=====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.\\
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<wrap left> {{drawio>integrator_circuit.svg}}</wrap>\\
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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 $i_\mathrm{n} = 0$)

$u_\mathrm{a}=-u_C=-\frac{1}{C}\int i_\mathrm{C}\,dt=-\frac{1}{RC}\int u_\mathrm{e}\,dt$\\
~~CLEARFIX~~
The integrated input voltage appears at the output. The product of resistance and capacitance has the character of a time constant:

$T_\mathrm{i}=RC$\\
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<wrap left> {{drawio>integrator_u-t-diagramme.svg}}\\
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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.\\

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====Experimental Tasks====

To analyze the behavior of the integrator, the following circuit is used:\\ 
<wrap left>
{{drawio>integrator_experiment.svg}}
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~~CLEARFIX~~

__Supply voltages (from power supply unit):__\\ 
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$UCC=+3~V, UEE~=-3~V$\\ 
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__Values of the components used:__\\ 
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$R1.3=10~kΩ, C1=10~nF$\\ 
~~CLEARFIX~~

  - Calculate the time constant $T_\mathrm{i}$ of the integrator from the given values. \\ \\ \\ \\ \\ \\ \\ \\
  - 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. \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
  - 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**\\ \\ <wrap left>{{drawio>oscilloscope_screen.svg}}</wrap>\\ \\ \\ Channel 1:$\frac {Volt}{Div}=$\\ \\ \\ Time basis: $\frac {T}{Div}=$ ~~CLEARFIX~~ \\
  - Build the circuit on the MEXLE-board. 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. \\ \\ \\ **C1 = 10 nF, f = 3 kHz**\\ \\ <wrap left>{{drawio>oscilloscope_screen.svg}}</wrap>\\ \\ \\ Channel 1: $\frac {Volt}{Div}=$\\ \\ Channel 2: $\frac {Volt}{Div}=$\\ \\ \\ Time basis: $\frac {T}{Div}=$ ~~CLEARFIX~~
  - Compare your measurement with the calculation from part 2 and the forecast from part 3. Explain your result. \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 

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====Test Questions - Integrator====
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