📚 CIE A-Level Physics: Non-inverting Amplifier Circuit Analysis and Gain Calculation | CIE A-Level 物理：同相放大器电路分析与增益计算
Operational amplifiers (op-amps) are one of the most important components in the CIE A-Level Physics syllabus. The non-inverting amplifier is a classic negative feedback configuration that you must be able to analyse quickly and accurately. This article explains the circuit, derives the gain formula, and gives you exam-style worked examples.
运算放大器（运放）是 CIE A-Level 物理考纲中最重要的元件之一。同相放大器是经典的负反馈结构，你必须能够快速、准确地对其进行分析。本文将解释电路结构、推导增益公式，并给出考试风格的例题。
1. Introduction to Operational Amplifiers | 运算放大器简介
An operational amplifier is a high-gain differential voltage amplifier. It has two inputs — the inverting input (−) and the non-inverting input (+) — and a single output.
运算放大器是一种高增益的差分电压放大器。它有两个输入端——反相输入端（−）和同相输入端（+）——以及一个输出端。
With no feedback (open-loop), the output is given by V_out = A₀(V₊ − V₋), where A₀ is the open-loop gain, typically 10⁵ or more. Because A₀ is so large, even a tiny difference between the inputs drives the output to saturation.
在没有反馈（开环）的情况下，输出电压为 V_out = A₀(V₊ − V₋)，其中 A₀ 是开环增益，通常为 10⁵ 或更大。由于 A₀ 很大，即使输入之间只有微小差值，也会使输出驱动到饱和状态。
2. The Non-inverting Amplifier Configuration | 同相放大器电路结构
The circuit diagram is straightforward: the input voltage V_in is connected directly to the non-inverting input (+). A potential divider made from two resistors R₁ and R_f supplies a fraction of the output voltage back to the inverting input (−).
电路图非常简单：输入电压 V_in 直接连接到同相输入端（+）。由两个电阻 R₁ 和 R_f 组成的分压器将输出电压的一部分反馈到反相输入端（−）。
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R₁ is connected from the inverting input to ground (0 V).
R₁ 连接在反相输入端与地（0 V）之间。
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R_f is the feedback resistor, connected from the output to the inverting input.
R_f 是反馈电阻，连接在输出端与反相输入端之间。
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The input signal sees the very high input impedance of the op-amp, so no significant current flows into the (+) terminal.
输入信号面对的是运放极高的输入阻抗，因此没有显著电流流入（+）输入端。
3. The Ideal Op-Amp Model | 理想运放模型
CIE A-Level questions assume an ideal op-amp unless stated otherwise. The ideal properties are:
除非另有说明，CIE A-Level 题目默认使用理想运放。理想性质如下：
| Ideal property | Consequence |
| Infinite open-loop gain A₀ | V₊ = V₋ (virtual short) |
| Infinite input impedance | No current enters the input terminals |
| Zero output impedance | Output voltage independent of load current |
For negative feedback, the most powerful concept is the virtual short: because A₀ is infinite and V_out is finite, V₊ − V₋ must be effectively zero. Hence V₊ = V₋.
对于负反馈，最强大的概念是虚短：由于 A₀ 无穷大而 V_out 有限，因此 V₊ − V₋ 必须近似为零。于是 V₊ = V₋。
4. Deriving the Voltage Gain Formula | 推导电压增益公式
In the non-inverting amplifier, V₊ = V_in. The virtual short forces V₋ = V_in.
在同相放大器中，V₊ = V_in。虚短迫使 V₋ = V_in。
The potential divider between output and ground gives the voltage at the inverting input:
输出端与地之间的分压器给出反相输入端电压：
V₋ = V_out × R₁ / (R₁ + R_f)
Since V₋ = V_in, we can write:
由于 V₋ = V_in，可以写出：
V_in = V_out × R₁ / (R₁ + R_f)
Rearranging to find the closed-loop voltage gain A_V:
整理得到闭环电压增益 A_V：
A_V = V_out / V_in = 1 + R_f / R₁
This is the key formula for the CIE A-Level non-inverting amplifier. Notice that the gain is always greater than 1, and it is positive, meaning the output is in phase with the input.
这是 CIE A-Level 同相放大器的关键公式。注意增益始终大于 1，且为正值，表示输出与输入同相。
5. Input and Output Impedance | 输入与输出阻抗
Because the input signal is applied directly to the non-inverting input of an ideal op-amp, no current flows into the amplifier. The input impedance is therefore extremely high (ideally infinite).
由于输入信号直接施加于理想运放的同相输入端，没有电流流入放大器。因此输入阻抗极高（理想情况下为无穷大）。
Negative feedback drastically reduces the output impedance. The op-amp adjusts its output to keep V₋ = V_in, so the output behaves like an ideal voltage source. This makes the non-inverting amplifier an excellent buffer for driving loads without affecting the source.
负反馈大大降低了输出阻抗。运放会不断调整其输出以保持 V₋ = V_in，因此输出端表现得像理想电压源。这使得同相放大器成为驱动负载而不影响信号源的优秀缓冲器。
6. Effect of Negative Feedback on Bandwidth | 负反馈对带宽的影响
Negative feedback exchanges gain for bandwidth. The product of the closed-loop gain and the closed-loop bandwidth is approximately constant (equal to the gain–bandwidth product, GBW).
负反馈以增益换取带宽。闭环增益与闭环带宽的乘积近似为常数（等于增益带宽积 GBW）。
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If you increase R_f / R₁, the gain increases but the bandwidth decreases.
如果增大 R_f / R₁，增益增大但带宽减小。
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If you reduce R_f / R₁, the gain approaches 1 and the bandwidth approaches the op-amp’s maximum.
如果减小 R_f / R₁，增益趋近于 1，带宽趋近于运放的最大值。
In A-Level exams, you may be asked to compare the gain of an amplifier with and without feedback, or to explain qualitatively how feedback affects bandwidth and stability.
在 A-Level 考试中，你可能会被要求比较有反馈和无反馈时放大器的增益，或定性解释反馈如何影响带宽与稳定性。
7. Saturation and Practical Output Limits | 饱和与实际输出限制
Real op-amps cannot output a voltage beyond their supply rails. If the required V_out = A_V × V_in exceeds the supply voltage, the op-amp saturates and the output is clamped to the maximum (or minimum) available voltage.
实际运放无法输出超过其电源轨的电压。如果所需的 V_out = A_V × V_in 超过电源电压，运放就会饱和，输出被限制在最大（或最小）可用电压。
For example, with ±9 V supplies, the output cannot exceed about +9 V or go below about −9 V. In practice, the maximum output may be 1–2 V below the supply rails, but CIE A-Level questions usually treat the rails as the limits.
例如，使用 ±9 V 电源时，输出不能超过约 +9 V，也不能低于约 −9 V。实际上，最大输出可能比电源轨低 1–2 V，但 CIE A-Level 题目通常将电源轨视为极限。
8. Worked Example: Calculating Gain and Output | 例题：计算增益与输出
Question. A non-inverting amplifier uses R₁ = 10 kΩ and R_f = 50 kΩ. The op-amp supplies are ±9 V.
题目。 一个同相放大器使用 R₁ = 10 kΩ 和 R_f = 50 kΩ。运放电源为 ±9 V。
(a) Calculate the voltage gain. (b) Find V_out when V_in = 0.20 V. (c) Determine whether the output saturates when V_in = 0.30 V.
（a）计算电压增益。（b）当 V_in = 0.20 V 时求 V_out。（c）判断当 V_in = 0.30 V 时输出是否饱和。
Solution (a) Using the gain formula:
解答（a） 使用增益公式：
A_V = 1 + R_f / R₁ = 1 + 50 / 10 = 6
Solution (b)
解答（b）
V_out = A_V × V_in = 6 × 0.20 = 1.20 V
Solution (c) When V_in = 0.30 V
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