Active Circuits

So far all circuits we have discussed are composed of passive components (resistors, capacitors and inductors) driven by current and voltage sources. In the future we will be considering active components such as bipolar junction transistors (BJT) and field-effect transistors (FET), operational amplifiers (Op-Amps), as well as voltage/current amplification circuits. These active components (as simple as single transistors and as complicated as some Op-Amps) can be considered as controlled current or voltage sources that generate current or voltage depending on the input current or voltage.

For the purpose of describing the overall function and performance of such components and circuits (instead of its internal structure and implementation), a general model can be used with the following three parameters:

• Gain $A$: The output voltage $v_{out}$ is related to the input voltage $v_{in}$ by $v_{out}=A v_{in}$, usually $A>1$.
• Input impedance (resistance) $r_{in}$: It is desirable to have a large $r_{in}$ so that little input current is drawn from the source, i.e., the source is minimally affected by the amplifier as a load. Ideally $r_{in}=\infty$.
• Output impedance (resistance) $r_{out}$: It is desirable to have a small $r_{out}$ 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:

$$v_{in}=i_s (R_s \vert\vert r_{in})=i_s \frac{R_s r_{in}}{R_s+r_{in}}$$

$$v_{out}=A v_{in} \frac{R_L}{R_L+r_{out}}= A i_s \frac{R_s r_{in}}{R_s+r_{in}} \frac{R_L}{R_L+r_{out}}$$

Ideally when $r_{in}=\infty$ and $r_{out}=0$, $v_{out}=A i_s R_s$.

Example 2:

$$v_{in}=v_s \frac{r_{in}}{R_s+r_{in}}$$

$$v_{out}=A v_{in} \frac{R_L}{R_L+r_{out}} =A v_s \frac{r_{in}}{R_s+r_{in}} \frac{R_L}{R_L+r_{out}}$$

The terms ``voltage gain'' $G_V$, ``current gain'' $G_I$, and ``power gain'' $G_P$ need to be specifically defined for different circuit configurations. In this case, they can be defined as below:

• Voltage gain $G_V$: defined as the ratio of the output voltage to the source voltage:
  
  $$G_V=\frac{v_{out}}{v_s}=A\frac{r_{in}}{R_s+r_{in}} \frac{R_L}{R_L+r_{out}}$$
  
  
  Ideally, when $r_{in}=\infty$, $r_{out}=0$, we have $v_{out}=A v_s$ and $G_V=A$.

• Current gain $G_I$: defined as the ratio of the output current to the input current:
  
  $$G_I=\frac{i_{out}}{i_{in}}=\frac{A v_{in}/(r_{out}+R_L)}{v_s/... ...(r_{out}+R_L)}{v_s/(R_s+r_{in})} =\frac{A r_{in}}{r_{out}+R_L}$$
  
  
  Ideally, when $r_{in}=\infty$, $r_{out}=0$, we have $G_I=\infty$.

• Power gain $G_P$: defined as the ratio of the power delivered to the load to that to the amplifier:
  
  $$G_P=\frac{v_{out}^2/R_L}{v_{in}^2/r_{in}}=G_V^2 \frac{r_{in}}{R_L}$$
  
  

The alternative definitions of these voltage, current, and power gains may be used, depending on the specific applications.

The voltage amplifier can be used as a component (a building block) in a larger circuit, such as two-port network with input port between terminals A and B and output port between terminals C and D. This network can be in turn described in terms of the three parameters, the open-circuit voltage gain, the input resistance and output resistance, as shown below:

• Input resistance $R_{in}$: This is the resistance between the two terminals A and B of the input port, while a load $R_L$ is connected to the output port between terminals C and D. $R_{in}$ can be obtained as the ratio:
  
  $$R_{in}=\frac{\mbox{ideal voltage source $v_s$ applied to the... ... port}}{\mbox{input current $i_{in}$ through the input port}}$$
  
  
  In general $R_{in}$ is affected by the load $R_L$.
• Output resistance $R_{out}$: According to the Thevenin's theorem, any one-port network can be treated as an ideal voltage source $v_T$ in series with a resistance $R_T$. We apply this theorem to the output port and define the output resistance as the Thevenin resistance
  
  $$R_{out}=R_T=\frac{\mbox{open-circuit voltage ($v_{oc}=v_T$)}}{\mbox{short-circuit current ($i_{sc}=v_T/R_T$)}}$$
  
  
  while a voltage source $v_s$ with an internal resistance $R_s$ is applied to the input port. In general, $R_{out}$ is affected by $R_s$ of the source.
• The open-circuit voltage gain $A_{oc}$: This is the ratio of the open-circuit output voltage $v_{oc}$ to an ideal voltage source $v_s$ without a load ($R_L=\infty$).
  
  $$A_{oc}=\frac{\mbox{open-circuit voltage ($v_{oc}$)}}{\mbox{ideal voltage source ($v_s=v_{AB}$)}}$$
  
  

Note that a non-ideal source with internal resistance $R_s$ is used in the definition of $R_{out}$ as it is affected by $R_s$, while an ideal source with $R_s=0$ is assumed in the definition of $R_{in}$ and $A_{oc}$. In case the source is not ideal with $R_s\ne 0$, we will use the voltage $v_{AB}$ appearing across the input port as the input voltage.

Example 3:

Find $R_{in}$, $R_{out}$, and $A_{oc}$ of this two-port network containing $R_1$ and $R_2$ as well as the amplifier modeled by $r_{in}$, $r_{out}$ and $A$.

• Input resistance: By inspection, the input resistance of this 2-port network can be found to be $R_{in}=R_1+r_{in}$.

• Output resistance: We assume the internal voltage source is Thevenin voltage $v_T$, and get the open-circuit voltage $v_{oc}=v_T R_2/(r_{out}+R_2)$ and the short-circuit current $i_{sc}=v_T/r_{out}$. The output resistance is
  
  $$R_{out}=\frac{v_{oc}}{i_{sc}}=\frac{r_{out}R_2}{r_{out}+R_2}=r_{out}\vert\vert R_2$$
  
  
  Alternatively, $R_{out}$ 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., $R_{out}=R_2 \vert\vert r_{out}$.

• Open-circuit voltage gain: This is the ratio of the voltage $V_{CD}$ across the output port to the voltage $V_{AB}$ across the input port, when the output port is an open circuit, i.e., $R_L=\infty$.
  

  $$v_{in} = V_{AB}\frac{r_{in}}{R_1+r_{in}}$$

  $$V_{CD} = A v_{in} \frac{R_2}{r_{out}+R_2}=A V_{AB}\frac{r_{in}}{R_1+r_{in}} \frac{R_2}{r_{out}+R_2}$$

  $$A_{oc} = \frac{V_{CD}}{V_{AB}}=\frac{A R_2 r_{in}}{(R_1+r_{in})(r_{out}+R_2)}$$

  
  

This 2-port network modeled as a voltage amplifier with $R_{in}$, $R_{out}$ and $A_{oc}$ can be used in more complicated circuits.

Example 4:

Find the parameters $R_{in}$, $R_{out}$ and $A_{oc}$ 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 $r_{out}$. Applying an ideal voltage source $v_s$ to the input, we get the voltage $v_{in}$ across $r_{in}$ and $v_1$ across $R_1$, respectively:
  
  $$v_{in}=v_s \frac{r_{in}}{R_1+r_{in}},\;\;\;\;\; v_1=v_s \frac{R_1}{R_1+r_{in}}$$
  
  
  The voltage across output port is therefore:
  
  $$v_{oc}=v_1+A v_{in}=v_s \frac{R_1}{R_1+r_{in}}+Av_s\frac{r_{in}}{R_1+r_{in}} =v_s \frac{R_1+A r_{in}}{R_1+r_{in}}$$
  
  
  The open-circuit voltage gain can be found as the ratio of the open-circuit output voltage to the input voltage:
  
  $$A_{oc}=\frac{v_{oc}}{v_s}=\frac{R_1+A r_{in}}{R_1+r_{in}}<A$$
  
  
  if $R_1=0$, the circuit is reduced to the original voltage amplifier and we have $A_{oc}=A$.
• Output resistance: We first find the short-circuit current $i_{sc}$ at the output port. Assume a voltage source $v_s$ with internal resistance $R_s$ is applied to the input port while the output port is short-circuited. Applying KVL to the two loops of the circuit, we get:
  
  $$v_s=(R_s+R_1+r_{in})i_{in}-R_1 i_{sc}$$
  
  
  
  
  $$A v_{in}=A r_{in} i_{in}=(r_{out}+R_1) i_{sc}-R_1 i_{in}$$
  
  
  Solving these two equations for the two unknowns $i_{in}$ and $i_{sc}$, we get
  
  $$i_{sc}=v_s \frac{Ar_{in}+R_1}{(1-A)R_1 r_{in} +r_{out}(R_s+r_{in}+R_1) +R_sR_1}$$
  
  
  The open-circuit output voltage $v_{oc}$ is
  
  $$v_{oc}=Ar_{in}i_{in}+R_1i_{in}=(Ar_{in}+R_1)\frac{v_s}{R_s+r_{in}+R_1}$$
  
  
  Now the output resistance can be found to be:
  
  $$R_{out}=\frac{v_{oc}}{i_{sc}} =\frac{(1-A)r_{in}R_1+r_{out}(... ...}+R_1} =r_{out}+\frac{(1-A) r_{in}R_1 +R_sR_1}{R_1+r_{in}+R_s}$$
  
  
  Note that $R_{out}$ is affected by internal resistance $R_s$ of the source. When $R_s=0$,
  
  $$R_{out}\stackrel{R_s=0}{\Longrightarrow} r_{out}-(A-1) r_{in}\vert\vert R_1 < r_{out}$$
  
  
  i.e., the output resistance is much reduced. Moreover, when $R_1=0$, $R_{out}=r_{out}$.
• Input resistance: This can be found by applying an ideal voltage source $v_s$ to the input port, while the output port is connected to a load $R_L$. The input resistance is $R_{in}=v_s/i_{in}$ where $i_{in}$ is the input current. Applying the KVL to the two loops of this circuit, we get
  
  $$v_s=(r_{in}+R_1) i_{in}-R_1 i_{out}$$
  
  
  
  
  $$A v_{in}=Ar_{in}i_{in}=(r_{out}+R_L+R_1)i_{out}-R_1i_{in}$$
  
  
  Solving these two equations for the two unknowns $i_{in}$ and $i_{out}$, we get
  
  $$i_{in}=v_s \frac{R_1+R_L+r_{out}}{(R_L+r_{out})(R_1+r_{in})+(... ...R_L\rightarrow \infty}{\Longrightarrow} \frac{v_s}{r_{in}+R_1}$$
  
  
  Now the input resistance can be found to be
  
  $$R_{in}=\frac{v_s}{i_{in}}=\frac{(R_L+r_{out})(R_1+r_{in})+(1-... ...krel{R_L\rightarrow \infty}{\Longrightarrow} R_1+r_{in}>r_{in}$$
  
  
  Note that $R_{in}$ is affected by the load $R_L$. When $R_L=\infty$, $R_{in}=R_1+r_{in}$, i.e., the input resistance is much increased. Moreover, if $R_1=0$, the circuit is reduced to the original voltage amplifier with $R_{in}=r_{in}$.

In summary, the resistor $R_1$ shared by both the input and output loops serves as a negative feedback:

$$v_s\uparrow \rightarrow i_{in}, v_{out}\uparrow \rightarrow v_1\uparrow \rightarrow i_{in}, v_{out}\downarrow$$

As the result, the voltage gain $A_{oc}$ is reduced but both the input and output resistances are improved, i.e., $R_{in}$ is increased and the $R_{out}$ is reduced.

Example 5: (Homework)

A voltage amplifier, denoted by the box (solid line) with three internal parameters $r_{in}$, $r_{out}$ and $A$, is used as a component in a two-port network as shown. Fine the input resistance $R_{in}$, output resistance $R_{out}$, and the open-circuit gain $A_{oc}$ of the two-port network. Note that when considering the input and output resistances of the network, you should take into account the load $R_L$ and the internal resistance $R_s$ of the source voltage.

Answer

Example 6: (Homework)

Two amplifiers with parameters $A_1$, $r_{i1}$, $r_{o1}$ and $A_2$, $r_{i2}$, $r_{o2}$, respectively, can be connected in cascade as shown in the figure. Given a voltage source $v_s$ in series with an internal resistance $R_s$, find the output voltage. To maximize the output $v_{out}$, how would you change the values of the six parameters?

Find the power gain $G_p$ of the system.

Answer

Example 7: (Homework)

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

Assume $v_s=1.5V$, $R_s=5 k\Omega$, and the input voltage is measured to be $v_{in}=1.25 V$; also, assume the two different load resistors used are $R_1=150 Omega$ and $R_2= 200 Omega$ respectively, with the two corresponding output voltage $v_1=18.75V$ and $v_2=20$. Find $R_{in}$, $R_{out}$ and $A_{oc}$.

Answer