Instead of a resistor at the transistor's collector, you might have seen a current source (active component) substituted in an audio amp's gain stage. Why go through the trouble of the added components? You'll find the active loading of a current source packs two big benefits for the gain stage: 1) a much bigger gain and 2) more current drive available to the next stage. A high gain is good for lowering distortion. And current drive is needed to supply the base currents of the output stage. Both circuits are placed inside of a small signal model of an audio amp to see how they perform.
Let's compare the two load types - resistor and current source - each in their own small signal model. The small signal model greatly simplifies an audio amp into three stages: a differential amplifier (GDIFF), a voltage gain stage (Q3) and an output buffer (EBUFF). Feedback resistors R2 and R1 close the loop on the amplifiers for a gain of 10x.
The BJT current source (top of page) produces a constant current out of Q4 regardless of the voltage at it's collector. How can you set its output current? Set the Io through RE4.
Io ≈ VD1 / RE4.
The gain stage with resistor has a couple of bragging rights - simplicity and low cost. This el cheapo design wins in the reduced parts count 1 to 5. But it performance may not be worth the cost savings.
As a quick review, the differential stage converts the differential input voltage (vin+ - vin-) to a current (GDIFF). The gain stage (Q3) converts this current back to a voltage at V(4). How? Ic3 flows through through the resistance at Q3's collector developing a voltage.
V(4) = Ic3 * RC3.
What is RC3? For the resistor load, its simply the resistor RC3. For the active load, RC3 is the effective resistance created by the constant current source. And rumor has it - its quite high.
Which gain stage has the better performance? To compare them, we'll measure the gain from differential input voltage V(1,2) to collector voltage V(4), or GAIN = V(4)/V(1,2).
CIRCUIT INSIGHT Run an AC ANALYSIS ANALYSIS. In separate plot windows, plot the gains for the two amplifiers.
What did you see? The resistor performed well with a respectable gain of of 30,000 V/V. However, as you might have guessed, the active load knocked the gain out of the park with 1,200,000 V/V! Why? the effective resistance of a BJT collector - when wired as a constant current source - is typically very high.
HANDS-ON DESIGN Let's try to increase the gain of both circuits. Its fairly obvious how to increase the resistor load's gain. Try increasing resistor RC3 by a factor of 2 or 3. Rerun the simulation and check the gain. How much did it increase? Its should scale roughly with the resistor.
Now try increasing the effective resistance of the active load. How? Try decreasing RE4 by a factor of 2 or so (this will increase the output current). Rerun the simulation. Did the gain increase or decrease? Try increasing RE4 by a factor of 2 above the original 650. What happened to the gain?
A valid question is this: why not just increase resistor load RC3 until you get a huge gain like the current source? Sounds good. However, a major function of this stage is providing base current to drive the next stage - the output buffer. Typically, the next stage is an emitter follower requiring some base current drive.
Let's look at the resistor's drive capability. Suppose you've got an output signal of 10V with 15V supply. That means with RC3=15k you've only got (15V-10V)/15k=0.333mA to drive the next stage. Furthermore, you may need a lot more gain, so you increase RC3 to 300k. Unfortunately, the available drive just got cut to 0.033 mA.
On the other hand, the current source performs head and shoulders above the resistor by two bench marks: 1) It has a much higher gain, and 2) It provides a beefy 1.000mA drive throughout its voltage range, about -15 to 14 V.