Übungsblatt 6 \\ \\ \\ \\ Bitte laden Sie das aufgefüllte PDF in ILIAS hoch. Details, Tipps und Tools zum Ausfüllen und Einfügen von Bildern finden Sie unter: \\ [[:tools_fuer_lehr_lern-veranstaltungen|Tools für Lehr/Lern-Veranstaltungen]] \\ \\ \\ \\ ^ Name ^ Vorname ^ Matrikelnummer ^ | $\quad\quad\quad\quad\quad\quad\quad\quad$ \\ (nbsp) \\ (nbsp)| $\quad\quad\quad\quad\quad\quad\quad\quad$| $\quad\quad\quad\quad\quad\quad\quad\quad$ | | (nbsp) \\ (nbsp) \\ (nbsp)| | | {{fa>pencil?32}} In the following, the transfer function of the differential amplifier is to be calculated. To do this, you should follow a few steps. - Derive the function $U_{\rm O} = f(U_{\rm I1}, U_{\rm I2})$ using superposition. - To do this, first draw an equivalent circuit in each case. - Briefly describe the resulting circuit. Which amplification circuit results in each case? - Then calculate the voltages $U_{\rm O1}$ and $U_{\rm O2}$, and from them $U_\rm O$. - Determine the function $U_{\rm O} = f(U_{\rm I1}, U_{\rm I2})$ or the resistance values ​​of the circuit shown. {{fa>pencil?32}} Derive the gain for the current-voltage converter, i.e. the transmission resistance. Use the procedure that we used for the other amplifiers. - Draw a circuit with the relevant voltages, currents, resistances, and the operational amplifier - What are you looking for? - What is the number of variables - What is the number of equations required? - Establishing the known equations. - Derivation of the transmission resistance. {{fa>pencil?32}} Derive the "amplification" for the voltage-to-current converter, i.e. the transfer conductance. Use the same procedure that we used for the other amplifiers. - Draw a circuit with the relevant voltages, currents, resistances, and the operational amplifier. - What are you looking for? - What is the number of variables? - What is the number of necessary equations? - Establishing the known equations. - Derivation of the transfer conductance. {{fa>pencil?32}} {{ elektronische_schaltungstechnik:erdbezogenelast.jpg?200|}} If the voltage-current converter is used as a current source, it must be ensured that the load has no contact with the ground. - Draw the voltage-to-current transformer with a load that is in contact with the ground. - Why is in this case the slope derived above no longer valid as an amplification factor? - Will the output current be higher or lower in this case?