二极管基础知识复习¶
Shockley 公式¶
William Shockley 于 1949/07 在 BSTJ 发表了 "The theory of p-n junctions in semiconductors and p-n junction transistors" https://ieeexplore.ieee.org/document/6773080 $\newcommand{\li}[1]{{i_\mathrm{#1}}}$ $\newcommand{\ui}[1]{{I_\mathrm{#1}}}$ $\newcommand{\lv}[1]{{v_\mathrm{#1}}}$ $\newcommand{\uv}[1]{{V_\mathrm{#1}}}$ $\newcommand{\sm}[2]{{{#1}_\mathrm{#2}}}$
$\li{D}=\ui{S}\Big(\exp\dfrac{\lv{D}}{n \uv{T}}-1\Big)\approx\ui{S}\exp\dfrac{\lv{D}}{n \uv{T}}$
$\uv{D} - \ui{D}$ 呈指数关系，如果 $n=1$，那 $\uv{D}$ 每增加 60mV，$\ui{D}$ 增大 10 倍，因为 $60 / 26\approx 2.3$, $e^{2.3}\approx 10$.
| $\ui{D}\propto \exp \uv{D}$ | $\ui{S}=0.1\mathrm{fA}$ | $\ui{S}=1\mathrm{fA}$ | $\ui{S}=10\mathrm{fA}$ |
|---|---|---|---|
| $\uv{D}=0.64$V | 0.00576 mA | 0.0576 mA | 0.576 mA |
| $\uv{D}=0.70$V | 0.0576 mA | 0.576 mA | 5.76 mA |
| $\uv{D}=0.76$V | 0.576 mA | 5.76 mA | 57.6mA |
Vt = 25.865
# import tabulate ?
print(' Id =', end='')
for I in [0.001, 0.01, 0.1, 1]:
print('%4.0f mA ' % (I * 1000), end='')
print()
#for Is in [1e-11, 1e-12, 1e-13, 1e-14, 1e-15]:
for Is in [1e-15, 1e-14, 1e-13, 1e-12, 1e-11]:
print('Is = %g A ' % Is, end='')
for I in [0.001, 0.01, 0.1, 1]:
print(" %4.0f mV" % (Vt * log(I / Is)), end='')
print()
Id = 1 mA 10 mA 100 mA 1000 mA Is = 1e-15 A 715 mV 774 mV 834 mV 893 mV Is = 1e-14 A 655 mV 715 mV 774 mV 834 mV Is = 1e-13 A 596 mV 655 mV 715 mV 774 mV Is = 1e-12 A 536 mV 596 mV 655 mV 715 mV Is = 1e-11 A 476 mV 536 mV 596 mV 655 mV
这个指数关系高度非线性，简单的工作点就不好算。解析解要用 Lambert W 函数，一般模拟电路书上不会讲。
ng.circ('''
V vcc 0 1V
R vcc d 1k
D d 0 D
.model D D(IS=1fA)
''')
ng.operating_point()
{'v#branch': array([-0.00031519]),
'd': array([0.68481116]),
'vcc': array([1.])}
近似估算 $\uv{D}\approx 0.7\,$V，$\ui{D}=\dfrac{1-0.7}{1\,\mathrm{k}\Omega}=0.3$mA，与仿真结果相去不远。
小信号模型¶
$\sm{r}{d}=\dfrac{\lv{d}}{\li{d}}$
小信号计算 $\Delta \lv{D}=5.2$mV @ $\ui{D}=1$mA, $\Delta \li{D}=\dfrac{5.2}{26}=0.2$mA.
大信号计算 $\Delta \li{D}= \ui{S}\Big(\exp\dfrac{\uv{D}+\Delta\lv{D}}{\uv{T}}-\exp\dfrac{\uv{D}}{\uv{T}}\Big)=\ui{D}\Big(\exp\dfrac{\Delta\lv{D}}{\uv{T}}-1\Big) = 0.2214$mA
利用 Taylor 展开，对于 PN 节的指数特性，一般教材认为 $\lv{i}\le 5$mV 算小信号。
$\sm{r}{d}=\dfrac{\Delta \lv{D}}{\Delta \li{D}}=\dfrac{\uv{T}}{\ui{D}}$，即 $\sm{r}{d}\propto \dfrac{1}{\ui{D}}$，换言之电流越大，$\sm{r}{d}$ 越小。
| $\uv{T}=26\mathrm{mV}$ | $\sm{r}{d}$ |
|---|---|
| $\ui{D}=0.1$mA | $\sm{r}{d}=260\,\Omega$ |
| $\ui{D}=1$mA | $\sm{r}{d}=26\,\Omega$ |
| $\ui{D}=10$mA | $\sm{r}{d}=2.6\,\Omega$ |
| $\ui{D}=100$mA | $\sm{r}{d}=0.26\,\Omega$ |
exp(5.2/26)-1
0.22140275816016985
print('Error %.2f %%' % (100 * (0.2/0.2214-1)))
Error -9.67 %
for vd in [0.1, 0.2, 0.5, 1, 2, 5, 10]:
Vt = 26
actual = exp(vd / Vt) - 1
estimate = vd / Vt
print('vd = %4.1f mV, error = %5.2f %%' % (vd, 100 * abs(estimate / actual-1)))
vd = 0.1 mV, error = 0.19 % vd = 0.2 mV, error = 0.38 % vd = 0.5 mV, error = 0.96 % vd = 1.0 mV, error = 1.91 % vd = 2.0 mV, error = 3.80 % vd = 5.0 mV, error = 9.31 % vd = 10.0 mV, error = 18.00 %
SPICE
ng.circ('''
D anode 0 D
V anode 0 DC 0.7V AC SINE(0.7 10m 1k)
.model D D(IS=10fA)
''')
ng.operating_point()
{'v#branch': array([-0.00567035]), 'anode': array([0.7])}
这里为了展示失真，特意取 $\lv{i}=10$mV.
ng.cmd('dc v 0.69 0.71 10u')
Vd = 1e3*ng.vector('anode')
Id = -1e3 * ng.vector('v#branch')
plt.plot(Vd, Id)
[<matplotlib.lines.Line2D at 0x7fe979511510>]
ng.cmd('tran 1u 3m')
t = ng.vector('time')
Vd = ng.vector('anode')
Id = -1e3 * ng.vector('v#branch')
fig, ax = plt.subplots()
plt.plot(t, Vd)
ax2 = ax.twinx()
plt.plot(t, Id, 'orange')
[<matplotlib.lines.Line2D at 0x7f7e8e154ed0>]
为了更明显看到非线性失真，用三角波做输入信号。
ng.circ('''
V anode 0 PULSE(0.69V 0.71V 0 1m 1m 1u 2m)
D anode 0 D
.model D D(IS=10fA)
''')
ng.cmd('tran 1u 6m')
plt.plot(ng.vector('time'), -1e3*ng.vector('v#branch'))
[<matplotlib.lines.Line2D at 0x7f1a55e0e890>]
小信号例子
$R=1\,$kΩ，$\uv{D}\approx 0.7\,$V，$\ui{D}\approx 0.3\,$mA, $\sm{r}{d}=\dfrac{\uv{T}}{\ui{D}}=86.7\,\Omega$
$\lv{i}=10\,$mV，$\lv{d}=\dfrac{\sm{r}{d}}{R+\sm{r}{d}} \lv{i}=\dfrac{86.7}{1000 + 86.7}\lv{i}=0.798\,$mV
ng.circ('''
V vcc 0 DC 1V SINE(1V 10mV 1k)
R vcc d 1k
D d 0 D
.model D D(IS=1fA)
''')
ng.operating_point()
{'v#branch': array([-0.00031519]),
'd': array([0.68481116]),
'vcc': array([1.])}
ng.cmd('tran 1u 3m')
t = ng.vector('time')
Vin = ng.vector('vcc')
Vd = ng.vector('d')
Id = -1e3 * ng.vector('v#branch')
fig, ax = plt.subplots()
# plt.plot(t, Vin)
plt.plot(t, 1e3*(Vd - 0.6848112))
#ax2 = ax.twinx()
#plt.plot(t, Vd, 'orange')
[<matplotlib.lines.Line2D at 0x7fedb3fbc810>]
Schottky 二极管¶
发光二极管 LED¶
ng.circ('''
I1 0 vd 10mA
D1 vd 0 LED
.model LED D(Is=1e-22 n=1.5)
''')
ng.operating_point()
{'vd': array([1.78668524])}
整流滤波电路¶
稳压二极管¶
温度补偿¶
