detecting low signal levels with a transistor
You're probably thinking of the common emitter circuit which is typically used for a switching transistor, i.e. with the emitter directly to ground. This is OK for switching, where two extreme states occur: one where there's no base current (base voltage is 0V) and one where there's a set base current of several mA to saturate the transistor.  

Now suppose you place a resistor between the emitter and ground, R2 in the schematic below:  

  

This resistor provides negative feedback, which stabilizes the circuit. A small increase in base current, for instance because of temperature changes,  will cause a large change in base voltage. That's because the extra base current isn't the only current flowing through R2, there's also \$I_C\$, which is \$H_{FE}\$ larger than \$I_B\$. This higher current will cause a large voltage drop across R2, in turn causing the base voltage to increase by the same amount.
Example: R2 = 10\$\Omega\$, \$H_{FE}\$ = 100, \$I_B\$ = 1mA. Then  


  \$ V_E = R2 \times (I_B + I_C) = 10\Omega \times (1mA + 100mA) = 1.01V \$


If for some reason the base current would rise by just 10\$\mu\$A, then  


  \$ V_E = R2 \times (I_B + I_C) = 10\Omega \times (1.01mA + 101mA) = 1.02V \$


so the base voltage will also rise by 10mV, which would counteract the current rise. This effect will be stronger when R2 is larger.  

edit
Now place a capacitor parallel to R2. That means that the impedance between emitter and ground is (much) lower than R2 for AC signals. So for AC the negative feedback is not as high, and the signal will be properly amplified, while the DC setting remains stable.