Floating_OpAmp.md

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OpAmp differential amplifier gain: Single-Ended Inputs vs. Double-Ended Input

The difference makes a difference ...

I'll show that the gain factor of an OpAmp differential amplifier will be different between single-ended operation and double-ended operation.

Single-ended operation is characterized by two input voltages, each referenced to ground (on the left side of the image above). In conrast, double-ended operation is characterized by one 'floating' input voltage, i.e. without reference to the ground potential (on the right side of the image above).

Note: The battery symbol has been used just to illustrate that the voltage source is potential-free and does not mean to be necessarily a battery component (it may be a sensor as well, for example).

Differential amplifier driven by single-ended inputs

Fig. 1: Differential amplifier driven by single-ended inputs \(V_{i_+}\) and \(V_{i_-}\) Fig. 2: The passive network modelling single-ended operation

The feedback-controlled steady-state will be reached as soon as the ground-referenced input voltages equal each other:

$$V_+ = V_-$$

Here, $$V_+$$ and $$V_-$$ drop along different current pathes, driven by $$V_{i_+}$$ and $$V_{i_-}$$, respectively, see Fig. 2.

Applying voltage divider rules, we get:

$$ \begin{align} V_{i_+} \frac{R_4}{R_3 + R_4} &= V_{i_-} : + : (V_o - V_{i_-}) \frac{R_1}{R_1 + R_2} \nonumber \\ V_{i_+} \frac{R_4}{R_3 + R_4} &= V_{i_-} \left(1 - \frac{R_1}{R_1 + R_2}\right) : + : V_o \frac{R_1}{R_1 + R_2} \nonumber \\ V_o \frac{R_1}{R_1 + R_2} &= V_{i_+} \frac{R_4}{R_3 + R_4} : - : V_{i_-} \left(1 - \frac{R_1}{R_1 + R_2}\right) \nonumber \\ V_o &= V_{i_+} \frac{R_1 + R_2}{R_3 + R_4} \frac{R_4}{R_1} : - : V_{i_-} \left(\frac{R_1 + R_2}{R_1} - 1\right) \nonumber \\ V_o &= V_{i_+} \left( \frac{R_1 + R_2}{R_3 + R_4} \frac{R_4}{R_1} \right) : - : V_{i_-} \left(\frac{R_2}{R_1}\right) \label{Eq:DiffAmp_SingleEnded} \end{align} $$

With $$R_1 = R_2 = R_3 = R_4$$ (i.e. unity gain) we get the net difference between the two gound-referenced input voltages:

\begin{equation} V_o = V_{i_+} ; - ; V_{i_-} \end{equation}

Differential amplifier driven by double-ended input

Fig. 3: Differential amplifier driven by double-ended input \(V_{d}\) Fig. 4: The passive network modelling double-ended operation

Again, the feedback-controlled steady-state will be reached as soon as the ground-referenced input voltages equal each other: \begin{equation}\label{Eq:DiffAmp_Base} V_+ = V_- \end{equation}

Perhaps you might expect that double-ended operation is equivalent to single-ended operation, given $$V_d = V_{i_+} - V_{i_-}$$. However, this is not the case.

The crucial difference to single-ended operation is that for double-ended operation $$V_+$$ and $$V_-$$ are not created by means of separate current pathes but rather by a common mesh current back to the floating voltage source $$V_d$$, see Fig. 4. This common mesh current is:

\begin{equation}\label{Eq:SingleEndedMeshCurrent} I = \frac{V_o + V_d}{R_1 + R_2 + R_3 + R_4} \end{equation}

From Fig. 4 we get:

$$ \begin{align*} V_+ &= I \cdot R_4\\ V_- &= I : (R_1 + R_3 + R_4) : - : V_d \end{align*} $$

Therefore, Eq. \ref{Eq:DiffAmp_Base} becomes:

$$ \begin{equation*} V_d = I : (R_1 + R_3) \end{equation*} $$

Furthermore, with Eq. \ref{Eq:SingleEndedMeshCurrent} we get:

$$ \begin{equation*} V_d = \frac{V_o + V_d}{R_1 + R_2 + R_3 + R_4} : (R_1 + R_3) \end{equation*} $$

Rearranging to $$V_o$$ yields:

$$ \begin{align} V_o &= V_d : \left(\frac{R_1 + R_2 + R_3 + R_4}{R_1 + R_3} : - : 1\right) \nonumber \\ V_o &= V_d : \frac{R_1 + R_2 + R_3 + R_4 - R_1 - R_3}{R_1 + R_3} \nonumber \\ V_o &= V_d : \frac{R_2 + R_4}{R_1 + R_3} \label{Eq:DiffAmp_DoubleEnded} \end{align} $$

Mentally replacing $$V_d$$ by $$V_{i_+} - V_{i_-}$$ in Eq. \ref{Eq:DiffAmp_DoubleEnded} we see that the result would not equal Eq. \ref{Eq:DiffAmp_SingleEnded}.

Hence, the hypothesis stated at the beginning has been proven: The difference of the input voltages \(V_{i_+} - V_{i_-}\) in single-ended operation will be amplified by another gain factor than a floating voltage \(V_d\) of equal magnitude.

One exeption holds for $$R_1 = R_2 = R_3 = R_4$$.

It should be noticed that the implementation of the OpAmp differential amplifier as an instrumentation amplifier won't be able to operate in a double-ended manner since their built-in voltage follower at the inputs $$V_{i_+}$$ and $$V_{i_-}$$ prevent the mesh current mentioned in Fig. 4 to flow.

References

Kenneth A. Kuhn, Floating Voltage Sources