High Gain Amplifier using BJT
Here's an answer that will achieve at least a gain of 1500 from just one common BJT. Details about the design below will, I think, make a clear example out of why you would not design one like this. But your instructor wants just a few parts and doesn't seem to care about anything else. So:







simulate this circuit &ndash; Schematic created using CircuitLab

I've used a 2N5550 because it can tolerate the required higher \$V_{CEO}\$ needed here. I've used a single \$90\:\textrm{V}\$ power supply so that I could get the necessary gain. I've assumed that \$Q_1\$ operates at room temperature and since it will only experience less than \$50\:\textrm{mW}\$, it probably won't be any higher than perhaps \$10\:^\circ\textrm{C}\$ above ambient, anyway. Not enough difference to matter here.

The design follows this logic:


I assumed temperature variation is modest enough to ignore.
Voltage gain will be \$A_V=\frac{r_o\vert\vert R_C}{r_e}=\frac{\left(r_o\vert\vert R_C\right)\cdot I_C}{V_T}\$.
Since \$A_V\ge 1500\$, it follows that: \$\left(r_o\vert\vert R_C\right)\cdot I_C\ge 39\:\textrm{V}\$
I chose to more than double this and set \$V_{CC}=90\:\textrm{V}\$ on the usual basis of setting the quiescent collector voltage "midway" and to allow room for the Early effect, too.
To further increase room for gain, I decided to lower the quiescent collector voltage to \$20\:\textrm{V}\$
I wanted to minimize distortion (which means reducing variation around the quiescent collector current), and taking into account #5 above, I decided the maximum output swing would be \$v_{pk}=5\:\textrm{V}\$
Decision #6 suggests that the maximum input swing is \$v_{pk}=\frac{5\:\textrm{V}}{1500}\approx 3\:\textrm{mV}\$


First thing is to pull up 2N5550 specs from an ORCAD model I found: \$V_A=100\:\textrm{V}\$, \$I_{SAT}=2.511\:\textrm{fA}\$, \$n_e=1.241\$, \$I_{KF}=.3495\$, and \$\beta_F=213.4\$.

Don't think that the \$\beta_F\$ is accurate, though. The \$n_e=1.241\$ and \$I_{KF}=.3495\$ are nominally \$n_e=1.5\$, \$I_{KF}=.4\$, which yields an adjustment factor of very close to 1 for typical collector currents. But given these values, and a rough idea of the collector current I'm expecting, the modified \$\beta\$ for purposes of estimating the base current will be about \$\beta_{ADJ}=\frac{213.4}{2.04}\approx 105\$ for our purposes below.



I probably should have used a different BJT to keep this simpler, but this illustrates an issue regarding some BJTs. These two values may vary a lot, but it usually works out that they don't combine to create such a difference.

The 2N5550 is one of the weird ones in this regard. It's adjusted \$\beta\$ varies more substantially over collector current than is common and its effective \$\beta\$ isn't what you'd imagine if only looking at one of its spice parameters.

But using this device actually makes an exaggerated example about why one does NOT depend upon \$\beta\$ for biasing a BJT like this. Not only does the circuit overly depend upon \$\beta\$ for the quiescent collector bias point, but the \$\beta\$ value of any one given physical unit of this part family can vary a lot depending on even modest changes in collector current.



From the above, I know that:

$$\begin{align*}R_C&amp; \ge \frac{\left(V_A+V_{C_Q}\right)\cdot 1500}{V_A+V_{C_Q}-1500\cdot V_T}\cdot\frac{V_T}{I_C}\\\\&amp;\ge 38.5\:\textrm{k}\Omega\end{align*}$$

In calculating the above, I decided to estimate \$I_C\approx 1.5\:\textrm{mA}\$. However, once it computed out a value for \$R_C\$, I then selected a higher standard value of \$R_C=45\:\textrm{k}\Omega\$ for it.

This choice now means I'm predicting a quiescent collector current of \$I_{C_Q}=\frac{90\:\textrm{V}-20\:\textrm{V}}{45\:\textrm{k}\Omega}=1\frac{5}{9}\:\textrm{mA}\$. I also therefore get \$r_o= \frac{V_A+V_{C_Q}}{I_C}=\frac{100\:\textrm{V}+20\:\textrm{V}}{I_C}\approx 77\:\textrm{k}\Omega\$, suggesting that my gain will be \$A_V=\frac{\left(r_o\vert\vert R_C\right)\cdot I_C}{V_T}\approx 1700\$.

The base current calculation was complicated by those factors. But we can now compute an estimated value as \$\frac{I_C}{\beta_{ADJ}}\approx 15\:\mu\textrm{A}\$. Using a rough estimate that \$V_{BE}\approx 700\:\textrm{mV}\$, I figure \$R_1=\frac{90\:\textrm{V}-700\:\textrm{mV}}{14\:\mu\textrm{A}}\approx 6\:\textrm{M}\Omega\$. I used a standard value close to that, but larger because I already used a lowish collector voltage and I don't want to push it any lower than that.

That completes the design.

Test it out in spice using an input source with no more than \$3\:\textrm{mV}\$ peak. Unloaded, the output should be about right, I think. Here's a model:


  .model 2N5550 NPN(Is=2.511f Xti=3 Eg=1.11 Vaf=100 Bf=213.4 Ne=1.241
  Ise=2.511f Ikf=.3495 Xtb=1.5 Br=3.24 Nc=2 Isc=0 Ikr=0 Rc=1 Cjc=4.883p
  Mjc=.3047 Vjc=.75 Fc=.5 Cje=18.79p Mje=.3416 Vje=.75 Tr=1.212n
  Tf=560.1p Itf=50m Vtf=5 Xtf=8 Rb=10)


(For high gain arrangements like this, by the way, there is also a small effect called the "Late effect" which might impinge. But I think the effect is likely small enough not to seriously impair the predicted gain.)



So that's what you get without any specifications to speak of.