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circuit_design:3_opamp_basic_circuits_i [2023/03/28 10:57]
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circuit_design:3_opamp_basic_circuits_i [2023/09/19 22:16] (aktuell)
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| ====== 3. Basic circuits of operational amplifiers I ====== | ====== 3 Basic circuits of operational amplifiers I ====== |
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| <callout>. | <callout>. |
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| ===== 3.5 Inverting Amplifier ===== | ===== 3.5 Inverting Amplifier ===== |
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| | The circuit of the inverting amplifier can be derived from that of the non-inverting amplifier (see <imgref pic8>). To do this, first consider the noninverting amplifier as a system with 3 connections (or as a quadripole): $U_\rm I$, $\rm GND$, and $U_\rm O$. These terminals can be rearranged - while keeping the output terminal $U_\rm O$. |
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| <WRAP><panel type="default"> <imgcaption pic6|Inverting Amplifier> </imgcaption> {{drawio>Invertierender_Verstärker_Schaltung.svg}} </panel></WRAP> | <WRAP><panel type="default"> <imgcaption pic6|Inverting Amplifier> </imgcaption> {{drawio>Invertierender_Verstärker_Schaltung.svg}} </panel></WRAP> |
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| <WRAP><imgcaption pic8|Converting non-inverting amplifier to inverting amplifier> (here: $U_{\rm E}=U_{\rm I}$ and $U_{\rm A} = U_{\rm O}$){{:circuit_design:inv_2_ninv.gif}}</imgcaption></WRAP> | Thus the voltage divider $R_1 + R_2$ is no longer between $U_\rm O$ and $\rm GND$, but between $U_\rm O$ and $U_\rm O$, see <imgref pic6>. |
| | In this circuit, the resistor $R_2$ is also called the negative feedback resistor. |
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| The circuit of the inverting amplifier can be derived from that of the non-inverting amplifier (see <imgref pic8>). To do this, first consider the noninverting amplifier as a system with 3 connections (or as a quadripole): $U_\rm I$, $\rm GND$, and $U_\rm O$. These terminals can be rearranged - while keeping the output terminal $U_\rm O$. Thus the voltage divider $R_1 + R_2$ is no longer between $U_\rm O$ and $\rm GND$, but between $U_\rm O$ and $U_\rm O$, see <imgref pic6>. | <WRAP><imgcaption pic8|Converting non-inverting amplifier to inverting amplifier. U_E is the input voltage (Eingangsspannung), U_A is the output voltage (Ausgangspannung).> (here: $U_{\rm E}=U_{\rm I}$ and $U_{\rm A} = U_{\rm O}$){{:circuit_design:inv_2_ninv.gif}}</imgcaption></WRAP> |
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| In this circuit, the resistor $R_2$ is also called the negative feedback resistor. | |
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| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
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| <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5BOJyWoVaAmTkDseBGADgIFYA2Sc-AFnJFNJAMgddIFMBaAggKBy4QRVpgrNMREGPo0QAVQD6AeQF4QPStNyYNBXPUw6QcpQEk1QruTmYarLjSl3WpxRYBKerUd08Eur5QJibswTBUCCRg+nC4CLi4MVIwpHwATtJ20vZZcmCQcqwscHwA5nnghZVgOOF89hBc2PnkUjwEujROwQSKNIqQigAeJJCdiiyD0H1DmIqY0KSDk4pg05BiK-IAOgDOFo3M3eBt2l09rEMDQ6MFExFws7BPK-OLy0ND649wW0O7PaqQSVFzMAI5VwKFR8ABGGjBLT0dRiTEgfGGCLA6DAYCsBBOYFycmIIDMADsAA4AVwALvsAMoASzK5IAhgAbOHHITYdCYajgMD0dGYggFcDiMT5UjoElSAAyAHs2QATfZK8n7ZR0mm0vgAd2CQU05yg5W8ZtNBSK5qNImk4gdOBSGWEokgKWknuN8HRmVEuWstlyxT9ht9XrB6LZxqdoidzCYPCYJT9GD9SBIASM5FxVEwtV6fvRR0cUmFUgd3S9xH6K1GLAI5EmXxe4zeGyeaB7WZWdZ+7c2n0UgIA0vwQR72s3wV0QBO+EA noborder}} </WRAP> | |
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| Before the voltage gain is determined, the node $\rm K1$ in <imgref pic6> is to be considered first. This is just larger than the ground potential by the voltage $U_\rm D$; thus, it lies on the potential difference $U_\rm D$. For a feedback amplifier with finite voltage supply, $U_\rm O$ can only be finite, and thus $U_{\rm D}= U_{\rm O} / A_{\rm D} \rightarrow 0$ (cf. [[https://wiki.mexle.org/circuit_design/3_opamp_basic_circuits_i#basic_equationgolden_rules|basic_equation of the operational amplifier]]), since $A_D \rightarrow \infty$ holds. Thus it can be seen that the node $\rm K1$ is __always__ at ground potential in the ideal operational amplifier. This property is called **virtual ground** because there is no direct short to ground. The op-amp regulates its output voltage $U_\rm O$ in such a way that the voltage divider sets a potential of $0~\rm V$ at node $\rm K1$. This can also be seen in the simulation by the voltage curve at $\rm K1$. | Before the voltage gain is determined, the node $\rm K1$ in <imgref pic6> is to be considered first. This is just larger than the ground potential by the voltage $U_\rm D$; thus, it lies on the potential difference $U_\rm D$. For a feedback amplifier with finite voltage supply, $U_\rm O$ can only be finite, and thus $U_{\rm D}= U_{\rm O} / A_{\rm D} \rightarrow 0$ (cf. [[https://wiki.mexle.org/circuit_design/3_opamp_basic_circuits_i#basic_equationgolden_rules|basic_equation of the operational amplifier]]), since $A_D \rightarrow \infty$ holds. Thus it can be seen that the node $\rm K1$ is __always__ at ground potential in the ideal operational amplifier. This property is called **virtual ground** because there is no direct short to ground. The op-amp regulates its output voltage $U_\rm O$ in such a way that the voltage divider sets a potential of $0~\rm V$ at node $\rm K1$. This can also be seen in the simulation by the voltage curve at $\rm K1$. |
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| | The following diagram shows again the interactive simulation. \\ |
| | $R_{\rm 1}$ and $R_{\rm 2}$ can be manipulated by the sliders. Hit ''Run / STOP'' to run |
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| | <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5BOJyWoVaAmTkDseBGADgIFYA2Sc-AFnJFNJAMgddIFMBaAggKBy4QRVpgrNMREGPo0QAVQD6AeQF4QPStNyYNBXPUw6QcpQEk1QruTmYarLjSl3WpxRYBKerUd08Eur5QJibswTBUCCRg+nC4CLi4MVIwpHwATtJ20vZZcmCQcqwscHwA5nnghZVgOOF89hBc2PnkUjwEujROwQSKNIqQigAeJJCdiiyD0H08ipjz0KSDk4pg05BiK-IAOgDOFo3M3eBt2l09rEMDQ6MFExFwfY-jinMLmEsrQ+svm8tDXZ7VSCSouZgBHKuBQqPgAIw04JaejqMSYkD4w0RYHQYDAVgIJzAuTkxBAZgAdgAHACuABd9gBlACWZQpAEMADbw45CbDoTDUcBgegYrEEArgcRifKkdCkqQAGQA9uyACb7ZUU-bKem0ul8ADuwSCmnOUHK3nNZoKRQtxpE0nEjpwKQywlEkBS0i9JvgGMyoly1lsuWK-qNfu94Ix7JNztEzuYTB4TBK-ow-tIgpwNhFmxoxLovX9GKOjikIqkju63uI-RWoxYBHIkyGL1mfQWHbQvaQRBW9d+sHgW0B+wA0vxQZ72i2IV0QFO+GApPoQAAxAi9H3wPQgDz8VfMehbnc4Pc8A+YPhAA noborder}} </WRAP> |
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| <WRAP column 100%> <panel type="danger" title="Notice: virtual ground"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%> For the ideal feedback amplifier, $U_\rm D \rightarrow 0$ holds. This means that the same voltage is always present at both inputs. if one of the two voltages is fixed, for example by connecting ground potential or even by a fixed voltage source, this property is called **virtual ground**. </WRAP></WRAP></panel> </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~ | <WRAP column 100%> <panel type="danger" title="Notice: virtual ground"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%> For the ideal feedback amplifier, $U_\rm D \rightarrow 0$ holds. This means that the same voltage is always present at both inputs. if one of the two voltages is fixed, for example by connecting ground potential or even by a fixed voltage source, this property is called **virtual ground**. </WRAP></WRAP></panel> </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~ |
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| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
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| <WRAP><panel type="default"> <imgcaption pic7|Inverting Amplifier - Animation> </imgcaption> | |
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| {{url>https://www.geogebra.org/material/iframe/id/hhxhcqbp/width/600/height/700/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false 400,500 noborder}} </panel></WRAP> | |
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| For the second derivation, the current flow through the resistors $R_1$ and $R_2$ of the unloaded voltage divider is to be considered. These two currents $I_1$ and $I_2$ are just equal. Thus: | For the second derivation, the current flow through the resistors $R_1$ and $R_2$ of the unloaded voltage divider is to be considered. These two currents $I_1$ and $I_2$ are just equal. Thus: |
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| This can also be converted into a "seesaw" or mechanical analog via **like triangles**. In the mechanical analog, the potentials are given by height. As in the electrical case with the ground potential, a height reference plane must be chosen in the mechanical picture. The electric currents correspond to forces (i.e., a momentum flux) - but the consideration of forces is not necessary here. [(Note1> To complete the mechanical analogue of the setup, one can assume that there is an external "force source". This always acts in such a way that it always lands on the height reference surface at the point corresponding to the virtual mass)]. | This can also be converted into a "seesaw" or mechanical analog via **like triangles**. In the mechanical analog, the potentials are given by height. As in the electrical case with the ground potential, a height reference plane must be chosen in the mechanical picture. The electric currents correspond to forces (i.e., a momentum flux) - but the consideration of forces is not necessary here. [(Note1> To complete the mechanical analogue of the setup, one can assume that there is an external "force source". This always acts in such a way that it always lands on the height reference surface at the point corresponding to the virtual mass)]. |
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| | <WRAP><panel type="default"> <imgcaption pic7|Inverting Amplifier - Animation> </imgcaption> |
| | {{url>https://www.geogebra.org/material/iframe/id/hhxhcqbp/width/600/height/700/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false 400,500 noborder}} </panel></WRAP> |
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| Now, if a certain height (voltage $U_\rm I$) is set, a certain height on the right side (voltage $U_\rm O$) is obtained via the force arm (resistor $R_1$) and load arm (resistor $R_2$). This is shown in <imgref pic7> above. In the figure, all points marked in red (<fc #ff0000>{{fa>circle?10}}</fc>) can be manipulated. Accordingly, the input voltage $U_{\rm I} = U_{\rm in}$ is adjustable and automatically results in a voltage $U_{\rm O}=U_{\rm out}$. In the circuit (figure below), the resistors $R_1$ and $R_2$ can be changed. | Now, if a certain height (voltage $U_\rm I$) is set, a certain height on the right side (voltage $U_\rm O$) is obtained via the force arm (resistor $R_1$) and load arm (resistor $R_2$). This is shown in <imgref pic7> above. In the figure, all points marked in red (<fc #ff0000>{{fa>circle?10}}</fc>) can be manipulated. Accordingly, the input voltage $U_{\rm I} = U_{\rm in}$ is adjustable and automatically results in a voltage $U_{\rm O}=U_{\rm out}$. In the circuit (figure below), the resistors $R_1$ and $R_2$ can be changed. |
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| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
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| ====== Exercises ====== | ====== Exercises ====== |