detecting low signal levels with a transistor

Question

This question made me wonder. Say I want to drive a LED when a small level audio signal is present. I understand how to do this with an opamp, but how about a transistor. You'd have to bias it so that it just doesn't conduct without the signal, so that any extra voltage on the base will cause collector current to flow, right? Can anybody please explain how you do that: bias the transistor so that it just doesn't conduct, independent of temperature variations and such? Or is this the wrong approach?

Answer 1

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.

Answer 2

What you ask is doable. Here is a circuit that should work, although I haven't tried it:

Q2 is biased just short of turning on. This is done by putting a little bit of current thru Q3 and using the resulting base voltage to bias Q2. This bias is additionally fed thru R5 so that the voltage on the base of Q2 will be a little lower if it does try to draw current. It also isolates the low impedance of the voltage source formed by Q3 from the audio signal. Since Q3 and Q2 are the same transistor type, they should have similar enough characteristics, including B-E voltage over temperature. This makes the biasing of Q2 reasonably immune to temperature changes.

When Q2 turns on, it will turn on Q1, which turns on the LED. Q1 adds more gain to the circuit so that a weak audio signal can eventually drive 10 mA thru the LED. R3 limits the base current of Q1 so that Q2 can saturate with no harm, as it will at the peaks of a reasonable audio signal. There will be about 420 µA thru R3 when Q2 is fully on. R1 is there to require some minimum current to turn on Q1 which might be necessary as Q2 is biased right at the edge of significant conduction. R1 will draw about 60 µA, leaving 360 µA thru the base of Q1 when Q2 is fully on. The LED will require up to 10 mA of collector current thru Q1, so this means Q1 must have a gain of at least 28, which is well within the capability of most small signal low voltage PNP transistors. You want Q1 to have some extra gain so that the LED will be fully on even when Q2 is only partially on.

C1 AC couples the audio input and prevents it from messing up the DC bias of Q2.