电路与电子学是电气工程和电子工程的基础学科,涵盖从基本电路分析到半导体器件原理的完整知识体系。本文从电路分析基础出发,逐步深入到半导体物理、基本电子器件(二极管、BJT、MOSFET)以及运算放大器和数字电路等应用领域,为读者构建系统化的电路与电子学知识框架。
电路中最基本的三个物理量是电压(V VV ,单位伏特 V)、电流(I II ,单位安培 A)和电阻(R RR ,单位欧姆 Ω \OmegaΩ )。它们之间的关系由欧姆定律 (Ohm's Law)描述:
V = I R V = IR
V = I R
数值示例 :一个 10 Ω 10\Omega1 0 Ω 的电阻两端施加 5 V 5V5 V 电压,则通过的电流为:
I = V R = 5 V 10 Ω = 0.5 A I = \frac{V}{R} = \frac{5V}{10\Omega} = 0.5A
I = R V = 1 0 Ω 5 V = 0 . 5 A
若将电压提升到 12 V 12V1 2 V ,则电流变为 1.2 A 1.2A1 . 2 A ;若电阻变为 100 Ω 100\Omega1 0 0 Ω 、电压仍为 5 V 5V5 V ,则电流降至 0.05 A 0.05A0 . 0 5 A (即 50 m A 50mA5 0 m A )。
基尔霍夫电流定律(KCL) :流入任意节点的电流之和等于流出该节点的电流之和,即 ∑ I i n = ∑ I o u t \sum I_{in} = \sum I_{out}∑ I i n = ∑ I o u t 。或者等价地,∑ I = 0 \sum I = 0∑ I = 0 (流入为正,流出为负)。
数值示例 :下图所示的节点有三个分支:I 1 = 2 A I_1 = 2AI 1 = 2 A 流入,I 2 = 0.5 A I_2 = 0.5AI 2 = 0 . 5 A 流入,I 3 I_3I 3 流出。由 KCL:
I 1 + I 2 = I 3 I_1 + I_2 = I_3
I 1 + I 2 = I 3
2 A + 0.5 A = 2.5 A = I 3 2A + 0.5A = 2.5A = I_3
2 A + 0 . 5 A = 2 . 5 A = I 3
基尔霍夫电压定律(KVL) :沿任意闭合回路,各元件电压降的代数和为零,即 ∑ V = 0 \sum V = 0∑ V = 0 。
数值示例 :一个包含电池 V S = 9 V V_S = 9VV S = 9 V 和两个串联电阻 R 1 = 100 Ω R_1 = 100\OmegaR 1 = 1 0 0 Ω 、R 2 = 200 Ω R_2 = 200\OmegaR 2 = 2 0 0 Ω 的回路。由 KVL:
V S − V R 1 − V R 2 = 0 V_S - V_{R1} - V_{R2} = 0
V S − V R 1 − V R 2 = 0
9 V − I R 1 − I R 2 = 0 9V - IR_1 - IR_2 = 0
9 V − I R 1 − I R 2 = 0
I = 9 V 100 Ω + 200 Ω = 9 V 300 Ω = 0.03 A = 30 m A I = \frac{9V}{100\Omega + 200\Omega} = \frac{9V}{300\Omega} = 0.03A = 30mA
I = 1 0 0 Ω + 2 0 0 Ω 9 V = 3 0 0 Ω 9 V = 0 . 0 3 A = 3 0 m A
因此 V R 1 = 0.03 × 100 = 3 V V_{R1} = 0.03 \times 100 = 3VV R 1 = 0 . 0 3 × 1 0 0 = 3 V ,V R 2 = 0.03 × 200 = 6 V V_{R2} = 0.03 \times 200 = 6VV R 2 = 0 . 0 3 × 2 0 0 = 6 V 。
元件
符号
单位
V-I 关系
关键特性
电阻 (R)
—‖—
Ω \OmegaΩ
V = I R V = IRV = I R
耗能元件
电容 (C)
—‖—
F(法拉)
I = C d V d t I = C\frac{dV}{dt}I = C d t d V
储能(电场)
电感 (L)
—ᗑ—
H(亨利)
V = L d I d t V = L\frac{dI}{dt}V = L d t d I
储能(磁场)
电容的充放电 :RC 电路的时间常数 τ = R C \tau = RCτ = R C ,表示电容充放电到稳态值的 63.2 % 63.2\%6 3 . 2 % 所需时间。
数值示例 :R = 10 k Ω R = 10k\OmegaR = 1 0 k Ω 、C = 1 μ F C = 1\mu FC = 1 μ F 的串联电路,时间常数:
τ = R C = 10 × 1 0 3 × 1 × 1 0 − 6 = 0.01 s = 10 m s \tau = RC = 10 \times 10^3 \times 1 \times 10^{-6} = 0.01s = 10ms
τ = R C = 1 0 × 1 0 3 × 1 × 1 0 − 6 = 0 . 0 1 s = 1 0 m s
电容电压随时间变化:V C ( t ) = V S ( 1 − e − t / τ ) V_C(t) = V_S(1 - e^{-t/\tau})V C ( t ) = V S ( 1 − e − t / τ ) 。充电到 V S V_SV S 的 90 % 90\%9 0 % 需要 t = − τ ln ( 0.1 ) ≈ 2.3 τ = 23 m s t = -\tau \ln(0.1) \approx 2.3\tau = 23mst = − τ ln ( 0 . 1 ) ≈ 2 . 3 τ = 2 3 m s 。
以节点电压为未知量,对每个非参考节点应用 KCL 建立方程。
示例 :下图电路包含电压源 V 1 = 10 V V_1 = 10VV 1 = 1 0 V 、R 1 = 2 Ω R_1 = 2\OmegaR 1 = 2 Ω 、R 2 = 5 Ω R_2 = 5\OmegaR 2 = 5 Ω 、R 3 = 10 Ω R_3 = 10\OmegaR 3 = 1 0 Ω 。设节点 0 为参考地,节点 1 电压已知为 V 1 = 10 V V_1 = 10VV 1 = 1 0 V ,对节点 2 列 KCL 方程:
V 2 − V 1 R 1 + V 2 R 2 + V 2 R 3 = 0 \frac{V_2 - V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_2}{R_3} = 0
R 1 V 2 − V 1 + R 2 V 2 + R 3 V 2 = 0
代入数值:
V 2 − 10 2 + V 2 5 + V 2 10 = 0 \frac{V_2 - 10}{2} + \frac{V_2}{5} + \frac{V_2}{10} = 0
2 V 2 − 1 0 + 5 V 2 + 1 0 V 2 = 0
乘以 10 101 0 :5 ( V 2 − 10 ) + 2 V 2 + V 2 = 0 5(V_2 - 10) + 2V_2 + V_2 = 05 ( V 2 − 1 0 ) + 2 V 2 + V 2 = 0
5 V 2 − 50 + 3 V 2 = 0 5V_2 - 50 + 3V_2 = 05 V 2 − 5 0 + 3 V 2 = 0
8 V 2 = 50 8V_2 = 508 V 2 = 5 0 ,V 2 = 6.25 V V_2 = 6.25VV 2 = 6 . 2 5 V
以网孔电流为未知量,对每个网孔应用 KVL。
在线性电路中,多个独立源共同作用产生的响应等于各独立源单独作用时产生的响应之和(此时其他独立源置零:电压源短路、电流源开路)。
数值示例 :电路两个电压源 V 1 = 12 V V_1 = 12VV 1 = 1 2 V 、V 2 = 6 V V_2 = 6VV 2 = 6 V 分别通过 R 1 = 4 Ω R_1 = 4\OmegaR 1 = 4 Ω 、R 2 = 2 Ω R_2 = 2\OmegaR 2 = 2 Ω 连接至 R L = 8 Ω R_L = 8\OmegaR L = 8 Ω 。
步骤
动作
计算方法
I R L I_{RL}I R L 结果
1
V 1 V_1V 1 单独作用,V 2 V_2V 2 短路
I ′ = 12 4 + 8 ∥ 2 I' = \frac{12}{4 + 8\|2}I ′ = 4 + 8 ∥ 2 1 2
0.96 A 0.96A0 . 9 6 A
2
V 2 V_2V 2 单独作用,V 1 V_1V 1 短路
I ′ ′ = 6 2 + 8 ∥ 4 I'' = \frac{6}{2 + 8\|4}I ′ ′ = 2 + 8 ∥ 4 6
0.48 A 0.48A0 . 4 8 A
3
叠加
I = I ′ + I ′ ′ I = I' + I''I = I ′ + I ′ ′
1.44 A 1.44A1 . 4 4 A
戴维南定理 :任何含源线性二端网络可以等效为一个电压源 V T h V_{Th}V T h 与一个电阻 R T h R_{Th}R T h 串联。
诺顿定理 :任何含源线性二端网络可以等效为一个电流源 I N I_NI N 与一个电阻 R N R_NR N 并联。其中 V T h = I N R T h V_{Th} = I_N R_{Th}V T h = I N R T h ,R T h = R N R_{Th} = R_NR T h = R N 。
数值示例 :求下图电路的戴维南等效。电路包含 V S = 24 V V_S = 24VV S = 2 4 V 、R 1 = 6 Ω R_1 = 6\OmegaR 1 = 6 Ω 、R 2 = 3 Ω R_2 = 3\OmegaR 2 = 3 Ω ,输出端为 AB:
开路电压 V T h = V S × R 2 R 1 + R 2 = 24 × 3 6 + 3 = 8 V V_{Th} = V_S \times \frac{R_2}{R_1 + R_2} = 24 \times \frac{3}{6+3} = 8VV T h = V S × R 1 + R 2 R 2 = 2 4 × 6 + 3 3 = 8 V
等效电阻 R T h = R 1 ∥ R 2 = 6 × 3 6 + 3 = 2 Ω R_{Th} = R_1 \| R_2 = \frac{6 \times 3}{6 + 3} = 2\OmegaR T h = R 1 ∥ R 2 = 6 + 3 6 × 3 = 2 Ω
诺顿电流 I N = V T h / R T h = 8 / 2 = 4 A I_N = V_{Th} / R_{Th} = 8 / 2 = 4AI N = V T h / R T h = 8 / 2 = 4 A
因此戴维南等效为 8 V 8V8 V 电压源串联 2 Ω 2\Omega2 Ω 电阻,诺顿等效为 4 A 4A4 A 电流源并联 2 Ω 2\Omega2 Ω 电阻。
RC 串联电路的零输入响应(电容初始电压 V 0 V_0V 0 ):
V C ( t ) = V 0 e − t / R C , I ( t ) = − V 0 R e − t / R C V_C(t) = V_0 e^{-t/RC}, \quad I(t) = -\frac{V_0}{R}e^{-t/RC}
V C ( t ) = V 0 e − t / R C , I ( t ) = − R V 0 e − t / R C
数值示例 :C = 100 μ F C = 100\mu FC = 1 0 0 μ F 初始充电到 10 V 10V1 0 V ,通过 R = 1 k Ω R = 1k\OmegaR = 1 k Ω 放电。
t tt
V C ( t ) V_C(t)V C ( t )
I ( t ) I(t)I ( t )
0 00
10.00 V 10.00V1 0 . 0 0 V
10.0 m A 10.0mA1 0 . 0 m A
τ = 0.1 s \tau = 0.1sτ = 0 . 1 s
3.68 V 3.68V3 . 6 8 V
3.68 m A 3.68mA3 . 6 8 m A
2 τ = 0.2 s 2\tau = 0.2s2 τ = 0 . 2 s
1.35 V 1.35V1 . 3 5 V
1.35 m A 1.35mA1 . 3 5 m A
3 τ = 0.3 s 3\tau = 0.3s3 τ = 0 . 3 s
0.50 V 0.50V0 . 5 0 V
0.50 m A 0.50mA0 . 5 0 m A
5 τ = 0.5 s 5\tau = 0.5s5 τ = 0 . 5 s
0.067 V 0.067V0 . 0 6 7 V
0.067 m A 0.067mA0 . 0 6 7 m A
RL 串联电路的时间常数 τ = L / R \tau = L/Rτ = L / R 。电流的零输入响应:
I L ( t ) = I 0 e − t R / L I_L(t) = I_0 e^{-tR/L}
I L ( t ) = I 0 e − t R / L
数值示例 :L = 100 m H L = 100mHL = 1 0 0 m H 、R = 10 Ω R = 10\OmegaR = 1 0 Ω ,时间常数:
τ = L / R = 0.1 / 10 = 0.01 s = 10 m s \tau = L/R = 0.1/10 = 0.01s = 10ms
τ = L / R = 0 . 1 / 1 0 = 0 . 0 1 s = 1 0 m s
RLC 串联电路的二阶微分方程:
L d 2 i d t 2 + R d i d t + i C = 0 L\frac{d^2i}{dt^2} + R\frac{di}{dt} + \frac{i}{C} = 0
L d t 2 d 2 i + R d t d i + C i = 0
特征方程 s 2 + R L s + 1 L C = 0 s^2 + \frac{R}{L}s + \frac{1}{LC} = 0s 2 + L R s + L C 1 = 0 。根据阻尼系数 ζ = R 2 C L \zeta = \frac{R}{2}\sqrt{\frac{C}{L}}ζ = 2 R L C 分为三种情况:
阻尼类型
条件
参数关系
响应特征
欠阻尼
ζ < 1 \zeta < 1ζ < 1
R < 2 L / C R < 2\sqrt{L/C}R < 2 L / C
振荡衰减
临界阻尼
ζ = 1 \zeta = 1ζ = 1
R = 2 L / C R = 2\sqrt{L/C}R = 2 L / C
最快无振荡
过阻尼
ζ > 1 \zeta > 1ζ > 1
R > 2 L / C R > 2\sqrt{L/C}R > 2 L / C
缓慢衰减
数值示例 :L = 1 H L = 1HL = 1 H 、C = 1 μ F C = 1\mu FC = 1 μ F 、R = 100 Ω R = 100\OmegaR = 1 0 0 Ω ,计算:
ω 0 = 1 L C = 1 1 × 1 0 − 6 = 1000 rad/s \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \times 10^{-6}}} = 1000 \text{ rad/s}
ω 0 = L C 1 = 1 × 1 0 − 6 1 = 1 0 0 0 rad/s
ζ = R 2 C L = 100 2 1 0 − 6 1 = 50 × 1 0 − 3 = 0.05 \zeta = \frac{R}{2}\sqrt{\frac{C}{L}} = \frac{100}{2}\sqrt{\frac{10^{-6}}{1}} = 50 \times 10^{-3} = 0.05
ζ = 2 R L C = 2 1 0 0 1 1 0 − 6 = 5 0 × 1 0 − 3 = 0 . 0 5
ζ = 0.05 < 1 \zeta = 0.05 < 1ζ = 0 . 0 5 < 1 ,属于欠阻尼振荡。振荡角频率:
ω d = ω 0 1 − ζ 2 = 1000 1 − 0.0025 ≈ 998.75 rad/s \omega_d = \omega_0\sqrt{1-\zeta^2} = 1000\sqrt{1-0.0025} \approx 998.75 \text{ rad/s}
ω d = ω 0 1 − ζ 2 = 1 0 0 0 1 − 0 . 0 0 2 5 ≈ 9 9 8 . 7 5 rad/s
正弦量 v ( t ) = V m cos ( ω t + ϕ ) v(t) = V_m \cos(\omega t + \phi)v ( t ) = V m cos ( ω t + ϕ ) 的相量表示为 V = V m ∠ ϕ \mathbf{V} = V_m \angle \phiV = V m ∠ ϕ 或 V m e j ϕ V_m e^{j\phi}V m e j ϕ 。
时域
相量域
v ( t ) = V m cos ( ω t + ϕ ) v(t) = V_m\cos(\omega t + \phi)v ( t ) = V m cos ( ω t + ϕ )
V = V m e j ϕ \mathbf{V} = V_m e^{j\phi}V = V m e j ϕ
i ( t ) = I m cos ( ω t + θ ) i(t) = I_m\cos(\omega t + \theta)i ( t ) = I m cos ( ω t + θ )
I = I m e j θ \mathbf{I} = I_m e^{j\theta}I = I m e j θ
d v / d t dv/dtd v / d t
j ω V j\omega\mathbf{V}j ω V
∫ v d t \int v\,dt∫ v d t
V / ( j ω ) \mathbf{V}/(j\omega)V / ( j ω )
元件
阻抗 Z ZZ
导纳 Y = 1 / Z Y = 1/ZY = 1 / Z
电阻 R RR
R RR
1 / R 1/R1 / R
电感 L LL
j ω L j\omega Lj ω L
1 / ( j ω L ) 1/(j\omega L)1 / ( j ω L )
电容 C CC
1 / ( j ω C ) 1/(j\omega C)1 / ( j ω C )
j ω C j\omega Cj ω C
数值示例 :50 H z 50Hz5 0 H z 交流电路中,C = 100 μ F C = 100\mu FC = 1 0 0 μ F 的容抗:
X C = 1 2 π f C = 1 2 π × 50 × 100 × 1 0 − 6 = 1 0.0314 ≈ 31.8 Ω X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50 \times 100 \times 10^{-6}} = \frac{1}{0.0314} \approx 31.8\Omega
X C = 2 π f C 1 = 2 π × 5 0 × 1 0 0 × 1 0 − 6 1 = 0 . 0 3 1 4 1 ≈ 3 1 . 8 Ω
功率类型
公式
单位
说明
瞬时功率
p ( t ) = v ( t ) i ( t ) p(t) = v(t)i(t)p ( t ) = v ( t ) i ( t )
W
随时间变化
有功功率
P = V I cos ϕ P = VI\cos\phiP = V I cos ϕ
W
实际消耗
无功功率
Q = V I sin ϕ Q = VI\sin\phiQ = V I sin ϕ
VAR
储能交换
视在功率
S = V I S = VIS = V I
VA
总功率
功率因数
P F = cos ϕ = P / S PF = \cos\phi = P/SP F = cos ϕ = P / S
—
效率指标
数值示例 :Z = 8 + j 6 Ω Z = 8 + j6\OmegaZ = 8 + j 6 Ω 的负载,电流 I = 5 A I = 5AI = 5 A ,则:
S = I 2 ∣ Z ∣ = 25 × 10 = 250 V A S = I^2|Z| = 25 \times 10 = 250VA
S = I 2 ∣ Z ∣ = 2 5 × 1 0 = 2 5 0 V A
P = I 2 R = 25 × 8 = 200 W P = I^2R = 25 \times 8 = 200W
P = I 2 R = 2 5 × 8 = 2 0 0 W
Q = I 2 X = 25 × 6 = 150 V A R Q = I^2X = 25 \times 6 = 150VAR
Q = I 2 X = 2 5 × 6 = 1 5 0 V A R
P F = cos ϕ = R / ∣ Z ∣ = 8 / 10 = 0.8 (滞后) PF = \cos\phi = R/|Z| = 8/10 = 0.8 \text{(滞后)}
P F = cos ϕ = R / ∣ Z ∣ = 8 / 1 0 = 0 . 8 (滞后)
纯硅(Si)在 T = 300 K T = 300KT = 3 0 0 K 时,本征载流子浓度 n i = 1.5 × 1 0 10 cm − 3 n_i = 1.5 \times 10^{10} \text{cm}^{-3}n i = 1 . 5 × 1 0 1 0 cm − 3 。通过掺杂形成两类半导体:
类型
掺杂元素
多数载流子
少数载流子
典型掺杂浓度
N 型
P、As、Sb
电子 (n ≈ N D n \approx N_Dn ≈ N D )
空穴 (p = n i 2 / n p = n_i^2/np = n i 2 / n )
1 0 15 − 1 0 18 cm − 3 10^{15} - 10^{18} \text{cm}^{-3}1 0 1 5 − 1 0 1 8 cm − 3
P 型
B、Al、Ga
空穴 (p ≈ N A p \approx N_Ap ≈ N A )
电子 (n = n i 2 / p n = n_i^2/pn = n i 2 / p )
1 0 15 − 1 0 18 cm − 3 10^{15} - 10^{18} \text{cm}^{-3}1 0 1 5 − 1 0 1 8 cm − 3
数值示例 :N 型硅掺杂 N D = 1 0 16 cm − 3 N_D = 10^{16} \text{cm}^{-3}N D = 1 0 1 6 cm − 3 ,则在 300 K 300K3 0 0 K 时:
n ≈ N D = 1 0 16 cm − 3 n \approx N_D = 10^{16} \text{cm}^{-3}
n ≈ N D = 1 0 1 6 cm − 3
p = n i 2 n = ( 1.5 × 1 0 10 ) 2 1 0 16 = 2.25 × 1 0 4 cm − 3 p = \frac{n_i^2}{n} = \frac{(1.5 \times 10^{10})^2}{10^{16}} = 2.25 \times 10^4 \text{cm}^{-3}
p = n n i 2 = 1 0 1 6 ( 1 . 5 × 1 0 1 0 ) 2 = 2 . 2 5 × 1 0 4 cm − 3
电子浓度是空穴浓度的约 4.4 × 1 0 11 4.4 \times 10^{11}4 . 4 × 1 0 1 1 倍。
PN 结交界面形成耗尽层(空间电荷区),建立内建电势差:
V 0 = k T q ln ( N A N D n i 2 ) V_0 = \frac{kT}{q}\ln\left(\frac{N_A N_D}{n_i^2}\right)
V 0 = q k T ln ( n i 2 N A N D )
数值示例 :N A = N D = 1 0 16 cm − 3 N_A = N_D = 10^{16} \text{cm}^{-3}N A = N D = 1 0 1 6 cm − 3 ,T = 300 K T = 300KT = 3 0 0 K 时:
V 0 = 0.0259 × ln ( 1 0 32 2.25 × 1 0 20 ) = 0.0259 × ln ( 4.44 × 1 0 11 ) ≈ 0.0259 × 26.82 ≈ 0.695 V V_0 = 0.0259 \times \ln\left(\frac{10^{32}}{2.25 \times 10^{20}}\right) = 0.0259 \times \ln(4.44 \times 10^{11}) \approx 0.0259 \times 26.82 \approx 0.695V
V 0 = 0 . 0 2 5 9 × ln ( 2 . 2 5 × 1 0 2 0 1 0 3 2 ) = 0 . 0 2 5 9 × ln ( 4 . 4 4 × 1 0 1 1 ) ≈ 0 . 0 2 5 9 × 2 6 . 8 2 ≈ 0 . 6 9 5 V
理想二极管方程(Shockley 方程):
I D = I S ( e V D / ( n V T ) − 1 ) I_D = I_S\left(e^{V_D/(nV_T)} - 1\right)
I D = I S ( e V D / ( n V T ) − 1 )
其中 V T = k T / q ≈ 25.9 m V V_T = kT/q \approx 25.9mVV T = k T / q ≈ 2 5 . 9 m V (室温),I S I_SI S 为反向饱和电流,n nn 为理想因子(1 ∼ 2 1 \sim 21 ∼ 2 )。
数值示例 :I S = 1 0 − 14 A I_S = 10^{-14}AI S = 1 0 − 1 4 A ,n = 1 n = 1n = 1 ,T = 300 K T = 300KT = 3 0 0 K ,V D = 0.7 V V_D = 0.7VV D = 0 . 7 V 时:
I D = 1 0 − 14 × ( e 0.7 / 0.0259 − 1 ) = 1 0 − 14 × ( e 27.03 − 1 ) ≈ 1 0 − 14 × 5.35 × 1 0 11 I_D = 10^{-14} \times \left(e^{0.7/0.0259} - 1\right) = 10^{-14} \times (e^{27.03} - 1) \approx 10^{-14} \times 5.35 \times 10^{11}
I D = 1 0 − 1 4 × ( e 0 . 7 / 0 . 0 2 5 9 − 1 ) = 1 0 − 1 4 × ( e 2 7 . 0 3 − 1 ) ≈ 1 0 − 1 4 × 5 . 3 5 × 1 0 1 1
I D ≈ 5.35 × 1 0 − 3 A = 5.35 m A I_D \approx 5.35 \times 10^{-3} A = 5.35mA
I D ≈ 5 . 3 5 × 1 0 − 3 A = 5 . 3 5 m A
电路类型
功能
关键参数
半波整流
只保留正半周期
效率约 40.6%
全波桥式整流
正半周期全部保留
效率约 81.2%
稳压管(Zener)
反向击穿区稳压
V Z V_ZV Z (如 3.3V、5.1V)
限幅器(Clipper)
限制电压幅度
限幅电平由偏置决定
钳位器(Clamper)
平移信号直流电平
时间常数 R C RCR C 需远大于周期
桥式整流数值示例 :输入 220 V r m s 220V_{rms}2 2 0 V r m s 交流经变压器降至 12 V r m s 12V_{rms}1 2 V r m s ,桥式整流后:
峰值电压 V p e a k = 12 × 2 − 2 × 0.7 ≈ 15.6 V V_{peak} = 12 \times \sqrt{2} - 2 \times 0.7 \approx 15.6VV p e a k = 1 2 × 2 − 2 × 0 . 7 ≈ 1 5 . 6 V (减去两个二极管压降)
滤波后直流约 15.6 V 15.6V1 5 . 6 V ,纹波取决于负载和滤波电容
BJT 是电流控制型器件,分为 NPN 和 PNP 两种。以 NPN 为例,三个区域:发射极(E)、基极(B)、集电极(C)。
核心电流关系:
I C = β I B I_C = \beta I_B
I C = β I B
I E = I B + I C = ( β + 1 ) I B I_E = I_B + I_C = (\beta + 1)I_B
I E = I B + I C = ( β + 1 ) I B
其中 β \betaβ 为电流增益。不同类型 BJT 的 β \betaβ 值范围:
类型
典型 β \betaβ
功率类型
典型应用
小信号通用
100 − 400 100 - 4001 0 0 − 4 0 0
NPN/PNP
放大器、开关
达林顿对
1000 − 20000 1000 - 200001 0 0 0 − 2 0 0 0 0
NPN
高增益放大器
功率管
10 − 100 10 - 1001 0 − 1 0 0
NPN/PNP
功率放大、电机驱动
设计电路:V C C = 12 V V_{CC} = 12VV C C = 1 2 V ,目标 I C = 1 m A I_C = 1mAI C = 1 m A ,β = 200 \beta = 200β = 2 0 0 。选择 R C = 5.6 k Ω R_C = 5.6k\OmegaR C = 5 . 6 k Ω ,R E = 1 k Ω R_E = 1k\OmegaR E = 1 k Ω :
V E = I E × R E ≈ I C × R E = 1 m A × 1 k Ω = 1 V V_E = I_E \times R_E \approx I_C \times R_E = 1mA \times 1k\Omega = 1V
V E = I E × R E ≈ I C × R E = 1 m A × 1 k Ω = 1 V
V B = V E + 0.7 V = 1.7 V V_B = V_E + 0.7V = 1.7V
V B = V E + 0 . 7 V = 1 . 7 V
设基极分压电流 I d i v = 10 × I B = 10 × 1 m A 200 = 50 μ A I_{div} = 10 \times I_B = 10 \times \frac{1mA}{200} = 50\mu AI d i v = 1 0 × I B = 1 0 × 2 0 0 1 m A = 5 0 μ A :
R 2 = V B I d i v = 1.7 V 50 μ A = 34 k Ω R_2 = \frac{V_B}{I_{div}} = \frac{1.7V}{50\mu A} = 34k\Omega
R 2 = I d i v V B = 5 0 μ A 1 . 7 V = 3 4 k Ω
R 1 = V C C − V B I d i v = 12 − 1.7 50 μ A = 206 k Ω R_1 = \frac{V_{CC} - V_B}{I_{div}} = \frac{12 - 1.7}{50\mu A} = 206k\Omega
R 1 = I d i v V C C − V B = 5 0 μ A 1 2 − 1 . 7 = 2 0 6 k Ω
模式
BE 结
BC 结
V C E V_{CE}V C E
用途
截止
反偏
反偏
≈ V C C \approx V_{CC}≈ V C C
开关断开
放大
正偏
反偏
0.2 V < V C E < V C C 0.2V < V_{CE} < V_{CC}0 . 2 V < V C E < V C C
线性放大
饱和
正偏
正偏
≈ 0.2 V \approx 0.2V≈ 0 . 2 V
开关闭合
MOSFET 是电压控制型器件,分为 N 沟道和 P 沟道,每种又分为增强型和耗尽型:
类型
符号
阈值电压 V t h V_{th}V t h
导电条件
NMOS 增强型
N-MOS
> 0 V >0V> 0 V
V G S > V t h V_{GS} > V_{th}V G S > V t h
PMOS 增强型
P-MOS
< 0 V <0V< 0 V
V G S < V t h V_{GS} < V_{th}V G S < V t h
NMOS 的 I-V 特性分为三个区域:
I D = { 0 , V G S < V t h (截止区) K n [ ( V G S − V t h ) V D S − V D S 2 2 ] , V D S < V G S − V t h (线性区/三极管区) K n 2 ( V G S − V t h ) 2 ( 1 + λ V D S ) , V D S ≥ V G S − V t h (饱和区/恒流区) I_D = \begin{cases}
0, & V_{GS} < V_{th} & \text{(截止区)} \\
K_n\left[(V_{GS} - V_{th})V_{DS} - \frac{V_{DS}^2}{2}\right], & V_{DS} < V_{GS} - V_{th} & \text{(线性区/三极管区)} \\
\frac{K_n}{2}(V_{GS} - V_{th})^2(1 + \lambda V_{DS}), & V_{DS} \geq V_{GS} - V_{th} & \text{(饱和区/恒流区)}
\end{cases}I D = ⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 0 , K n [ ( V G S − V t h ) V D S − 2 V D S 2 ] , 2 K n ( V G S − V t h ) 2 ( 1 + λ V D S ) , V G S < V t h V D S < V G S − V t h V D S ≥ V G S − V t h (截止区) (线性区 / 三极管区) (饱和区 / 恒流区)
其中 K n = μ n C o x W L K_n = \mu_n C_{ox} \frac{W}{L}K n = μ n C o x L W 为跨导参数,λ \lambdaλ 为沟道长度调制系数。
数值示例 :NMOS 参数 K n = 200 μ A / V 2 K_n = 200\mu A/V^2K n = 2 0 0 μ A / V 2 、V t h = 1 V V_{th} = 1VV t h = 1 V 、V G S = 3 V V_{GS} = 3VV G S = 3 V :
V D S V_{DS}V D S
区域
I D I_DI D
0.5 V 0.5V0 . 5 V
线性区
200 × [ ( 3 − 1 ) × 0.5 − 0. 5 2 / 2 ] = 175 μ A 200 \times [(3-1) \times 0.5 - 0.5^2/2] = 175 \mu A2 0 0 × [ ( 3 − 1 ) × 0 . 5 − 0 . 5 2 / 2 ] = 1 7 5 μ A
1.0 V 1.0V1 . 0 V
线性区
200 × [ ( 2 × 1 ) − 1 / 2 ] = 300 μ A 200 \times [(2 \times 1) - 1/2] = 300 \mu A2 0 0 × [ ( 2 × 1 ) − 1 / 2 ] = 3 0 0 μ A
2.5 V 2.5V2 . 5 V
饱和区
200 / 2 × ( 3 − 1 ) 2 = 400 μ A 200/2 \times (3-1)^2 = 400 \mu A2 0 0 / 2 × ( 3 − 1 ) 2 = 4 0 0 μ A
5.0 V 5.0V5 . 0 V
饱和区
400 μ A 400 \mu A4 0 0 μ A (忽略 λ \lambdaλ )
跨导 g m = ∂ I D ∂ V G S = K n ( V G S − V t h ) = 2 K n I D g_m = \frac{\partial I_D}{\partial V_{GS}} = K_n(V_{GS} - V_{th}) = \sqrt{2K_n I_D}g m = ∂ V G S ∂ I D = K n ( V G S − V t h ) = 2 K n I D
数值示例 :I D = 1 m A I_D = 1mAI D = 1 m A ,K n = 200 μ A / V 2 K_n = 200\mu A/V^2K n = 2 0 0 μ A / V 2 ,则:
g m = 2 × 200 × 1 0 − 6 × 1 0 − 3 = 4 × 1 0 − 7 ≈ 632 μ S g_m = \sqrt{2 \times 200 \times 10^{-6} \times 10^{-3}} = \sqrt{4 \times 10^{-7}} \approx 632 \mu S
g m = 2 × 2 0 0 × 1 0 − 6 × 1 0 − 3 = 4 × 1 0 − 7 ≈ 6 3 2 μ S
特性
BJT
MOSFET
控制方式
电流控制
电压控制
输入阻抗
低(k Ω k\Omegak Ω 级)
极高(M Ω M\OmegaM Ω 级)
增益
β \betaβ (50 − 500 50-5005 0 − 5 0 0 )
g m g_mg m (1 − 50 m S 1-50mS1 − 5 0 m S )
噪声
较低
较高
功耗
较大(需基极电流)
极小(静态)
开关速度
较慢(米勒效应)
较快
集成度
低
极高
理想运算放大器满足:
输入阻抗 Z i n → ∞ Z_{in} \to \inftyZ i n → ∞ (输入电流为 0 00 )
输出阻抗 Z o u t → 0 Z_{out} \to 0Z o u t → 0
开环增益 A O L → ∞ A_{OL} \to \inftyA O L → ∞
带宽 → ∞ \to \infty→ ∞
由此推出两个重要规则:虚断 (输入端不取电流)和虚短 (同相端与反相端电压相等,当反馈存在时)。
反相放大器 :
V o u t = − R f R i n V i n V_{out} = -\frac{R_f}{R_{in}}V_{in}
V o u t = − R i n R f V i n
数值示例 :R i n = 1 k Ω R_{in} = 1k\OmegaR i n = 1 k Ω ,R f = 10 k Ω R_f = 10k\OmegaR f = 1 0 k Ω ,V i n = 0.5 V V_{in} = 0.5VV i n = 0 . 5 V :
V o u t = − 10 k 1 k × 0.5 = − 5 V V_{out} = -\frac{10k}{1k} \times 0.5 = -5V
V o u t = − 1 k 1 0 k × 0 . 5 = − 5 V
输入 V i n V_{in}V i n
输出 V o u t V_{out}V o u t
增益
0.1 V 0.1V0 . 1 V
− 1.0 V -1.0V− 1 . 0 V
− 10 -10− 1 0
0.5 V 0.5V0 . 5 V
− 5.0 V -5.0V− 5 . 0 V
− 10 -10− 1 0
1.0 V 1.0V1 . 0 V
− 10.0 V -10.0V− 1 0 . 0 V
− 10 -10− 1 0
同相放大器 :
V o u t = ( 1 + R f R i n ) V i n V_{out} = \left(1 + \frac{R_f}{R_{in}}\right)V_{in}
V o u t = ( 1 + R i n R f ) V i n
数值示例 :R i n = 1 k Ω R_{in} = 1k\OmegaR i n = 1 k Ω ,R f = 4 k Ω R_f = 4k\OmegaR f = 4 k Ω ,V i n = 0.5 V V_{in} = 0.5VV i n = 0 . 5 V :
V o u t = ( 1 + 4 k 1 k ) × 0.5 = 2.5 V V_{out} = \left(1 + \frac{4k}{1k}\right) \times 0.5 = 2.5V
V o u t = ( 1 + 1 k 4 k ) × 0 . 5 = 2 . 5 V
电路类型
增益公式
输入阻抗
输出相位
反相放大器
− R f / R i n -R_f/R_{in}− R f / R i n
R i n R_{in}R i n
反相 18 0 ∘ 180^\circ1 8 0 ∘
同相放大器
1 + R f / R i n 1 + R_f/R_{in}1 + R f / R i n
∞ \infty∞
同相
电压跟随器
1 11
∞ \infty∞
同相
求和放大器
− ( R f R 1 V 1 + R f R 2 V 2 ) -\left(\frac{R_f}{R_1}V_1 + \frac{R_f}{R_2}V_2\right)− ( R 1 R f V 1 + R 2 R f V 2 )
R 1 , R 2 R_1, R_2R 1 , R 2
反相
差分放大器
R f R i n ( V 2 − V 1 ) \frac{R_f}{R_{in}}(V_2 - V_1)R i n R f ( V 2 − V 1 )
中等
取决于输入
积分器
− 1 R C ∫ V i n d t -\frac{1}{RC}\int V_{in}\,dt− R C 1 ∫ V i n d t
R RR
反相
微分器
− R C d V i n d t -RC\frac{dV_{in}}{dt}− R C d t d V i n
R RR
反相
型号
类型
带宽
压摆率
典型应用
LM741
通用
1.5 M H z 1.5MHz1 . 5 M H z
0.5 V / μ s 0.5V/\mu s0 . 5 V / μ s
一般用途
TL081
JFET 输入
3 M H z 3MHz3 M H z
13 V / μ s 13V/\mu s1 3 V / μ s
音频、高速
LM358
双通道
1 M H z 1MHz1 M H z
0.6 V / μ s 0.6V/\mu s0 . 6 V / μ s
电池供电
NE5532
低噪声
10 M H z 10MHz1 0 M H z
9 V / μ s 9V/\mu s9 V / μ s
专业音频
OP07
精密
0.6 M H z 0.6MHz0 . 6 M H z
0.3 V / μ s 0.3V/\mu s0 . 3 V / μ s
精密测量
LM324
四通道
1.2 M H z 1.2MHz1 . 2 M H z
0.5 V / μ s 0.5V/\mu s0 . 5 V / μ s
多通道应用
逻辑门
符号
布尔表达式
真值表(A B → Y A\;B \to YA B → Y )
AND
Y = A ⋅ B Y = A \cdot BY = A ⋅ B
Y = A ⋅ B Y = A \cdot BY = A ⋅ B
00 → 0 , 01 → 0 , 10 → 0 , 11 → 1 00\to0, 01\to0, 10\to0, 11\to10 0 → 0 , 0 1 → 0 , 1 0 → 0 , 1 1 → 1
OR
Y = A + B Y = A + BY = A + B
Y = A + B Y = A + BY = A + B
00 → 0 , 01 → 1 , 10 → 1 , 11 → 1 00\to0, 01\to1, 10\to1, 11\to10 0 → 0 , 0 1 → 1 , 1 0 → 1 , 1 1 → 1
NOT
Y = A ‾ Y = \overline{A}Y = A
Y = A ‾ Y = \overline{A}Y = A
0 → 1 , 1 → 0 0\to1, 1\to00 → 1 , 1 → 0
NAND
Y = A ⋅ B ‾ Y = \overline{A \cdot B}Y = A ⋅ B
Y = A B ‾ Y = \overline{AB}Y = A B
00 → 1 , 01 → 1 , 10 → 1 , 11 → 0 00\to1, 01\to1, 10\to1, 11\to00 0 → 1 , 0 1 → 1 , 1 0 → 1 , 1 1 → 0
NOR
Y = A + B ‾ Y = \overline{A + B}Y = A + B
Y = A + B ‾ Y = \overline{A+B}Y = A + B
00 → 1 , 01 → 0 , 10 → 0 , 11 → 0 00\to1, 01\to0, 10\to0, 11\to00 0 → 1 , 0 1 → 0 , 1 0 → 0 , 1 1 → 0
XOR
Y = A ⊕ B Y = A \oplus BY = A ⊕ B
Y = A B ‾ + A ‾ B Y = A\overline{B} + \overline{A}BY = A B + A B
00 → 0 , 01 → 1 , 10 → 1 , 11 → 0 00\to0, 01\to1, 10\to1, 11\to00 0 → 0 , 0 1 → 1 , 1 0 → 1 , 1 1 → 0
CMOS 逻辑利用互补的 NMOS 和 PMOS 晶体管。以 CMOS 反相器为例:
PMOS 连接 V D D V_{DD}V D D (电源),NMOS 连接 GND
输入为高时:NMOS 导通,PMOS 截止,输出为低
输入为低时:PMOS 导通,NMOS 截止,输出为高
静态功耗对比 :
逻辑系列
每门功耗
传播延迟
扇出能力
TTL(双极)
10 m W 10mW1 0 m W
10 n s 10ns1 0 n s
10 101 0
CMOS(早期)
1 m W 1mW1 m W (静态)
50 n s 50ns5 0 n s
50 505 0
CMOS(现代)
< 10 n W <10nW< 1 0 n W (静态)
< 1 n s <1ns< 1 n s
> 100 >100> 1 0 0
触发器类型
特性
激励方程
SR 触发器
置位-复位
Q n + 1 = S + R ‾ Q n Q_{n+1} = S + \overline{R}Q_nQ n + 1 = S + R Q n
D 触发器
数据锁存
Q n + 1 = D Q_{n+1} = DQ n + 1 = D
JK 触发器
万能触发器
Q n + 1 = J Q n ‾ + K ‾ Q n Q_{n+1} = J\overline{Q_n} + \overline{K}Q_nQ n + 1 = J Q n + K Q n
T 触发器
翻转
Q n + 1 = T ⊕ Q n Q_{n+1} = T \oplus Q_nQ n + 1 = T ⊕ Q n
类型
效率
纹波
复杂性
典型应用
线性稳压器(78xx)
30 − 50 % 30-50\%3 0 − 5 0 %
极低
简单
低噪声模拟
开关稳压器(Buck)
80 − 95 % 80-95\%8 0 − 9 5 %
较高
中等
降压转换
开关稳压器(Boost)
80 − 95 % 80-95\%8 0 − 9 5 %
较高
中等
升压转换
开关稳压器(Buck-Boost)
75 − 90 % 75-90\%7 5 − 9 0 %
较高
较复杂
升降压
线性稳压器效率示例 :输入 12 V 12V1 2 V ,输出 5 V 5V5 V ,负载 1 A 1A1 A :
效率 = P o u t P i n = 5 V × 1 A 12 V × 1 A = 41.7 % \text{效率} = \frac{P_{out}}{P_{in}} = \frac{5V \times 1A}{12V \times 1A} = 41.7\%
效率 = P i n P o u t = 1 2 V × 1 A 5 V × 1 A = 4 1 . 7 %
剩余 58.3 % 58.3\%5 8 . 3 % 的功率转化为热量,需要散热片。相比之下,Buck 转换器效率可达 90 % 90\%9 0 % 以上。
graph TD
A[电路分析基础] --> B[基本定律 KCL/KVL]
A --> C[基本元件 R/L/C]
A --> D[分析方法]
D --> E[节点/网孔分析]
D --> F[叠加/戴维南/诺顿]
D --> G[正弦稳态分析]
G --> H[相量法]
G --> I[功率分析]
A --> J[半导体物理]
J --> K[PN结二极管]
K --> L[BJT晶体管]
K --> M[MOSFET]
L --> N[模拟电路基础]
M --> N
N --> O[运放电路]
N --> P[电源电路]
N --> Q[数字电路基础]
Sedra, A.S., Smith, K.C., Microelectronic Circuits , 7th ed., Oxford University Press, 2015
Horowitz, P., Hill, W., The Art of Electronics , 3rd ed., Cambridge University Press, 2015
Nilsson, J.W., Riedel, S.A., Electric Circuits , 10th ed., Pearson, 2015
Razavi, B., Fundamentals of Microelectronics , 2nd ed., Wiley, 2014
Neamen, D.A., Microelectronics: Circuit Analysis and Design , 4th ed., McGraw-Hill, 2010
李翰荪,《电路分析基础》,高等教育出版社
Kittel, C., Introduction to Solid State Physics , 8th ed., Wiley, 2005