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# 分压器

${i}_{\text{1}}={i}_{\text{2}}\phantom{\rule{2em}{0ex}}$ 我们直接称之为 $i$

$v=i\phantom{\rule{0.167em}{0ex}}\text{R}\phantom{\rule{2em}{0ex}}$ 欧姆定律
${v}_{in}=i\phantom{\rule{0.167em}{0ex}}\left(\text{R1}+\text{R2}\right)$

$i={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{1}{\text{R1}+\text{R2}}$

${v}_{out}=i\phantom{\rule{0.167em}{0ex}}\text{R2}$

${v}_{out}=\left({v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{1}{\text{R1}+\text{R2}}\right)\text{R2}$

### 示例 - 使用分压器方程来求出 ${v}_{out}$‍

${v}_{out}={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{\text{R2}}{\text{R1}+\text{R2}}$

${v}_{out}=12\phantom{\rule{0.167em}{0ex}}\text{V}\cdot \frac{3\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}{1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }+3\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}$
${v}_{out}=12\phantom{\rule{0.167em}{0ex}}\text{V}\cdot \frac{3\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}{4\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}$
${v}_{out}=12\phantom{\rule{0.167em}{0ex}}\text{V}\cdot \frac{3}{4}=9\phantom{\rule{0.167em}{0ex}}\text{V}$

$i=\frac{{v}_{in}}{\text{R}1+\text{R}2}=\frac{12\phantom{\rule{0.167em}{0ex}}\text{V}}{1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }+3\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}=\frac{12\phantom{\rule{0.167em}{0ex}}\text{V}}{4\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}=3\phantom{\rule{0.167em}{0ex}}\text{mA}$

$p=i\phantom{\rule{0.167em}{0ex}}v=3\phantom{\rule{0.167em}{0ex}}\text{mA}\cdot 12\phantom{\rule{0.167em}{0ex}}\text{V}=36\phantom{\rule{0.167em}{0ex}}\text{mW}$

### 分压器实践问题

#### 问题 1

${v}_{in}=6\phantom{\rule{0.167em}{0ex}}\text{V}$, $\text{R}1=50\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$, $\text{R}2=10\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$

${v}_{out}=\phantom{\rule{1em}{0ex}}$
$\text{V}$

#### 问题 2

$\text{R}1=90\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$, $\text{R}2=10\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$

${v}_{in}=\phantom{\rule{1em}{0ex}}$
$\text{V}$

#### 问题 3

${v}_{in}=5\phantom{\rule{0.167em}{0ex}}\text{V}$, ${v}_{out}=2\phantom{\rule{0.167em}{0ex}}\text{V}$, $\text{R}1=30\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$

$\text{R}2=\phantom{\rule{1em}{0ex}}$
$\mathrm{\Omega }$

#### 问题4 - 挑战

${v}_{in}=1\phantom{\rule{0.167em}{0ex}}\text{V}$, ${v}_{out}=\frac{{v}_{in}}{2}$

$\text{R}1=\phantom{\rule{1em}{0ex}}$
$\mathrm{\Omega }\phantom{\rule{2em}{0ex}}$
$\text{R}2=\phantom{\rule{1em}{0ex}}$
$\mathrm{\Omega }$

## 检查假设（有难度）

### 在中值附近操作分压器

${\text{R}}_{\text{L}}=10\phantom{\rule{0.167em}{0ex}}\text{R2}$
$\text{R2}$${\text{R}}_{\text{L}}$ 是并联关系。使用并联电阻公式可以得到2个并联电阻的组合值 $\text{R2}\phantom{\rule{0.167em}{0ex}}||\phantom{\rule{0.167em}{0ex}}{\text{R}}_{\text{L}}$
$\text{R2}\phantom{\rule{0.167em}{0ex}}||\phantom{\rule{0.167em}{0ex}}{\text{R}}_{\text{L}}=\frac{\text{R2}\cdot {\text{R}}_{\text{L}}}{\text{R2}+{\text{R}}_{\text{L}}}=\frac{\text{R2}\cdot 10\phantom{\rule{0.167em}{0ex}}\text{R2}}{\text{R2}+10\phantom{\rule{0.167em}{0ex}}\text{R2}}=\frac{10}{11}\phantom{\rule{0.167em}{0ex}}\text{R2}=0.91\phantom{\rule{0.167em}{0ex}}\text{R2}$

$10×$ 倍的负载电阻的作用是将分压器底部的电阻降低大约 $9\mathrm{%}$。这个额外的负载对分压器的输出电压有什么影响？如果没有负载，预期的输出是$0.5\phantom{\rule{0.167em}{0ex}}{v}_{in}$。现在我们算出有负载电阻时的输出电压。
${v}_{out}={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}{\text{R}1+0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}$

${v}_{out}={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{0.91}{1+0.91}$
${v}_{out}={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{0.91}{1.91}=0.48\phantom{\rule{0.167em}{0ex}}{v}_{in}$

$\frac{0.48}{0.50}=0.96=96\mathrm{%}$

### 接近极值附近操作分压器

#### 案例1：${v}_{out}=90\mathrm{%}$‍ ${v}_{in}$‍

${v}_{out}$ = ${v}_{in}$$90\mathrm{%}$。期待的输出是 $0.90\phantom{\rule{0.167em}{0ex}}{v}_{in}$

$\frac{{v}_{out}}{{v}_{in}}=0.90=\frac{\text{R}2}{\text{R}1+\text{R}2}$
$0.90\phantom{\rule{0.167em}{0ex}}\left(\text{R}1+\text{R}2\right)=\text{R}2$
$0.90\phantom{\rule{0.167em}{0ex}}\text{R}1=\text{R}2-0.90\phantom{\rule{0.167em}{0ex}}\text{R}2$
$0.90\phantom{\rule{0.167em}{0ex}}\text{R}1=0.10\phantom{\rule{0.167em}{0ex}}\text{R}2$
$\text{R}2=\frac{0.90\phantom{\rule{0.167em}{0ex}}\text{R}1}{0.10}=9\phantom{\rule{0.167em}{0ex}}\text{R}1$
$\text{R}2$$\text{R}1$$9$ 倍大小。

$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}{\text{R}1+0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}$

$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91\phantom{\rule{0.167em}{0ex}}\left(9\phantom{\rule{0.167em}{0ex}}\text{R}1\right)}{\text{R}1+\text{0}.91\phantom{\rule{0.167em}{0ex}}\left(9\phantom{\rule{0.167em}{0ex}}\text{R}1\right)}$

$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91\phantom{\rule{0.167em}{0ex}}\left(9\right)}{1+0.91\phantom{\rule{0.167em}{0ex}}\left(9\right)}=\frac{8.19}{9.19}=0.89$

$\frac{0.89}{0.90}=0.99$

#### 案例 2：${v}_{out}$‍  = ${v}_{in}$‍  的 $10\mathrm{%}$‍

${v}_{out}$ =${v}_{in}$$10\mathrm{%}$。期望的输出是 $0.10\phantom{\rule{0.167em}{0ex}}{v}_{in}$
$\text{R}1$$\text{R}2$表示为一个$10\mathrm{%}$的分压器。
$\frac{{v}_{out}}{{v}_{in}}=0.10=\frac{\text{R}2}{\text{R}1+\text{R}2}$
$0.10\phantom{\rule{0.167em}{0ex}}\left(\text{R}1+\text{R}2\right)=\text{R}2$
$0.10\phantom{\rule{0.167em}{0ex}}\text{R}1=\text{R}2-0.10\phantom{\rule{0.167em}{0ex}}\text{R}2$
$0.10\phantom{\rule{0.167em}{0ex}}\text{R}1=0.90\phantom{\rule{0.167em}{0ex}}\text{R}2$
$\text{R}1=\frac{0.90\phantom{\rule{0.167em}{0ex}}\text{R}2}{0.10}=9\phantom{\rule{0.167em}{0ex}}\text{R}2$
$\text{R}1$$\text{R}2$$9$ 倍。

$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}{\text{R}1+0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}$

$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}{9\phantom{\rule{0.167em}{0ex}}\text{R}2+\text{0}.91\phantom{\rule{0.167em}{0ex}}\text{R}2}$

$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91}{9+0.91}=\frac{0.91}{9.91}=0.092$

$\frac{0.092}{0.10}=0.92$

### 负载分压器的结论

• 接近中值时，输出电压降低$5\mathrm{%}$
• 在其范围的顶部附近，错误将大幅下降，大约为$1\mathrm{%}$
• 接近其范围的底部时，误差大约是中档的两倍。电压实际输出比预想的少了 $8\mathrm{%}$

## 总结

${v}_{out}={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{\text{R2}}{\text{R1}+\text{R2}}$