Active Loads in Amplifier Circuits
Discrete Semiconductor Devices and Circuits
We know that the current in a series circuit may be calculated with this formula:
We also know that the voltage dropped across any single resistor in a series circuit may be calculated with this formula:
Combine these two formulae into one, in such a way that the I variable is eliminated, leaving only ER expressed in terms of Etotal, Rtotal, and R.
Suppose we were to compare the performance of two voltage divider circuits side-by-side. The circuit on the left has one variable resistor (R2), while the circuit on the right has two variable resistors (R1 and R2). The right-hand circuit’s resistors are ganged together in such a way that as one resistance increases, the other will decrease by the same amount, keeping the circuit’s total resistance constant:
Knowing that the voltage output by a voltage divider is described by the following formula, determine which voltage divider circuit yields the greatest change in output voltage for a given change in R2‘s resistance.
The purpose of a current mirror circuit is to maintain constant current through a load despite changes in that load’s resistance:
If we were to crudely model the transistor’s behavior as an automatically-varied rheostat - constantly adjusting resistance as necessary to keep load current constant - how would you describe this rheostat’s response to changes in load resistance?
In other words, as Rload increases, what does Rtransistor do - increase resistance, decrease resistance, or remain the same resistance it was before? How does the changing value of Rtransistor affect total circuit resistance?
An interesting technique to achieve extremely high voltage gain from a single-stage transistor amplifier is to substitute an active load for the customary load resistor (located at the collector terminal):
Usually, this äctive load” takes the form of a current mirror circuit, behaving as a current regulator rather than as a true current source.
Explain why the presence of an active load results in significantly more voltage gain than a plain (passive) resistor. If the active load were a perfect current regulator, holding collector current absolutely constant despite any change in collector-base conductivity for the main amplifying transistor, what would the voltage gain be?