Differences
This shows you the differences between two versions of the page.
| Both sides previous revisionPrevious revisionNext revision | Previous revisionLast revisionBoth sides next revision | ||
| learn:courses:real-analog-chapter-5:start [2017/01/19 23:24] – Martha | learn:courses:real-analog-chapter-5:start [2018/11/13 18:30] – [Real Analog: Chapter 5] Martha | ||
|---|---|---|---|
| Line 1: | Line 1: | ||
| ====== Real Analog: Chapter 5 ====== | ====== Real Analog: Chapter 5 ====== | ||
| - | * {{ : | + | [[{}/learn/courses/real-analog-chapter-4/start|Back to Chapter |
| - | * {{ :learn:courses:real-analog-chapter-5: | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| + | --> | ||
| + | * Lecture Material: | ||
| + | * {{ : | ||
| + | * {{ : | ||
| + | * [[http:// | ||
| + | * [[http:// | ||
| + | * Chapter 5 Video: | ||
| + | * [[http:// | ||
| + | * {{ : | ||
| + | * {{ : | ||
| + | * {{ : | ||
| + | * {{ : | ||
| + | * {{ : | ||
| + | * {{ : | ||
| + | * {{ : | ||
| + | * {{ : | ||
| + | * {{ : | ||
| + | * {{ : | ||
| + | * {{ : | ||
| * {{ : | * {{ : | ||
| - | * {{ : | + | * {{ : |
| + | <-- | ||
| ====== 5. Introduction and Chapter Objectives ====== | ====== 5. Introduction and Chapter Objectives ====== | ||
| // | // | ||
| Line 75: | Line 84: | ||
| The above constraint is based on pure inequalities – in general, the output voltage range will be somewhat less than the range specified by V- and V+. The margin between the output and the supply voltages will vary depending on the specific op-amp. Any attempt to drive the output voltage beyond the range specified by the supply voltages will cause the output to // | The above constraint is based on pure inequalities – in general, the output voltage range will be somewhat less than the range specified by V- and V+. The margin between the output and the supply voltages will vary depending on the specific op-amp. Any attempt to drive the output voltage beyond the range specified by the supply voltages will cause the output to // | ||
| - | |||
| - | Similarly it makes sense that the power supply voltages will constrain the range of allowable input voltages, as provided below: | ||
| **Input Voltage Constraint**: | **Input Voltage Constraint**: | ||
| Line 108: | Line 115: | ||
| * The voltages at the input terminals are the same: $v_n = v_p$ | * The voltages at the input terminals are the same: $v_n = v_p$ | ||
| * The output voltage is constrained to be between the positive and negative power supply voltages: $V^- < v_{OUT} < V^+$ | * The output voltage is constrained to be between the positive and negative power supply voltages: $V^- < v_{OUT} < V^+$ | ||
| - | * The input voltages are constrained to be between | + | * Nothing is known about the current out of the op-amp, $i_{OUT}$ |
| * All voltages on the above diagram are relative to the same reference. | * All voltages on the above diagram are relative to the same reference. | ||
| Line 134: | Line 141: | ||
| $$V_{OUT} = K \left(v_p - v_n \right) = K \cdot \Delta v_{in} | $$V_{OUT} = K \left(v_p - v_n \right) = K \cdot \Delta v_{in} | ||
| - | Where in $\Delta v_{in}$ is the difference between the voltages at the input terminals and //K// is a very large number. (Values of //K// for typical commercially available operational amplifiers can be on the order of $10^6$$ or higher.) Since the output voltage is constrained to be less than the supply voltages, | + | Where in $\Delta v_{in}$ is the difference between the voltages at the input terminals and //K// is a very large number. (Values of //K// for typical commercially available operational amplifiers can be on the order of $10^6$ or higher.) Since the output voltage is constrained to be less than the supply voltages, |
| $$V^- < K \cdot \Delta v_{in} < V^+$$ | $$V^- < K \cdot \Delta v_{in} < V^+$$ | ||
| Line 197: | Line 204: | ||
| ==== Exercises ==== | ==== Exercises ==== | ||
| - | - An operational amplifier has a gain K = 10,000. The voltage supplies are $V+ = 20V$ and $V-= -10V$. Determine the output voltage if the voltage difference between the input terminals $\left( v_p - v_n \right) is: | + | - An operational amplifier has a gain K = 10,000. The voltage supplies are $V+ = 20V$ and $V-= -10V$. Determine the output voltage if the voltage difference between the input terminals $\left( v_p - v_n \right)$ is: |
| - 1mV | - 1mV | ||
| - 2mV | - 2mV | ||
| Line 518: | Line 525: | ||
| * The effect of a non-zero output resistance on an operational amplifier’s operation is that the power output of the amplifier is limited. Thus, a realistic operational amplifier will not be able to provide any arbitrary current to a load. Whether this has a significant effect on the overall circuit’s operation is primarily dependent upon the value of the load resistance. | * The effect of a non-zero output resistance on an operational amplifier’s operation is that the power output of the amplifier is limited. Thus, a realistic operational amplifier will not be able to provide any arbitrary current to a load. Whether this has a significant effect on the overall circuit’s operation is primarily dependent upon the value of the load resistance. | ||
| * The effect of a finite op-amp gain is that the voltage difference across the input terminals may not be identically zero. | * The effect of a finite op-amp gain is that the voltage difference across the input terminals may not be identically zero. | ||
| + | |||
| + | |||
| + | [[{}/ | ||
| + | [[{}/ | ||