from matplotlib import pyplot as plt
import numpy as np


def make_current_graph():
    v = [400, 600, 800, 1000, 1500, 2000, 2500, 3000, 3500, 4000, 4500, 5000, 5500, 6000]
    i = [75.5, 79.8, 84.7, 88, 94, 97.4, 100.4, 102.8, 105.7, 107.8, 109.2, 112, 114.2, 115]

    fig, ax = plt.subplots(1, 1)
    ax.plot(v, i, linewidth=2, marker='x')
    ax.set_xlabel("Voltage (V)")
    ax.set_ylabel("Current (uA)")
    plt.savefig('opto-150.png')
    plt.show()


if __name__ == '__main__':
    # This script calculates the voltage vs time of a capacitor (representing an electrostatic actuator) which is
    # initially completely discharged. The capacitor is connected to a voltage source v through an opto-coupler OPTO-150
    # The scripts uses a crude forward finite differences scheme to calculate the voltage over time on the capacitor
    # (i.e. actuator) taking into account: 1) the current through the OPTO-150, which decreases when the capacitor
    # charges, and 2) the current spent in the voltage feedback reading circuit, which is proportional to the voltage
    # on the capacitor. This can be use to estimate how fast an actuator of a given size (i.e. capacitance) can be
    # charged by the hvps-x
    v = 4000    # enter the voltage set point of the hvps-x
    c = 1.4E-9    # enter the capacitance connected to the output in F
    Rmon = 500E6    # This is the monitoring resistor (R13). 250M in the initial release. Will be increased to 500M.
    dt = 1E-3       # time step of the FD calculation
    tf = 200E-3     # time at which the simulation stops


    time = np.arange(0, tf, dt)
    u = np.zeros_like(time)
    for i in range(len(u)-1):
        v_opto = v - u[i]
        i_opto = (14.6 * np.log(v_opto) - 13.3) / 1E6
        i_meas = u[i] / Rmon
        i_charge = i_opto - i_meas
        u[i+1] = u[i] + dt * i_charge / c

    fig, ax = plt.subplots(1, 1)
    ax.plot(time, u)
    ax.set_xlabel('time (s)')
    ax.set_ylabel('Voltage (V)')
    plt.savefig('fd.png')
    plt.show()