===== Rectangular-to-Triangle Signal Conversion - Integrator ===== ==== Background Information ==== The operation of an OPV in the linear operating range can be enforced by means of circuitry by feeding back the output signal, i.e., returning it to the inverting input (- input). In the circuit shown, the negative feedback is provided by a capacitor.\\ \\ \\ {{drawio>mexlefirst_public:integrator_circuit.svg}} \\ \\ Analysis of the circuit:\\ Negative feedback $\Rightarrow u_\mathrm{d} = 0 \Rightarrow i_R = \frac{u_\mathrm{e}}{R}$ $i_R=i_C$ (because OPV input current $i_\mathrm{n} = 0$) $u_\mathrm{a}=-u_C=-\frac{1}{C}\int i_\mathrm{C}\,dt=-\frac{1}{RC}\int u_\mathrm{e}\,dt$\\ \\ \\ \\ \\ \\ The integrated input voltage appears at the output. The product of resistance and capacitance has the character of a time constant: $T_\mathrm{i}=RC$\\ \\ \\ {{drawio>mexlefirst_public:integrator_u-t-diagramme.svg}}\\ \\ \\ \\ \\ \\ The figure shows the output voltage of an integrator with a square wave voltage at the input. The output voltage at the start $u_\mathrm{a}(t=0)$ depends on the charge state of the capacitor when switched on.\\ \\ \\ \\ \\ \\ ==== Experimental Tasks ==== To analyze the behavior of the integrator, the following circuit is used:\\ \\ \\ {{drawio>mexlefirst_public:integrator_experiment.svg}} \\ \\ \\ __Supply voltages (from power supply unit):__ \\ \\ $UCC=+3~V, UEE~=-3~V$\\ \\ __Values of the components used:__ \\ \\ $R1.3=10~kΩ, C1=10~nF$\\ \\ \\ \\ \\ \\ \\ \\ - task-type-result Calculate the time constant $T_\mathrm{i}$ of the integrator from the given values. - task-type-result Assumption: the capacitor is initially uncharged. A voltage $u_\mathrm{e}=+3~V$ is applied to the input. How long does it take for the output voltage to reach $u_\mathrm{Tr}=-3~V$? Document your calculation. - task-type-result Roughly sketch the voltage curves that you expect at the TR output when you apply a bipolar square wave signal to the $u_\mathrm{e}$ input. \\ \\ **Output TR** \\ \\ {{drawio>mexlefirst_public:oscilloscope_screen.svg}} \\ \\ \\ Channel 1:$\frac {Volt}{Div}=$ \\ \\ \\ Time basis: $\frac {T}{Div}=$ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ - task-type-oscilloscope Build the circuit on the MEXLE-board. **Please use the level shifting circuit at the input of the circuit.** Make sure that the jumper at the bottom of the op-amp is set to the left so that the op-amp is supplied with +/- 3V. Connect channel 1 on the oscilloscope to $U_\mathrm{e}$ and channel 2 to TR. Connect the function generator to the $U_\mathrm{e}$ input. Set to square wave (bipolar) with a frequency of 3kHz and a voltage of 3 V (amplitude). Switch on the power supply. Take a photo of the oscilloscope screen image. \\ \\ \\ **C1 = 10 nF, f = 3 kHz** \\ \\ {{drawio>mexlefirst_public:oscilloscope_screen.svg}} \\ \\ \\ Channel 1: $\frac {Volt}{Div}=$ \\ \\ Channel 2: $\frac {Volt}{Div}=$ \\ \\ \\ Time basis: $\frac {T}{Div}=$ \\ \\ \\ \\ \\ \\ \\ \\ - task-type-freetext Compare your measurement with the calculation from part 2 and the forecast from part 3. Explain your result. - task-type-single-choice (single choice test) Two resistors $R_1 = 1\,\mathrm{k\Omega}$ and $R_2 = 2\,\mathrm{k\Omega}$ are connected in series to a $9\,\mathrm{V}$ source. What is the voltage across $R_2$? * $3\,\mathrm{V}$ * $4.5\,\mathrm{V}$ * $6\,\mathrm{V}$ * $9\,\mathrm{V}$ - task-type-multiple-choice (multiple choice test) Two resistors $R_1 = 100\,\Omega$ and $R_2 = 220\,\Omega$ are connected in parallel to a DC voltage source. Which statements are correct? * The same voltage drops across both resistors. * The same current flows through both resistors. * The total resistance is smaller than $100\,\Omega$. * $R_\mathrm{sum} = R_1 + R_2 = 320\,\Omega$ * The larger resistor carries the larger current. * The total current equals the sum of the two branch currents. ==== Test Questions - Integrator ====