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

📚 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 p 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（−）。

• 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。

• 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。

• 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）。

ack 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。

---

广告 / Sponsored

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）。

• 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。

• 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。

utput cannot exceed about +9 V or go below about −9 V. In practice, the maximum output may be 1– 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

Published by TutorHao | A-Level Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）