📚 Operational Amplifiers: Working Principles and Key Parameters | 运算放大器：工作原理与关键参数
Operational amplifiers (op-amps) are fundamental building blocks in analogue electronics and a core topic in the CIE A-Level Physics syllabus. This article explains how op-amps work, the key parameters you must memorise, and how to analyse the three standard configurations: inverting, non-inverting and comparator circuits.
运算放大器（op-amp）是模拟电子学中的基本构件，也是 CIE A-Level 物理考纲中的核心专题。本文将讲解运算放大器的工作原理、必须牢记的关键参数，以及反相、同相和比较器三种标准电路的分析方法。
1. What Is an Operational Amplifier? | 什么是运算放大器？
An operational amplifier is a high-gain direct-coupled voltage amplifier with two inputs — one inverting (−) and one non-inverting (+) — and a single output. Its gain is so large that in most circuits negative feedback is used to control the overall behaviour.
运算放大器是一种高增益直流耦合电压放大器，具有两个输入端——一个反相端（−）和一个同相端（+）——以及一个输出端。其增益非常大，因此大多数电路中都要通过负反馈来控制整体行为。
Inverting input (−): a voltage applied here produces an output that is inverted (180° phase shift).
反相输入端（−）：在此处施加电压，输出端会产生反相（180° 相位差）的信号。
Non-inverting input (+): a voltage applied here produces an output that is in phase with the input.
同相输入端（+）：在此处施加电压，输出端会产生与输入同相的信号。
The op-amp is powered by dual supplies, typically +Vs and −Vs, and the output cannot exceed these supply rails.
运算放大器通常采用双电源供电，即 +Vs 和 −Vs，输出电压不能超过这两个电源轨。
2. The Ideal Op-Amp Model | 理想运算放大器模型
CIE exam questions often begin with the ideal op-amp assumptions. You must know these four properties and use them in circuit analysis.
CIE 考试题通常从理想运算放大器的假设出发。你必须掌握这四个特性，并在电路分析中运用它们。
| Property | Ideal value | Consequence |
| Open-loop voltage gain A₀ | Infinity | Even tiny input differences drive output to saturation |
| Input impedance | Infinity | No current flows into either input |
| Output impedance | Zero | Output voltage independent of load |
| Bandwidth | Infinity | Amplifies all frequencies equally |
3. The "Golden Rule" — Virtual Earth | 黄金法则——虚地
When negative feedback is applied and the op-amp is not saturated, the output will adjust itself so that the voltage difference between the two inputs becomes zero. Since the non-inverting input is often connected to 0 V (earth), the inverting input is said to be at "virtual earth" — it is at 0 V but no current flows into it.
当施加负反馈且运算放大器未饱和时，输出会自动调整，使两个输入端之间的电压差变为零。由于同相输入端通常连接到 0 V（地），反相输入端便被称为处于"虚地"——它的电位是 0 V 但没有电流流入。
V₋ = V₊ (virtual short) and I₊ = I₋ = 0
This is the single most important rule for solving A-Level op-amp circuit problems. It allows you to write current equations at the inverting node as if it were a real earth.
这是解决 A-Level 运算放大器电路问题最重要的一条规则。它允许你将反相节点处的电流方程当作真的接地节点来书写。
4. The Inverting Amplifier | 反相放大器
The inverting amplifier is the most tested configuration in CIE A-Level Physics. The input signal is applied through resistor R₁ to the inverting input, and the feedback resistor R₂ is connected between the output and the inverting input. The non-inverting input is grounded.
反相放大器是 CIE A-Level 物理中考查最多的电路。输入信号通过电阻 R₁ 加到反相输入端，反馈电阻 R₂ 连接在输出端和反相输入端之间，同相输入端接地。
Using the virtual earth concept:
利用虚地概念：
Since V₋ = 0, the current through R₁ is I₁ = Vin / R₁. No current enters the op-amp, so the same current flows through R₂. Thus Vout = −I₁R₂ = −(Vin / R₁)R₂.
由于 V₋ = 0，通过 R₁ 的电流为 I₁ = Vin / R₁。没有电流进入运算放大器，因此同样的电流流过 R₂。于是 Vout = −(Vin / R₁)R₂。
Voltage gain = Vout / Vin = −R₂ / R₁
The negative sign indicates a 180° phase inversion. The gain depends only on the two resistors, not on the op-amp itself — this is the beauty of negative feedback.
负号表示产生了 180° 的相位反转。增益只取决于两个电阻的比值，而与运算放大器本身无关——这正是负反馈的妙处。
5. The Non-Inverting Amplifier | 同相放大器
In the non-inverting configuration, the input signal is connected directly to the non-inverting input (+), and the feedback network (R₁ and R₂) is connected between the output, the inverting input (−) and earth. The gain is always greater than or equal to 1.
在同相放大电路中，输入信号直接连接到同相输入端（+），反馈网络（R₁ 和 R₂）连接在输出端、反相输入端（−）和地之间。增益始终大于或等于 1。
Because V₊ = V₋ = Vin, the potential divider formed by R₁ and R₂ gives:
因为 V₊ = V₋ = Vin，由 R₁ 和 R₂ 构成的分压器给出：
Vin = Vout × R₁ / (R₁ + R₂)
Voltage gain = Vout / Vin = 1 + R₂ / R₁
Notice there is no minus sign — the output is in phase with the input. This circuit has a much higher input impedance than the inverting amplifier, because the signal sees the op-amp's input directly.
注意这里没有负号——输出与输入同相。该电路的输入阻抗远高于反相放大器，因为信号直接进入运算放大器的输入端。
6. The Op-Amp as a Comparator | 运算放大器用作比较器
If no negative feedback is applied, the op-amp operates in open-loop mode and acts as a comparator. The output swings to either +Vs or −Vs (positive or negative saturation), depending on which input is at a higher potential.
如果未施加负反馈，运算放大器工作在开环模式，充当比较器。输出会摆向 +Vs 或 −Vs（正向或负向饱和），具体取决于哪个输入端的电位更高。
If V₊ > V₋, then Vout = +Vs (positive saturation).
若 V₊ > V₋，则 Vout = +Vs（正向饱和）。
If V₊ < V₋, then Vout = −Vs (negative saturation).
若 V₊ < V₋，则 Vout = −Vs（负向饱和）。
Comparators are used to convert analogue signals into digital logic levels, for example in temperature-sensing circuits where an LDR or thermistor forms part of a potential divider.
比较器常用于将模拟信号转换为数字逻辑电平，例如在温度传感电路中，光敏电阻或热敏电阻构成分压器的一部分。
7. Saturation and Output Limits | 饱和与输出限制
No real op-amp can output a voltage beyond its supply rails. When the required output exceeds the supply voltage, the op-amp "saturates" — the output stops changing and remains at approximately the supply voltage. In CIE questions, the maximum output is typically taken as the supply voltage Vs (sometimes slightly less).
真实的运算放大器不能输出超过其电源轨的电压。当所需的输出电压超过电源电压时，运算放大器会"饱和"——输出不再变化，而是保持在接近电源电压的数值。在 CIE 考题中，最大输出通常取为电源电压 Vs（有时略低）。
Saturation is a key exam trap: students calculate a gain and multiply it by Vin ue like 15 V, but the supply is only ±9 V. The correct answer is +9 V (or −9 V), not 15 V.
饱和是考题中常见的陷阱：学生计算增益后乘以 Vin，得出 15 V 之类的结果，但电源只有 ±9 V。正确答案是 +9 V（或 −9 V），而不是 15 V。
8. Key Parameters: Gain, Bandwidth and Slew Rate | 关键参数：增益、带宽与压摆率
Beyond the ideal model, real op-amps have finite parameters that A-Level students should be able to interpret.
在理想模型之外，真实的运算放大器具有有限的参数，A-Level 学生应当能够理解这些参数的含义。
| Parameter | Definition | Typical effect |
| Open-loop gain | Gain without feedback, usually 10⁵–10⁶ | Too high for linear use; needs feedback |
| Input impedance | Resistance between inputs, typically MΩ or higher | Prevents loading of the signal source |
| Output impedance | Resistance seen at the output, typically < 100 Ω | Allows driving external loads |
| Bandwidth | Frequency range over which gain is constant | Gain–bandwidth product is constant |
| Slew rate | Maximum rate of change of output voltage (V/µs) | Limits high-frequency large-signal performance |
9. Negative Feedback and Its Benefits | 负反馈及其优点
Negative feedback is the secret behind the op-amp's versatility. By feeding a fraction of the output back to the inverting input, the enormous open-loop gain is traded for predictable, stable behaviour.
负反馈是运算放大器多功能性的秘诀。将输出的一部分反馈到反相输入端，用巨大的开环增益换来了可预测、稳定的行为。
Gain is determined only by external resistors, so it is accurate and stable.
增益仅由外部电阻决定，因此精确且稳定。
Bandwidth is increased compared to the open-loop configuration.
与开环配置相比，带宽得到了扩展。
Distortion is reduced and the frequency response becomes more uniform.
失真被减小，频率响应变得更加均匀。
The input–output relationship becomes linear over a wide range.
输入与输出的关系在很大范围内呈现线性。
Vout = A₀(V₊ − V₋) — with feedback, A₀(V₊ − V₋) is forced toward the required Vout
10. Exam-Style Worked Example | 考试风格例题
an inverting amplifier with R₁ = 2.0 kΩ and R₂ = 12 kΩ. The supply voltages are ±9 V. If Vin = +0.50 V, calculate Vout.
一个运算放大器连接成反相放大器，R₁ = 2.0 kΩ，R₂ = 12 kΩ，电源电压为 ±9 V。若 Vin = +0.50 V，求 Vout。
The theoretical gain is −R₂/R₁ = −12/2 = −6. Therefore Vout = −6 × 0.50 = −3.0 V. Since −3.0 V lies within the supply rails, no saturation occurs — the output is −3.0 V.
理论增益为 −R₂/R₁ = −12/2 = −6，因此 Vout = −6 × 0.50 = −3.0 V。由于 −3.0 V 在电源电压范围内，不会发生饱和——输出即为 −3.0 V。
If instead Vin ctual output limits at −9 V (negative saturation). This example illustrates the importance of always checking saturation.
若改为 Vin = +2.5 V，理论输出为 −15 V。但电源只有 ±9 V，因此实际输出被限制在 −9 V（负向饱和）。这个例子说明了检查饱和的重要性。
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
Students lose marks in op-amp questions for predictable reasons.
学生在运算放大器题目中失分的原因往往是可以预见的。
Forgetting virtual earth: V₋ is not exactly 0 V — it is forced to be equal to V₊. In a standard inverting amplifier this is 0 V, but the key is that no current flows into the input.
忘记虚地：V₋ 并不是真正的 0 V——它被强制等于 V₊。在标准反相放大器中这个值是 0 V，但关键在于没有电流流入输入端。
Mixing up the gain formulas: inverting gain is −R₂/R₁, non-inverting gain is 1 + R₂/R₁. Never swap them.
混淆增益公式：反相增益是 −R₂/R₁，同相增益是 1 + R₂/R₁。千万不要混淆。
Forgetting saturation: always check whether the calculated output exceeds the supply voltage.
忘记饱和：务必检查计算得出的输出电压是否超过电源电压。
Ignoring input impedance: in a comparator, if both inputs are connected through resistors, use potential divider equations to find V₊ and V₋ first.
忽略输入阻抗：在比较器中，如果两个输入端都通过电阻连接，先用分压公式求出 V₊ 和 V₋。
Writing units incorrectly: gain has no units, but Vout is in volts and currents are in amperes.
单位书写错误：增益没有单位，Vout 的单位是伏特，电流的单位是安培。
12. Summary of Essential Equations | 必备公式小结
These are the only equations you need for CIE A-Level op-amp questions.
以下是 CIE A-Level 运算放大器题目中你唯一需要的公式。
Inverting amplifier: Vout = −(R₂ / R₁) Vin
Non-inverting amplifier: Vout = (1 + R₂ / R₁) Vin
Comparator: Vout = +Vs if V₊ > V₋, else Vout = −Vs
Open-loop gain: Vout = A₀(V₊ − V₋)
Master these, practise drawing the circuits from memory, and always check saturation before writing your final answer.
掌握这些公式，练习凭记忆画出电路图，并在写出最终答案之前始终检查是否饱和。
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