Non linear ODE of a diode clipper

Skreaming

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Nov 30, 2014
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Hey guys I'm stuck on a problem.


The non linear ODE of the diode clipper stage in a guitar distortion unit is:

dVo = (Vi - Vo)/(RC) - 2 *(Is/C)*(sinh(Vo/Vt));
dt
Where Resistance R = 2200 ohms
Capacitance C = 0.01uF
Saturation Current Is = 2.52nA
thermal Voltage Vt = 45.3mV
Input Voltage Vi
Output Voltage Vo


Background:
basics methods solve equations of form:

dv/dt = v' = f(t,v) (1)

in the case of a linear constant coefficient differential equation, (1) can be written in state space form:

v'(t) = Av(t) + Bu(t) (2)




I have to implement using the trapezoidal rule:

v[n] = v[n-1] +(T/2)*(v'[n]+n[n-1])

where v[n] is the system state at discrete time n, and T is the sampling interval.




I don't even know where to begin or if what I have written even makes sense.

I've been looking at this for 8 hours and getting literally NOWHERE.
Any pointers or questions are welcomed. thanks
 
The non linear ODE of the diode clipper stage in a guitar distortion unit is:

dVo = (Vi - Vo)/(RC) - 2 *(Is/C)*(sinh(Vo/Vt));
dt
Is the above equation meant to be as follows?

. . . . .\(\displaystyle \dfrac{dV_0}{dt}\, =\, \dfrac{V_i\, -\, V_0}{RC}\, -\, 2 \cdot \left(\dfrac{Is}{C}\right) \cdot \sinh\left(\dfrac{V_0}{V_t}\right)\)

If not, kindly please reply with corrections. Thank you! ;)
 
Is the above equation meant to be as follows?

. . . . .\(\displaystyle \dfrac{dV_0}{dt}\, =\, \dfrac{V_i\, -\, V_0}{RC}\, -\, 2 \cdot \left(\dfrac{Is}{C}\right) \cdot \sinh\left(\dfrac{V_0}{V_t}\right)\)

If not, kindly please reply with corrections. Thank you! ;)


That's it :)
 
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