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--- lab_electrical_engineering:rectangular-to-triangle_signal_conversion_integrator [2026/06/17 11:05] – mexleadmin
+++ lab_electrical_engineering:rectangular-to-triangle_signal_conversion_integrator [2026/06/17 13:21] (current) – mexleadmin
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 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>\\
-<wrap left> {{drawio>mexlefirst_intern:integrator_circuit.svg}} </wrap>\\
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 Analysis of the circuit:\\
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 $u_\mathrm{a}=-u_C=-\frac{1}{C}\int i_\mathrm{C}\,dt=-\frac{1}{RC}\int u_\mathrm{e}\,dt$\\
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+~~CLEARFIX~~
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 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}}\\
-<wrap left> {{drawio>mexlefirst_intern:integrator_u-t-diagramme.svg}}\\
 </wrap>\\
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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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 ~~PAGEBREAK~~ ~~CLEARFIX~~
 ====Experimental Tasks====
 To analyze the behavior of the integrator, the following circuit is used:\\
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 <wrap left>
-{{drawio>mexlefirst_intern:integrator_experiment.svg}}
+{{drawio>integrator_experiment.svg}}
 </wrap>
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+~~CLEARFIX~~
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 __Supply voltages (from power supply unit):__\\
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 $R1.3=10~kΩ, C1=10~nF$\\
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+~~CLEARFIX~~
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-  - 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>mexlefirst_intern:oscilloscope_screen.svg}}</wrap>\\ \\ \\ Channel 1:$\frac {Volt}{Div}=$\\ \\ \\ Time basis: $\frac {T}{Div}=$\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
-  - 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**\\ \\ <wrap left>{{drawio>mexlefirst_intern:oscilloscope_screen.svg}}</wrap>\\ \\ \\ Channel 1: $\frac {Volt}{Div}=$\\ \\ Channel 2: $\frac {Volt}{Div}=$\\ \\ \\ Time basis: $\frac {T}{Div}=$\\ \\ \\ \\ \\ \\ \\ \\ \\
-  - Compare your measurement with the calculation from part 2 and the forecast from part 3. Explain your result.
+  - 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. \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
+<WRAP hide>
 ====Test Questions - Integrator====
+ </WRAP>