this is the equation for a capacitor discharging through a diode.

assumptions:
C = capacitor in pF
Id = Is*(e^(V/(n*Vt))-1) = diode current
Is = saturation current of diode = ~6300pA for 1N4148
n = device property = ~2 for 1N4148
Vt = thermal voltage = KT/q = ~26mV
V = voltage across capacitor/diode
Vi = starting voltage on cap in mV
t = time in ms
x = ratio of start to end voltage


voltage on cap versus time:

V = -n*Vt*ln[(e^((-Is*t)/(C*n*Vt)))*((e^(-Vi/(n*Vt)))-1)+1]


time to reach a certain percentage of start voltage:

t = -(C*n*Vt/Is)*ln[((e^(-Vi*x/(n*Vt)))-1)/((e^(-Vi/(n*Vt)))-1)]

for e^(Vi/n*Vt) >> 1 -> Vi > ~2.5*n*Vt = ~130mV, this can be approximated as:

t = -(C*n*Vt/Is)*ln[(1-(e^((-Vi*x)/(n*Vt))))]

for e^(Vi*x/n*Vt) >> 1 -> Vi*x > ~2.5*n*Vt = ~130mV, this can be approximated as:

t = -(C*n*Vt/Is)*(e^((-Vi*x)/(n*Vt)))

for Vi*x < ~0.1*n*Vt = ~5.2mV, this can be approximated as:

t = -(C*n*Vt/Is)*ln[(Vi*x)/(n*Vt)]