C3 · BJT

The Small-Signal Model: Up Close, Even a Curve Is a Line

To small swings around the operating point, a transistor looks like a linear element. In this unit you shrink the signal amplitude yourself and find the size below which the output becomes a clean copy of the input.

Shrink the signal to find a clean copy

The faint curve is the input, the swing of the base-emitter voltage, and the bold curve is the output, the swing of the collector current. When the signal is large, the output is distorted, with sharp peaks and flattened troughs. Shrink the amplitude to find the size at which the output takes the same sine shape as the input.

Signal amplitude vbevbe = 80 mV

Small-signal relation · ic = gm vbe

gm = Ic/VT ≈ 192 mA/V

rπ = β/gm ≈ 0.52 kΩ

The amplitude is comparable to the thermal voltage. The bend of the exponential curve shows up fully, so the output's peaks rise far higher than its troughs fall.

Large signal · distorted

Zoom in at the operating point and the curve is a line

The collector current curves upward exponentially with the base-emitter voltage. Still, cut out and magnify just a very narrow stretch around the operating point and any smooth curve looks like its tangent, a straight line. A small signal is exactly a swing that stays within this narrow stretch, and within it input and output are related by a straight line: they are linear.

The slope of the tangent, gm = Ic/VT

The slope of that tangent is the transconductance gm. Because an exponential stays proportional to itself when differentiated, the slope is the operating-point collector current divided by the thermal voltage, gm = Ic/VT. So the larger the bias current chosen in C2, the steeper the tangent and the larger gm, and the same small input produces a larger output current. The operating point is, in effect, the knob for gain.

The small-signal equivalent

Now the transistor can be redrawn as a single linear box. On the input side it looks like an input resistance rπ = β/gm, the base voltage divided by the base current, and on the output side it is a current source with ic = gm vbe = β ib. A DC supply, whose value never changes, has no swing as far as the signal is concerned, so for AC it is treated as ground. This equivalent circuit is the tool for calculating gain in the next units, and it holds only while the signal stays within the narrow stretch.

Observeic = gm vbe

The output current is the transconductance times the input voltage.

Choosegm = Ic / ?

The transconductance is the operating current over the thermal voltage.

Fill inrπ = β / ?

The input resistance is β over the transconductance.

On your owngm rπ = ?

Multiplying the two brings back the current gain β.

Back to the first screen

When the signal was large, the output was a distorted waveform with sharp peaks and flattened troughs: the bend of the exponential curve showing through. As you shrank the amplitude far below the thermal voltage, the curve looked straight over that narrow stretch, and the output became a clean sine copying the input. The slope of that straight line is gm, and with it the transistor becomes a linear box holding only an input resistance rπ and a current source gm vbe. The large DC is the stand that holds the position, and only the small swing on top of it is the signal to be calculated from here on.

To a small signal around the operating point, a transistor is a linear element. Magnify the exponential curve Ic(Vbe) at the Q point and it looks like a straight line of slope gm = Ic/VT (the transconductance), so a small vbe produces ic = gm vbe . The transistor becomes a small-signal equivalent model made of an input resistance rπ = β/gm and a current source gm vbe (gm rπ = β). It holds only while the signal is small enough for the curve to look straight.

What comes next

Now you know how to turn a transistor into a linear box. In the next unit you put this box into a real circuit, the common-emitter amplifier. When the input signal swings the base, the current source drives current through the collector resistor and makes a large output voltage, and you calculate that voltage gain directly for the first time with the small-signal model.