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Differential Voltage Amplifiers#

Definition: Finite-Gain Differential Voltage Amplifier

A finite-gain differential voltage amplifier is a \(5\)-terminal network\(\mathcal{A}\) consisting of the following:

  • an inverting input terminal (\(-\)) whose voltage is \(v_{-}\) and whose in-flowing current is \(i_{-}\);
  • a non-inverting input terminal (\(+\)) whose voltage is \(v_{+}\) and whose in-flowing current is \(i_{+}\);
  • an output terminal (\(\text{out}\)) whose voltage is \(v_{\text{out}}\) and whose in-flowing current is \(i_{\text{out}}\);
  • a negative power supply terminal (\(EE\)) whose voltage is \(-v_{\text{EE}}\);
  • a positive power supply terminal (\(CC\)) whose voltage is \(+v_{\text{CC}}\).

Definition: Differential Voltage

We call \(v_d \overset{\text{def}}{=} v_{+} - v_{-}\) the differential voltage.

Definition: Saturation Voltages

We call \(-v_{EE}\) the negative saturation voltage and \(+v_{CC}\) the positive saturation voltage.

The voltage \(v_{\text{out}}\) is given as

\[v_{\text{out}} = \begin{cases} -v_{EE} & \text{if } A v_d \in (-\infty; -v_{EE}) & \text{(negative saturation region)} \\ A v_d & \text{if } A v_d \in [-v_{EE};v_{CC}] & \text{(linear region)} \\ +v_{CC} & \text{if } A v_d \in (+v_{CC};+\infty) & \text{(positive saturation region)}\end{cases}\]

for some constant \(A \in \mathbb{R}\).

Definition: Voltage Gain

The constant \(A\) is known as the voltage gain.

Notation

The symbol for a differential voltage amplifier with finite saturation is the following:

Finite-Gain DVA Symbol

The saturation voltages model the fact that amplification requires energy and so differential voltage amplifiers cannot amplify beyond certain limits imposed by the energy restrictions.

Physical differential voltage amplifiers operated in their linear region can have huge voltage gains, on the order of \(10^5\). Moreover, their differential voltage is typically very small, so we can essentially model the voltage gain as being infinite in comparison.

Definition: Infinite-Gain Differential Voltage Amplifier

An infinite-gain differential voltage amplifier is a \(5\)-terminal network\(\mathcal{A}\) consisting of the following:

  • an inverting input terminal (\(-\)) whose voltage is \(v_{-}\) and whose in-flowing current is \(i_{-}\);
  • a non-inverting input terminal (\(+\)) whose voltage is \(v_{+}\) and whose in-flowing current is \(i_{+}\);
  • an output terminal (\(\text{out}\)) whose voltage is \(v_{\text{out}}\) and whose in-flowing current is \(i_{\text{out}}\);
  • a negative power supply terminal (\(EE\)) whose voltage is \(-v_{\text{EE}}\);
  • a positive power supply terminal (\(CC\)) whose voltage is \(+v_{\text{CC}}\).

Definition: Differential Voltage

We call \(v_d \overset{\text{def}}{=} v_{+} - v_{-}\) the differential voltage.

Definition: Saturation Voltages

We call \(-v_{EE}\) the negative saturation voltage and \(+v_{CC}\) the positive saturation voltage.

The transfer characteristic is what characterizes \(v_{\text{out}}\) in terms of \(v_d\):

\[\left\vert \begin{aligned} v_{\text{out}} &= -v_{\text{EE}} & \text{if } v_d \lt 0 & \qquad \text{(negative saturation region)} \\ v_{\text{out}} &\in [-v_{\text{EE}}; +v_{\text{CC}}] & \text{if } v_d = 0 & \qquad \text{(linear region)} \\ v_{\text{out}} &= +v_{\text{CC}} & \text{if } v_d \gt 0 & \qquad \text{(positive saturation region)} \end{aligned}\right.\]

Notation

The following symbol is used for infinite-gain differential voltage amplifiers:

Infinite-Gain DVA Symbol

The graph of the transfer characteristic for an infinite-gain differential voltage amplifier looks like the following:

Infinite-Gain DVA Transfer Characteristic

Four-Terminal DVAs#

In practice, the saturation voltages of a DVA are constant and usually equal but opposite in sign, i.e. \(v_{EE} = v_{CC} = V_{\text{sat}}\). We can model this by connecting the power supply terminals to time-invariant ideal voltage sources:

DVA Connected Power Rails

When this is the case, all the voltages are given with reference to the common connection point of these ideal voltage sources.

Notation: Four-Terminal DVAs

The above configurations are denoted by the following symbols:

Four-Terminal DVA Symbols

Non-Inverting Differential Voltage Amplifiers#

Definition: Non-Inverting Differential Voltage Amplifier

A differential voltage amplifier is non-inverting if its voltage gain \(A\) is positive:

\[A \gt 0\]

Example: Non-Inverting Differential Voltage Amplifier via Op-Amp

A non-inverting differential voltage amplifier can be constructed using an ideal operational amplifier and Ohmic resistors:

Non-Inverting DVA via Op-Amp

As long as the ideal op-amp is operated in its linear region, the above circuit behaves like a non-inverting differential voltage amplifier with the following voltage gain:

\[A = \frac{v_{\text{out}}}{v_{\text{in}}} = 1 + \frac{R_0}{R_1}\]

To ensure that the ideal op-amp is indeed operated in its linear region, we need to have \(v_{\text{in}} \in \left[-\frac{R_1}{R_0 + R_1}V_{\text{sat}}; +\frac{R_1}{R_0 + R_1}V_{\text{sat}}\right]\).

Non-Inverting DVA Example Graph

We can see this by analyzing the network.

Linear region:

When the ideal op-amp is operated in its linear region, we know that \(v_d = 0\), i.e. \(v_{-} = v_{+}\).

According to Kirchhoff's voltage law, noting that \(v_{+} = v_{\text{in}}\), we have the following:

\[\left\vert\begin{aligned}v_{\text{in}} &= R_1 i_1 \\ v_{\text{out}} &= R_1 i_1 + R_0 i_0\end{aligned}\right.\]

However, \(i_{-}\) is zero, since we have an ideal operational amplifier. Therefore, Kirchhoff's current law tells us that \(i_0 = i_1\). Substituting this into the second equation and taking the ratio of the two equations, we obtain the following:

\[\frac{v_{\text{out}}}{v_{\text{in}}} = \frac{R_1 i_1 + R_0 i_1}{R_1 i_1} = \frac{(R_1 + R_0) i_1}{R_1 i_1} = 1 + \frac{R_0}{R_1}\]

Saturation regions:

When the ideal op-amp is operated outside its linear region, we know that \(v_d \ne 0\). Recall that \(v_d = v_{+} - v_{-}\).

According to Kirchhoff's voltage law, and given \(v_{+} = v_{\text{in}}\), we have the following:

\[\left\vert\begin{aligned}v_{\text{in}} &= R_1 i_1 + v_d \\ v_{\text{out}} &= R_1 i_1 + R_0 i_0\end{aligned}\right.\]

Rearranging the first equation, we get an expression for \(i_1\):

\[i_1 = \frac{1}{R_1}(v_{\text{in}}-v_d)\]

However, \(i_{-}\) is zero, since we have an ideal operational amplifier. Therefore, Kirchhoff's current law tells us that \(i_0 = i_1\). We can thus substitute the above expression into the second equation:

\[v_{\text{out}} = (R_1 + R_0)i_1 = \left(1 + \frac{R_0}{R_1}\right)(v_{\text{in}}-v_d)\]

By performing some simple algebraic manipulations, we can obtain an expression for \(v_d\):

\[v_d = v_{\text{in}} - \frac{R_1}{R_1 + R_0}v_{\text{out}}\]

When the ideal op-amp is operated in its negative saturation region, we have \(v_d \lt 0\) and \(v_{\text{out}} = -V_{\text{sat}}\):

\[v_d = v_{\text{in}} - \frac{R_1}{R_1 + R_0}(-V_{\text{sat}}) \lt 0\]

By performing some equivalent transformations, we get the following:

\[v_{\text{in}} \lt -\frac{R_1}{R_1 + R_0} V_{\text{sat}}\]

Therefore, we know that the ideal op-amp is operated in its negative saturation region whenever \(v_{\text{in}} \lt -\frac{R_1}{R_1 + R_0} V_{\text{sat}}\).

By contrast, when the ideal op-amp is operated in its positive saturation region, we have \(v_d \gt 0\) and \(v_{\text{out}} = +V_{\text{sat}}\):

\[v_d = v_{\text{in}} - \frac{R_1}{R_1 + R_0}(V_{\text{sat}}) \gt 0\]

By performing some equivalent transformations, we get the following:

\[v_{\text{in}} \gt \frac{R_1}{R_1 + R_0} V_{\text{sat}}\]

Therefore, we know that the ideal op-amp is operated in its positive saturation region whenever \(v_{\text{in}} \gt \frac{R_1}{R_1 + R_0} V_{\text{sat}}\).

Inverting Differential Voltage Amplifiers#

Definition: Non-Inverting Differential Voltage Amplifier

A differential voltage amplifier is inverting if its voltage gain \(A\) is negative:

\[A \lt 0\]

Example: Inverting Differential Voltage Amplifier

An inverting differential voltage amplifier can be constructed using an ideal operational amplifier and Ohmic resistors:

Inverting DVA via Op-Amp

As long as the ideal op-amp is operated in its linear region, the above circuit behaves like an inverting differential voltage amplifier with the following voltage gain:

\[A = \frac{v_{\text{out}}}{v_{\text{in}}} = -\frac{R_0}{R_1}\]

To ensure that the ideal op-amp is indeed operated in its linear region, we need to have \(v_{\text{in}} \in \left[-\frac{R_1}{R_0}V_{\text{sat}}; +\frac{R_1}{R_0}V_{\text{sat}}\right]\).

Inverting DVA Example Graph

We can see this by analyzing the network.

Linear region:

When the ideal op-amp is operated in its linear region, we know that \(v_d = 0\), i.e. \(v_{-} = v_{+}\).

According to Kirchhoff's voltage law, we have the following:

\[\left\vert\begin{aligned}v_{\text{in}} &= R_1 i_1 \\ v_{\text{out}} &= -R_0 i_0\end{aligned}\right.\]

However, \(i_{-}\) is zero, since we have an ideal operational amplifier. Therefore, Kirchhoff's current law tells us that \(i_0 = i_1\). Taking the ratio of the two previous equations, we obtain the following:

\[\frac{v_{\text{out}}}{v_{\text{in}}} = \frac{R_1 i_1}{-R_0 i_0} = -\frac{R_1 i_1}{R_0 i_1} = -\frac{R_1}{R_0}\]

Saturation regions:

When the ideal op-amp is operated outside its linear region, we know that \(v_d \ne 0\).

According to Kirchhoff's voltage law, we have the following:

\[\left\vert\begin{aligned}v_{\text{in}} &= R_1 i_1 - v_d \\ v_{\text{out}} &= -v_d -R_0 i_0\end{aligned}\right.\]

Rearranging the first equation, we get an expression for \(i_1\):

\[i_1 = \frac{1}{R_1}(v_{\text{in}}+v_d)\]

However, \(i_{-}\) is zero, since we have an ideal operational amplifier. Therefore, Kirchhoff's current law tells us that \(i_0 = i_1\). We can thus substitute the above expression into the second equation:

\[v_{\text{out}} = -v_d - \frac{R_0}{R_1}(v_{\text{in}}+v_d)\]

By performing some simple algebraic manipulations, we can obtain an expression for \(v_d\):

\[v_d = -\frac{1}{R_1 + R_0}\left(R_1 v_{\text{out}} + R_0 v_{\text{in}}\right)\]

When the ideal op-amp is operated in its negative saturation region, we have \(v_d \lt 0\) and \(v_{\text{out}} = -V_{\text{sat}}\):

\[v_d = -\frac{1}{R_1 + R_0}\left(-R_1 V_{\text{sat}} + R_0 v_{\text{in}}\right) \lt 0\]

By performing some equivalent transformations, we get the following:

\[v_{\text{in}} \gt \frac{R_1}{R_0} V_{\text{sat}}\]

Therefore, we know that the ideal op-amp is operated in its negative saturation region whenever \(v_{\text{in}} \gt \frac{R_1}{R_0} V_{\text{sat}}\).

By contrast, when the ideal op-amp is operated in its positive saturation region, we have \(v_d \lt 0\) and \(v_{\text{out}} = +V_{\text{sat}}\):

\[v_d = -\frac{1}{R_1 + R_0}\left(R_1 V_{\text{sat}} + R_0 v_{\text{in}}\right) \gt 0\]

By performing some equivalent transformations, we get the following:

\[v_{\text{in}} \lt -\frac{R_1}{R_0} V_{\text{sat}}\]

Therefore, we know that the ideal op-amp is operated in its positive saturation region whenever \(v_{\text{in}} \lt -\frac{R_1}{R_0} V_{\text{sat}}\).