\beta)R_E = 4727 + (183)(600) = 114527 \Omega\$

Solving for the input resistance:

\$A_v = 50 = \frac{182(6000) || (5000)}{4727 + (183)(600)}(\frac{R_i}{R_i+100})\$

Giving \$R_i=92.02 \Omega\$

\$R_i = R_{thev} || R_{ib}\$

Some algebraic manipulation results in

\$R_{thev} = \frac{-R_i R_{ib}}{R_i – R_{ib}} = 92 \Omega\$

I write a loop around the Emitter-Base loop as:

\$ -Vcc(\frac{R_2}{R_2 + R_1}) + R_{thev} + 0.7 + I_E R_E = 0 \$

Selecting \$R_2 = 6000 \Omega\$, \$R_1\$ is solved for \$50 000 \Omega\$

Simulating the circuit and running a dynamic DC analysis shows that the bias conditions are enforced:

However I am confused because when I work out \$R_i\$ as:

\$ R_{i} = R_{thev} || R_{ib} = \frac{1}{\frac{1}{6000} + \frac{1}{50 000} + \frac{1}{114527.27}} = 5117 \Omega \$

And furthermore the thevenin resistance of \$R_1, R_2\$

\$R_{Thev} = \frac{R_1 R_2}{R_1 + R_2} = 5357 \Omega\$

I would expect these values to be the same as what I worked them out to be previously from the voltage gain equation (\$92 \Omega \$)? In simulation this method seems to work however I can't understand how the input and thevenin resistances have seemingly changed value. Could anyone explain what is happening here?

I'd also be interested to know how more experienced designers would approach this simple design problem. I find that my courses at university are very theoretical and so I tend to over complicate the designs. In practice I suspect that designers tend to take more of a heuristic approach to circuit design rather than solve equations like this.