## Active Components and Circuits

All circuits we have discussed so far are only composed of passive components (resistors, capacitors and inductors) driven by current and/or voltage sources. Later we will consider circuits containing active components such as bipolar junction transistors (BJT), field-effect transistors (FET), operational amplifiers (Op-Amps) containing many transistors, and voltage amplifiers. These active components can be considered as controlled voltage or current sources as functions (typically linear) of the input voltage or current.

• We first consider modeling the overall function and performance of an active component (instead of its internal structure and implementation which may be very complicated) by the following three parameters:
• Voltage gain : The output voltage is related to the input voltage by , usually .
• Input impedance (resistance) : It is desirable to have a large so that little input current is drawn from the source, i.e., the source is minimally affected by the amplifier as a load. Ideally .
• Output impedance (resistance) : It is desirable to have a small so that little voltage drop across this resistance will result when the load of the amplifier draws a current from the amplifier, i.e., the load will minimally affect the output voltage of the amplifier.

Example 1: Consider the circuit below containing an active component, a voltage amplifier, model by the three parameters , and , dirven by either a current source or a voltage with internal resistance :

• We next consider a circuit containing an active circuit such as a transistor, an op-amp, or a voltage amplifier. This circuit can be modeled as a two-port network with input port between terminals A and B and output port between terminals C and D, and described in terms of the three parameters, the open-circuit voltage gain, the input resistance and output resistance:
• Input resistance : This is the resistance between the two terminals A and B of the input port, while a load is connected to the output port between terminals C and D. can be obtained as the ratio:

In general is affected by the load .
• Output resistance : According to Thevenin's theorem, any one-port network can be treated as an ideal voltage source in series with a resistance . We apply this theorem to the output port and define the output resistance as the Thevenin resistance

while a voltage source with an internal resistance is applied to the input port. In general, is affected by of the source.
• The open-circuit voltage gain : This is the ratio of the open-circuit output voltage (without a load () to an ideal voltage source

Note that a non-ideal source with internal resistance is used in the definition of as it is affected by , while an ideal source with is assumed in the definition of and . In case the source is not ideal with , we will use the voltage appearing across the input port as the input voltage.

The performance of such a circuit containing active components can also be described by terms such as voltage gain'' , current gain'' , and power gain'' , which can be found based on the specific circuits. They are defined as below:

• Voltage gain : defined as the ratio of the output voltage to the source voltage:

For the output of the voltage amplifier to be as high as possible, we would like to have ideally and , so that the output voltage is maximized and the gain is .

• Current gain : defined as the ratio of the output current to the input current:

Ideally, when , , we have .

• Power gain : defined as the ratio of the power delivered to the load to that to the amplifier:

Example 3:

Find , , and of this two-port network containing and as well as the amplifier modeled by , and .

• Input resistance: By inspection, the input resistance of this 2-port network can be found to be .

• Output resistance: We assume the internal voltage source is the Thevenin voltage , and get the open-circuit voltage and the short-circuit current . The output resistance is

Alternatively, can be found as the resistance between the two terminals C and D of the output port when the voltage source of the amplifier is turned off (short-circuit), i.e., .

• Open-circuit voltage gain: This is the ratio of the voltage across the output port to the voltage across the input port, when the output port is an open circuit, i.e., .

This 2-port network modeled as a voltage amplifier with , and can be used in more complicated circuits.

Example 4:

A 2-port network with a voltage aplifier on the left can be modeled by the circuit on the right. Find the parameters , and of the two-port network with the voltage amplifier embedded.

• Open-circuit voltage gain: As the output port is open circuit, the output current is zero and so is the voltage drop across . Applying an ideal voltage source to the input, we get the voltage across and across , respectively:

The open-circuit voltage across output port is therefore:

The open-circuit voltage gain can be found as the ratio of the open-circuit output voltage to the input voltage:

if , the circuit is reduced to the original voltage amplifier and we have .

• Output resistance:

We first find the short-circuit current at the output port. Assume a voltage source with internal resistance is applied to the input port while the output port is short-circuited. Applying KVL to the two loops of the circuit, we get:

Solving these two equations for the two unknowns and , we get

The open-circuit output voltage is

Now the output resistance can be found to be:

Note that is affected by internal resistance of the source. When ,

i.e., the output resistance is much reduced. Moreover, when , .

• Input resistance: This can be found by applying an ideal voltage source to the input port, while the output port is connected to a load . The input resistance is where is the input current. Applying the KVL to the two loops of this circuit, we get

Solving these two equations for the two unknowns and , we get

Now the input resistance can be found to be

Note that is affected by the load . When , , i.e., the input resistance is much increased. Moreover, if , the circuit is reduced to the original voltage amplifier with .
In summary, the resistor shared by both the input and output loops serves as a negative feedback:

As the result, the voltage gain is reduced but both the input and output resistances are improved, i.e., is increased and the is reduced.

Example 5: (Homework)

A voltage amplifier, denoted by the inner box (solid line) with three internal parameters , and , is used as a component in a two-port network, denoted by the outer box (dashed line). Its open-circuit output voltage is . Find the following three parameters of the two-port network.

Note that all output between C and D of the output port is fed back to the input port between A and B: or , i.e., it is a negative feedback.

Then simplify the three results above by making reasonable approximations based on the assumptions that , .

• The input resistance , where is the source voltage applied across A and B, is the current through the input port, when a load is connected to the output port between C and D.
• The output resistance , where and are the open-circuit voltage and short-circuit current when an ideal source voltage (with ) is applied to the input port.
• The open-circuit gain .

Example 6: (Homework)

Two amplifiers with parameters , , and , , , respectively, can be connected in cascade as shown in the figure. Given a voltage source in series with an internal resistance , find the output voltage. To maximize the output , how would you change the values of the six parameters?

Find the power gain of the system.

Example 7: (Homework)

The input and output resistances and , as well as the voltage gain of a two-port network can be obtained experimentally. First, connect an ideal voltage source (a new battery with very low internal resistance) in series with a resistor , and then connect load of two different resistances to the output port. Now the three parameters can be derived from the known values of , and the two measurements of the load voltage , corresponding to the two resistance values used.

Assume , , and the input voltage is measured to be ; also, assume the two different load resistors used are and respectively, with the two corresponding output voltage and . Find , and .