您的示例不起作用，因为您的函数采用了三个参数a、b和c（尽管未使用c），但您仅通过scipy.optimize.curve_fit的p0参数给予了两个初始猜测。
除此之外，您不能期望函数$f（x）= a当取两边的对数时，得到$\log{f} = \log{\exp{-bx}- 1} + log{a}$，它不像直线的形式，$y = mx + c$。移除$-1$将给予一条斜率为$-B$且截距为$\log{a}$的直线：$\log{f} = -bx + \log{a}$.
下面的代码能完成您想要的任务吗？

import numpy as np
from matplotlib import pyplot as plt
from scipy.optimize import curve_fit

voltage = np.array([0.015, 0.060, 0.108, 0.162, 0.214, 0.268, 0.313, 0.351, .381, 0.417, 0.455, 0.517, 0.545, 0.582, 0.604])
current = np.array([0.0200, 0.0800, 0.2000, 0.6200, 1.6900, 4.8400, 11.5800, 24.5000, 44.3400, 90.4000, 183.3000, 505.5000, 744.3800, 1149.0500, 1346.6000]) / 1e6

def fexp(x, a, b):
    return a * np.exp(b * x)

def fexp_yshift(x, a, b):
    return a * (np.exp(b * x) - 1)

popt, pcov = curve_fit(fexp, xdata=voltage, ydata=current, p0=[3.651e-9, 19.77])
popt_shift, pcov_shift = curve_fit(fexp_yshift, xdata=voltage, ydata=current, p0=[3.651e-9, 19.77])

x = np.linspace(np.min(voltage), np.max(voltage), 1000)
fit = fexp(x, *popt)
fit_shift = fexp_yshift(x, *popt_shift)

plt.plot(voltage, current, "k.", label="Measurement")
plt.plot(x, fit, "m--", label="Model")
plt.plot(x, fit_shift, "c-.", label="Model with Shift")
plt.xlabel("Voltage $V$ / V")
plt.ylabel("Current $I$ / A")
plt.grid()
plt.legend(loc='upper left')
plt.show()

plt.plot(voltage, current, "k.", label="Measurement")
plt.plot(x, fit, "m--", label="Model")
plt.plot(x, fit_shift, "c-.", label="Model with Shift")
plt.xlabel("Voltage $V$ / V")
plt.ylabel("Current $I$ / A")
plt.yscale('log')
plt.grid()
plt.legend(loc='upper left')
plt.show()

输出量：

对不起，我不知道如何添加数学公式堆栈溢出后。

EDIT：我刚刚意识到这并没有回答如何提高拟合的质量。要得到更好的拟合，直接用直线拟合电流的对数：

import numpy as np
from matplotlib import pyplot as plt
from scipy.optimize import curve_fit

voltage = np.array([0.015, 0.060, 0.108, 0.162, 0.214, 0.268, 0.313, 0.351, .381, 0.417, 0.455, 0.517, 0.545, 0.582, 0.604])
current = np.array([0.0200, 0.0800, 0.2000, 0.6200, 1.6900, 4.8400, 11.5800, 24.5000, 44.3400, 90.4000, 183.3000, 505.5000, 744.3800, 1149.0500, 1346.6000]) / 1e6
current_log = np.log10(current)

def line(x, m, c):
    return m * x + c

popt, pcov = curve_fit(line, xdata=voltage, ydata=current_log)
fit = line(voltage, *popt)

plt.plot(voltage, current_log, "k.", label="Measurement")
plt.plot(voltage, fit, "m--", label="Model")
plt.xlabel("Voltage $V$ / V")
plt.ylabel("Logarithmic Current $\log{(I / \mathrm{A})}$")
plt.grid()
plt.legend(loc='upper left')
plt.show()

输出量：