What is the maximum voltage gain that this amplifier can provide?
I can't provide you a direct answer -- only you know what's being asked of you -- but it's kind of painful to leave you in a lurch, too, squirming over the question of the quiescent current. So I'll draw out some simple logic and leave it to you to work things out for yourself, from there.

The highest voltage gain that can be hoped for with a BJT uses a grounded emitter. Without wasting time right now on DC biasing questions, the circuit will look something like this very simple behavioral example:







simulate this circuit &ndash; Schematic created using CircuitLab

We expect the magic biasing network to set things up such that, without a signal input being applied, the DC biasing point will place \$V_\text{C}\$ at a nominal, quiescent point we'll call \$V_{\text{C}_\text{Q}}\$.

The quiescent collector current is then:

$$I_{\text{C}_\text{Q}}=\frac{V_\text{CC}-V_{\text{C}_\text{Q}}}{R_\text{C}}$$

We know also then that the quiescent dynamic emitter resistance is approximately:

$$r_e\approx \frac{V_T}{I_{\text{C}_\text{Q}}}$$

(Technically, it's smaller by the factor \$\frac{\beta}{\beta+1}\$, but who is counting? And since I also didn't take into account the Ohmic \$r_b^{'}\$ or \$r_e^{'}\$, which probably have a larger impact anyway, you'll just have to accept this simplification for now.)

The instantaneous voltage gain at that quiescent collector voltage is then:

$$A_v = \frac{R_\text{C}}{r_e}\approx \frac{R_\text{C}}{\left[\frac{V_T}{I_{\text{C}_\text{Q}}}\right]}=\frac{R_\text{C}\:I_{\text{C}_\text{Q}}}{V_T}=\frac{R_\text{C}}{V_T}\cdot\frac{V_\text{CC}-V_{\text{C}_\text{Q}}}{R_\text{C}}=\frac{V_\text{CC}-V_{\text{C}_\text{Q}}}{V_T}$$

So, you can see that the maximum voltage gain doesn't depend on the quiescent collector current. The maximum instantaneous voltage gain depends on where you set the quiescent collector voltage. And if you set it very low, such that the BJT is already well into saturation, then only slightly more than half of the input signal will experience that high gain. The other half will be mostly clipped away, as it drives the BJT only still deeper into saturation. Even if you avoid this problem, pushing towards maximum voltage gain means serious distortion if the input signal is anything but insubstantially small.

So what's the answer to your question? Up to you.



The above takes a simplified view. I'm assuming the device temperature is stable and I didn't deal with the Early Effect or \$r_o\$. Nor did I take into account the three Ohmic resistances associated with any realistic device or any of the charge storage issues, which may or may not be important in some circumstance. And I'm not interested in attempting any of that here. I just wanted to get a basic point across.