Ideal operational amplifiers
引入| English | 释义 | 拼音 |
|---|---|---|
| operational amplifier/ˌɒpəˈreɪʃənl ˈæmplɪfaɪə/ | 运算放大器 | yùn suàn fàng dà qì |
| ideal operational amplifier/aɪˈdɪəl ˌɒpəˈreɪʃənl ˈæmplɪfaɪə/ | 理想运算放大器 | lǐ xiǎng yùn suàn fàng dà qì |
| negative feedback/ˈneɡətɪv ˈfiːdbæk/ | 负反馈 | fù fǎn kuì |
| virtual short/ˈvɜːtʃuːəl ʃɔːt/ | 虚短路 | xū duǎn lù |
| inverting amplifier/ɪnˈvɜːtɪŋ ˈæmplɪfaɪə/ | 反相放大器 | fǎn xiāng fàng dà qì |
| non-inverting amplifier/nɒn ɪnˈvɜːtɪŋ ˈæmplɪfaɪə/ | 同相放大器 | tóng xiāng fàng dà qì |
| voltage follower | 电压跟随器 | diàn yā gēn suí qì |
| summing amplifier | 求和放大器 | qiú hé fàng dà qì |
| saturation/ˌsætʃəˈreɪʃn/ | 饱和 | bǎo hé |
From 10 mV to 2 V
- A microphone produces 10 mV; a speaker amplifier wants 2 V. Something must multiply by 200, with a way to set exactly how much.
- That something is the operational amplifier. The whole lesson rests on one trick: feedback makes a device with enormous gain behave like an exact, mild one.
The ideal op-amp and its two golden rules
- An operational amplifier 运算放大器 amplifies the difference between its inputs. The ideal operational amplifier 理想运算放大器 has infinite gain, infinite input resistance and zero output resistance.
- With negative feedback 负反馈 to the $-$ input: (1) no current enters either input; (2) the inputs sit at the same voltage, a virtual short 虚短路. With a gain above $10^{5}$, volts at the output need only microvolts between the inputs.
The virtual-short rule says the two input voltages are equal. Why?
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The output can only stay inside the rails if the fed-back difference is essentially zero - A × (v+ − v−) finite with A → ∞ requires v+ ≈ v−.
The inverting amplifier
- Inverting amplifier 反相放大器: input through $R_{in}$ to the $-$ node, feedback $R_{f}$, $+$ grounded. Rule 2 holds the $-$ node at 0 V, a virtual ground; rule 1 sends all of $\dfrac{v_{in}}{R_{in}}$ through $R_{f}$.
- So $\dfrac{v_{out}}{v_{in}} = -\dfrac{R_{f}}{R_{in}}$. The chip's huge gain cancels out, and the source sees exactly $R_{in}$.
$R_{in} = 10\,\text{k}\Omega$, $R_{f} = 100\,\text{k}\Omega$: gain −10, so 0.2 V in gives −2 V out. The 20 μA input current flows on through $R_{f}$: $20\,\mu\text{A} \times 100\,\text{k}\Omega = 2\,\text{V}$.
An inverting amplifier has R_in = 4.7 kΩ and R_f = 47 kΩ. v_in = +0.15 V gives what output?
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Gain = −47/4.7 = −10, so the output is −10 × 0.15 = −1.5 V, comfortably inside ±15 V.
Non-inverting gain and the follower
- Non-inverting amplifier 同相放大器: signal at $+$, divider at $-$: gain $1 + \dfrac{R_{f}}{R_{in}}$, never below 1. Its input draws no current at all.
- With $R_{f} = 0$ the gain is exactly 1: the voltage follower 电压跟随器, a buffer that lets a weak source drive a heavy load.
$R_{in} = 10\,\text{k}\Omega$, $R_{f} = 30\,\text{k}\Omega$: gain $1 + 3 = 4$, so 0.5 V in gives 2 V out.
A non-inverting amplifier has R_in = 10 kΩ and R_f = 47 kΩ. What is its gain?
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$1 + \dfrac{47}{10} = 5.7$. A non-inverting gain can never fall below 1.
The virtual ground adds signals
- Feed two inputs, each through its own resistor, into the same virtual ground: their currents add in $R_{f}$. This is the summing amplifier 求和放大器.
- $v_{out} = -\left(\dfrac{R_{f}}{R_{1}}v_{1} + \dfrac{R_{f}}{R_{2}}v_{2}\right)$. Audio mixers work exactly this way.
$R_{1} = R_{2} = 10\,\text{k}\Omega$, $R_{f} = 100\,\text{k}\Omega$, $v_{1} = 0.2\,\text{V}$, $v_{2} = -0.05\,\text{V}$: $v_{out} = -(2 - 0.5) = -1.5\,\text{V}$.
Match each circuit to its gain.
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All four come from the same two golden rules applied to different feedback networks.
A summing amplifier has R_f = 20 kΩ, R_1 = 10 kΩ and R_2 = 5 kΩ. With v_1 = 0.3 V and v_2 = 0.1 V, what is v_out, in volts?
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$v_{out} = -\left(\dfrac{20}{10} \times 0.3 + \dfrac{20}{5} \times 0.1\right) = -(0.6 + 0.4) = -1.0\,\text{V}$.
Saturation ends the linear story
- Saturation 饱和: the output cannot leave the supply rails, and usually stops 1 to 2 V inside them. A ±15 V op-amp asked for −30 V stops near −14 V; beyond that the virtual short is simply false.
- Always check $|\text{gain} \times \text{input}|$ against the rails before quoting an output.
Gain −100 with $v_{in} = 0.3\,\text{V}$ asks for −30 V; the real output saturates near −14 V. "The output saturates" is the correct answer, not −30 V.
A gain −100 inverting stage with ±15 V supplies receives 0.3 V. The best answer is —
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The linear answer −30 V exceeds the rails, so the output stage clips near −14 V and the virtual short breaks down.
Cascaded stages multiply
- Two inverting stages multiply their gains: $(-2) \times (-3) = +6$. Each stage keeps its own feedback loop, so each has its own virtual short.
- Check every stage against the rails, not just the last: 0.4 V → −0.8 V → +2.4 V stays well inside ±15 V.
Two inverting stages with gains −2 and −3 are cascaded. The overall gain is —
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Cascaded gains multiply - (−2) × (−3) = +6, two inversions restore the original polarity.
Three common slips
- Losing the minus sign of the inverting stage, or "correcting" it away.
- Using the virtual short without negative feedback. No feedback path to the $-$ input, no virtual short.
- Reporting a voltage beyond the rails. The answer is "it saturates", not the impossible number.
Without any feedback path, the two inputs of an op-amp still sit at the same voltage.
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The virtual short is produced by negative feedback. Without it, any small input difference drives the output straight to a rail.
Recap
- Negative feedback gives the golden rules: no input current, no input voltage difference.
- Inverting gain $-\dfrac{R_{f}}{R_{in}}$; non-inverting $1 + \dfrac{R_{f}}{R_{in}}$; summing adds weighted inputs.
- The output is limited by the rails; check every stage.