C5 · BJT

Gain and Impedance: An Amplifier Feels Its Neighbors

An amplifier sits between a source and a load. Load it heavily or lightly yourself and see how the ideal gain of the last unit is trimmed in practice, and what decides that trimming.

Change the load and watch the gain get trimmed

The dashed peaks are the ideal output with no load, and the bold curve is the actual output vout. Shrink the load resistor RL to load it heavily. The actual output shrinks inside the dashed envelope. Make the load larger and it returns toward the ideal.

Load resistor RLRL = 0.20 kΩ

Actual voltage gain Av

Av = -gm (Rc || RL) = -32

Rout = Rc = 1 kΩ   ·   |Av|max = 192

The load resistor is smaller than Rc. In parallel with Rc the effective resistance drops a lot, so the actual gain falls to a fraction of the ideal.

Heavy load · gain cut

Input impedance, what the source sees

From the standpoint of the source that gives the signal, looking into the amplifier you see a single resistance. In the common-emitter this input impedance is the small-signal input resistance rπ in parallel with the bias resistors. The source always has its own internal resistance Rs, which forms a divider with the input impedance. If the input impedance is small, only part of the source voltage reaches the base and signal is lost from the very start.

Output impedance, what the load sees

Conversely, looking into the amplifier from the output, the collector, the load sees yet another resistance. The transistor’s output is ideally a current source of infinite resistance, so the output impedance the load sees is effectively the collector resistor Rc. So when you attach a load resistor RL, it sits beside Rc, that is, in parallel. The smaller the output impedance, the less the voltage is trimmed for the same load.

The load trims the gain

Since the load RL parallels the collector resistor Rc, the effective resistance the signal current sees is Rc || RL. So the actual voltage gain is Av = -gm (Rc || RL), always less than the unloaded -gm Rc. The heavier the load, that is, the smaller RL, the smaller the parallel value and the more the gain is trimmed. To chain several stages for a large gain, the output impedance of a stage must be small enough compared with the input impedance of the next so as not to lose signal. That is why, in designing an amplifier, impedance matters as much as gain.

ObserveRout = Rc

The output impedance is the collector resistor.

Choosevbe = vs · rπ / (rπ + ?)

The input enters in the ratio of the input resistance over the source resistance plus it.

Fill inAv = -gm (Rc || ?)

The actual gain follows the parallel of Rc and the load.

On your ownRc || RL = Rc RL / (Rc + ?)

The parallel value is the product over the sum.

Back to the first screen

The heavier you loaded it, the more the actual output shrank inside the dashed envelope, and lightening the load returned it toward the dashes. That dashed envelope is the ideal unloaded gain -gm Rc, and the actual output was always the smaller -gm (Rc || RL). What set the trimming was the resistance the load saw at the collector, the parallel of the output impedance Rc with the load RL. Likewise at the input, the input impedance and the source resistance form a divider that splits the signal. An amplifier never lives alone, and its gain is always decided together with its neighboring impedances. This completes one full turn around the BJT amplifier.

An amplifier never stands alone. The input impedance Rin (≈ rπ) is what the source sees and forms a divider with the source resistance Rs that trims the input; the output impedance Rout (= Rc) is what the load sees, and the load RL sits in parallel with Rc. So the actual voltage gain Av = -gm (Rc || RL) is always less than the unloaded -gm Rc. The heavier the load, the more it is trimmed. To cascade stages without losing signal, the Rout of one stage must be small enough compared with the Rin of the next.

What comes next

The BJT governed a large current with a small current. The next group meets another transistor, the MOSFET, which opens and closes its channel not with current but with voltage, the electric field applied to a gate alone. Starting from the fact that almost no current flows into the gate, giving a very large input impedance, its first unit looks at the structure and the threshold voltage.