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Image_1767350962090.png

解答图片中的问题。

Python

python
from PIL import Image, ImageEnhance
img = Image.open('/mnt/data/Image_1767350962090.png')
img.size


(1211, 996)

Python

python
import matplotlib.pyplot as plt, numpy as np
plt.figure(figsize=(12,8))
plt.imshow(img)
plt.axis('off')
plt.show()

Suy nghĩ

Exploring RL=32kΩ misprint and power calculation

The text indeed says RL = 32kΩ, but this doesn’t align with typical LM386 use, which works with 8-32Ω loads. Maybe it’s a misprint or simplified for small power output, but 32kΩ doesn’t work well. Still, I'll follow the question and compute the maximum output power based on this setup.

Determining maximum output power with load connection

If the load is connected directly to the output pin (assuming a coupling capacitor isn't specified), the load would receive both DC and AC. However, as the question mentions "maximum output power" (likely referring to AC power), I'll calculate using the typical formula for AC power with a coupling capacitor, ignoring DC.

Interpreting load connection and type classification

The phrase "若 V_O 处接负载电阻 RL 后接地" seems to imply a direct load connection at Vo and then to ground, which is typical for LM386 with a coupling capacitor. Although the question doesn't explicitly mention the capacitor, it likely refers to an OTL configuration, as this type requires an external coupling capacitor. So, the answer appears to be OTL.

Calculating max output power for OTL configuration

For OTL with a capacitor-coupled RL, the maximum output power depends on the maximum VRMS value, which is calculated as (Vcc/2)/√2. When RL is 32kΩ, I get 0.5625mW of power. If RL is 32Ω instead, this would be 0.5625W. Accounting for DC vs AC components, total power includes both DC and AC contributions.

Clarifying power amplifier type and output power calculations

For "最大输出功率", this usually refers to AC output power, like when using a coupling capacitor — so I'd expect 0.5625mW for the AC power. The amplifier type, based on the question setup, hints at OTL, assuming an external coupling capacitor. In part 3, the calculation for maximum output power would be 0.5625mW for 32k.

Describing voltage-series negative feedback

The feedback sample is taken from the output (shunt), and it's mixed in series with the signal at the inverting input (v-). This creates a voltage-series feedback configuration. Assuming v+ increases and v- is grounded, this feedback opposes the change. The system uses a series mixing configuration, reflected as "电压串联负反馈".

Understanding feedback configuration in LM386

The inverting input (v-) is connected to the base of T1, with R1 to ground (50k). The feedback likely goes to the base of T4, but the output is actually through R6, feeding into the emitter of T4. In the LM386, the feedback configuration is a voltage shunt, with a feedback resistor network forming a voltage divider. This configuration allows for gain adjustment.

Analyzing LM386 feedback classification

It looks like the feedback configuration in the LM386 involves the output being fed back to the input stage, possibly in series or shunt. The feedback path consists of resistors like R6, R5, and R4. The gain is influenced by the resistor between pins 1 and 8, with capacitors bypassing it for higher gain. The network at pin1 seems to connect to T4’s base and emitter.

Feedback path and output power calculation

The feedback path via R6 creates a voltage-series feedback system, affecting the emitter voltage and causing series mixing. At the differential input, feedback influences the v- input through emitter referencing—confirming a voltage-series configuration.

For output power calculation with Vcc = 12V, the peak voltage across the load (RL) results in 0.5625 mW, and if the DC component is considered, 1.6875 mW. I'll keep this calculation clear and concise.

Thought for 3m 2s

Trả lời

这张图是 **单电源音频功率放大器（LM386 一类）**的内部等效电路。按题目三问：

1) 瞬时极性法判断反馈及反馈组态

取一个瞬时：令输入端使输出端 vOv_O 瞬时“上升”（输出端取“+”）。

  • 输出电压 vOv_O 的上升，经 R6(15kΩ)R_6(15k\Omega) 反馈到左侧增益/反馈网络（到(1)端及其内部通道）。
  • 该反馈作用到输入级，使等效的“反相端”电位随之上升，从而使差模输入 (v+−v−)(v_+ - v_-) 变小，抵消原来使 vOv_O 上升的变化。

因此这是 负反馈。

从“取样/混合”看：

  • 在输出端取样的是 电压 vOv_O（电压取样）；
  • 在输入端以差分方式等效为 串联叠加/抵消（使 v+−v−v_+ - v_- 变小）。

➡️ 反馈组态：电压串联负反馈（也常称“电压—串联”负反馈）。

2) 接法对应 OTL / OCL / BTC 哪一种？

题设“在 vOv_O 处接负载电阻 RLR_L 后接地”，且该电路是 单电源 VCCV_{CC}，输出静态工作点通常在 约 VCC/2V_{CC}/2。

  • OCL：通常需要 ± 双电源（输出静态在 0V），这里不符合；
  • BTC/BTL：负载应接在两路输出之间（桥接），这里是单端对地，不符合；
  • 单电源、单端输出对地要避免直流流过负载，典型做法是 输出串联电容耦合负载 ⇒ OTL。

➡️ 该功放属于 OTL。

3) VCC=12V,  RL=32kΩV_{CC}=12V,\; R_L=32k\Omega 时最大输出功率 PomP_{om}

OTL 单电源、输出静态在 VCC/2V_{CC}/2 时，理想情况下最大不失真正弦摆幅约为：

  • 输出交流峰值：  Vo,pk≈VCC/2=6V\;V_{o,pk} \approx V_{CC}/2 = 6V
  • 输出交流有效值：  Vo,rms=Vo,pk2=62≈4.24V\;V_{o,rms} = \dfrac{V_{o,pk}}{\sqrt2}=\dfrac{6}{\sqrt2}\approx 4.24V

最大输出功率：

Pom=Vo,rms2RL=(VCC22)2RL=VCC28RLP_{om}=\frac{V_{o,rms}^2}{R_L} =\frac{\left(\frac{V_{CC}}{2\sqrt2}\right)^2}{R_L} =\frac{V_{CC}^2}{8R_L}

代入 VCC=12V,  RL=32kΩV_{CC}=12V,\;R_L=32k\Omega：

Pom=1228×32000=144256000=5.625×10−4  W=0.5625  mWP_{om}=\frac{12^2}{8\times 32000} =\frac{144}{256000} =5.625\times 10^{-4}\;W =0.5625\;mW

➡️ Pom≈0.56 mWP_{om}\approx 0.56\,\text{mW}（按理想最大正弦摆幅计算）。

注：若负载真的是“直接接在 vOv_O 对地且没有输出隔直电容”，则还会有约 VCC/2V_{CC}/2 的直流分量导致额外直流功耗；但题第(2)问已指向 OTL（默认输出电容耦合），所以通常按上面的交流输出功率计算。

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