Exam question: measure amplifier \$h_{oe}\$ using a signal generator and oscilloscope

First, translate the specifications into constraint equations.

For the static power dissipation:

Assume, for now, that \$I_{R2} \ge 10 \cdot I_B = \dfrac{I_C}{10}\$ for the worst case \$\beta = 100 \$.

The supply current is then:

\$I_{PS} = I_C + 11 \cdot I_B = 1.11 \cdot I_C \$

The static power constraint then becomes:

\$\rightarrow I_C < \dfrac{25mW}{1.11 \cdot 10V} = 2.25mA\$

The bias equation:

The BJT bias equation is:

\$I_C = \dfrac{V_{BB} - V_{EE} - V_{BE}}{\frac{R_{BB}}{\beta} + \frac{R_{EE}}{\alpha}} \$

For this circuit, we have:

\$V_{BB} = 10V \dfrac{R_2}{R_1 + R_2}\$

\$V_{EE} = 0V\$

\$V_{BE} = 0.6V\$

\$R_{BB} = R_1||R_2\$

\$R_{EE} = R_E\$

So, the bias equation for this circuit is:

\$I_C = \dfrac{10V \frac{R_2}{R_1 + R_2} - 0.6V}{\frac{R_1||R_2}{\beta} + \frac{R_E}{\alpha}} \$

Now, you want less than 5% variation in \$I_C\$ for \$100 \le \beta \le 800\$. After a bit of algebra, find that this requires:

\$ \rightarrow R_E > 0.165 \cdot R_1||R_2 \$

Output swing:

The positive clipping level can be shown to be:

\$v^+_O = 3V = I_C \cdot R_C||R_L \$

The negative clipping level can be shown to be about:

\$v^-_O = -3V = I_C(R_C + R_E) - 9.8V \rightarrow 6.8V = I_C(R_E + R_C)\$

Put all this together:

Choose, for example, \$I_C = 1mA \$ then:

\$R_C||10k\Omega = 3k\Omega \rightarrow R_C = 4.3k\Omega\$

\$R_E + R_C = 6.8k\Omega \rightarrow R_E = 2.5k\Omega \$

Thus, \$V_E = 2.5V\$ and \$V_B = 3.1V\$

Then,

\$R_2 = \dfrac{V_B}{10 \cdot I_B} = \dfrac{3.1V}{100\mu A} = 31k\Omega \$

\$R_1 = \dfrac{10 - V_B}{11 \cdot I_B} = \dfrac{6.9}{110\mu A} = 62.7k\Omega \$

Now, check

\$0.165 \cdot R_1||R_2 = 3.42k \Omega > R_E \$

So, this doesn't meet the bias stability constraint equation we established earlier.

So run through this again (use a spreadsheet!) with larger \$I_C\$ until you've met the bias stability constraint equation.

If you can't meet the constraint with \$I_C < 2.25mA \$, you may need to increase current through the base voltage divider, e.g., \$I_{R2} = 20 \cdot I_B \$ and work through the static power constraint again.

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As the the correctness of the clipping level calculations above has been questioned, I simulated the circuit using values calculated from the above except that \$I_C \$ was increased to \$2mA\$ for the calculation.

The DC solution:

Driving the amplifier with a 500mV 1kHz sine wave:

Note the clipping levels are precisely +3V and -3V as designed. The variation in \$I_C\$ is just over 5% over the range of \$\beta\$ so the next step would be to increase the multiple of base current through R2 to e.g., 20 and plug in the numbers (which does result in meeting all the constraints).

The DC collector current is determined by \$R_E\$:

\$I_C = \alpha \dfrac{9.4V}{R_E} \approx \dfrac{9.4V}{R_E}\$

Since you require \$I_C < 1.25mA \$, the constraint equation is:

\$R_E > \dfrac{9.4V}{1.25mA} = 7.52k\Omega\$

The second requirement, maximum output voltage swing, without any other constraint, doesn't fix the collector resistor value.

We have:

\$ V_{o_{max}} = 19.8V - I_C(R_C + R_E)\$

But, the voltage across \$R_E\$ is fixed at 9.4V so:

\$V_{o_{max}} = 10.4V - I_C R_C\$

\$V_{o_{min}} = -I_C * R_C||R_L\$

If you stare at this a bit, you'll see that maximum output voltage swing is 10.4V but this requires that the product \$I_C R_C = 0\$* which is absurd.

Now, if we also require symmetric clipping, then, by inspection:

(1) \$V_{o_{max}} - V_{o_{min}} = 2 I_C (R_C||R_L)\$

(2) \$10.4V = I_C(R_C + R_C||R_L) \$

Looking at (1), note that, for maximum swing, we get more "bang for the buck" by increasing \$I_C \$ rather than \$R_C \$.

Since we have an upper limit on \$I_C\$, (2) becomes:

\$R_C + R_C||R_L = \dfrac{10.4V}{1.25mA} = 8.32k \Omega\$

which can be solved for \$R_C\$.

*unless \$R_L\$ is an open circuit