Measurement Technology › Differential Signals

Amplifying Differential Signals

This chapter discusses how differential signals can be amplified. For this we are looking for amplifier circuits that we can connect to a bridge circuit. The A/D converter needs a ground-referenced signal at its input. So the amplifier must convert a differential signal into a ground-referenced signal.

To amplify differential signals, we use a differential amplifier. If the output voltage of the half bridge with PT100 is connected to the inputs of the differential amplifier, the following circuit diagram results:

Bridge circuit with PT100 at the differential amplifier: φ1 and φ2 of the bridge at the inputs via R_1, feedback via R_2

To derive the transfer function, let us assume that the ground-referenced voltages U₁ and U₂ shown are applied at the inputs of the differential amplifier (on the red lines). These voltages correspond to the potentials from the previous chapter. Then the output voltage is:

\[ \begin{gathered} \text{Differential amplifier: } U_{\mathrm{Aus,OP}} = \frac{R_2}{R_1} \cdot (U_2 - U_1) \\[6pt] U_1 = \varphi_1 - \varphi_0 \text{ with } \varphi_0 = 0\,\mathrm{V} \rightarrow U_1 = \varphi_1 \\[6pt] U_2 = \varphi_2 - \varphi_0 \text{ with } \varphi_0 = 0\,\mathrm{V} \rightarrow U_2 = \varphi_2 \\[6pt] U_S = \varphi_2 - \varphi_1 = U_2 - U_1 \\[6pt] U_{\mathrm{Aus,OP}} = \frac{R_2}{R_1} \cdot (U_2 - U_1) = \frac{R_2}{R_1} \cdot U_S \approx \frac{R_2}{R_1} \cdot U_0 \cdot \frac{0.4\,\frac{\Omega}{°\mathrm{C}} \cdot T}{200\,\Omega} \\[6pt] H = \frac{U_{\mathrm{Aus,OP}}}{T} \approx \frac{R_2}{R_1} \cdot U_0 \cdot \frac{0.4\,\frac{\Omega}{°\mathrm{C}}}{200\,\Omega} = v \cdot U_0 \cdot \frac{0.4\,\frac{\Omega}{°\mathrm{C}}}{200\,\Omega} \text{ with } v = \frac{R_2}{R_1} \end{gathered} \]

The transfer function can now be adjusted with the resistor values. We can create any slope of the characteristic curve. The output voltage of the op-amp is ground-referenced and can therefore be digitised by an ADC.

Example

Bridge circuit with PT100 at the differential amplifier

\[ \begin{gathered} \text{Given:} \\[6pt] \text{Temperature range: } T = [0\,°\mathrm{C} \ldots 100\,°\mathrm{C}] \\[6pt] \text{Reference voltage of the ADC: } U_{\mathrm{Ref}} = 5\,\mathrm{V} \\[6pt] U_0 = 5\,\mathrm{V} \end{gathered} \]

\[ \begin{gathered} \textbf{Solution with approximation:} \\[6pt] H = \frac{U_{\mathrm{Aus,OP}}}{T} \approx v \cdot U_0 \cdot \frac{0.4\,\frac{\Omega}{°\mathrm{C}}}{200\,\Omega} \\[6pt] \text{Calculating the output voltage of the op-amp:} \\[6pt] U_{\mathrm{Aus,OP}} = H \cdot T = v \cdot U_0 \cdot \frac{0.4\,\frac{\Omega}{°\mathrm{C}}}{200\,\Omega} \cdot T \\[6pt] v = \frac{U_{\mathrm{Ref}}}{U_{S,\mathrm{max}}} = \frac{5\,\mathrm{V}}{U_0 \cdot \frac{0.4\,\frac{\Omega}{°\mathrm{C}}}{200\,\Omega} \cdot 100\,°\mathrm{C}} = \frac{5\,\mathrm{V}}{1\,\mathrm{V}} = 5 \\[6pt] \text{Solution: e.g. } R_1 = 1\,\mathrm{k\Omega} \text{ and } R_2 = 5\,\mathrm{k\Omega} \end{gathered} \]

\[ \begin{gathered} \textbf{Solution without approximation:} \\[6pt] U_S = U_0 \cdot \frac{0.4\,\frac{\Omega}{°\mathrm{C}} \cdot T}{200\,\Omega + 0.4\,\frac{\Omega}{°\mathrm{C}} \cdot T} \\[6pt] \text{Calculating the output voltage of the op-amp:} \\[6pt] U_{\mathrm{Aus,OP}} = v \cdot U_S = v \cdot U_0 \cdot \frac{0.4\,\frac{\Omega}{°\mathrm{C}} \cdot T}{200\,\Omega + 0.4\,\frac{\Omega}{°\mathrm{C}} \cdot T} \\[6pt] v = \frac{U_{\mathrm{Ref}}}{U_{S,\mathrm{max}}} = \frac{5\,\mathrm{V}}{U_0 \cdot \frac{0.4\,\frac{\Omega}{°\mathrm{C}} \cdot 100\,°\mathrm{C}}{200\,\Omega + 0.4\,\frac{\Omega}{°\mathrm{C}} \cdot 100\,°\mathrm{C}}} = \frac{5\,\mathrm{V}}{0.83\,\mathrm{V}} = 6 \\[6pt] \text{Solution: e.g. } R_1 = 1\,\mathrm{k\Omega} \text{ and } R_2 = 6\,\mathrm{k\Omega} \end{gathered} \]

Simulation

The approximation is quite drastic, and the approximated result is rather far from the real result. We use it for rough estimates in practice and for calculating exam problems.