ACCEPTED SOLUTION
Accepted Solutions
With the diode voltage Vanode versus Vcathode (and diode current Id running from anode to cathode) the diode model could be:
(Where Is is the leakage current, generally below 1 pA, k is Boltzmann's constant, e is the electron charge and T is absolute temperature, around 300 K at room.)
The capacitor current Ic depends on the voltage across it Ud, as follows:
now the diode voltage depends on the input voltage Ui as:
Ud=Uc-Ui
And finally Ic+Id=0, which leads to the differential equation:
The solution is:
Which brings in a(n integration) constant C0 (which is a voltage).
This expression can be worked to an expression for Uc:
To find C0 we use the initial condition set: at t=0, Uc=0, but also the integral is 0. So we get:
So the expression for Uc simplifies to:
With Ui known,
(where A=1 V and f=1*kHz)
the integral can be solved, but not symbolically, and then Ui can be subtracted from it to get Ud.
Remember that Ud was defined upside-down. In fact V2 = -Ud, so we can write:
Lets take some values and plot:
We see the voltage across the capacitor Uc slowly growing more negative as time progresses. This moves V2 further above Ui.
Success!
Luc
Hi,
I can also do simulation, but that's not what I'm interested in now about simulation, but how to solve the circuit analytically using mathcad.
Anyway, your simulation set-up and result is not yet right (see below - here assuming forward voltage drop of the diode = 0V):
Prime 11 VDi=on=0
Prime 11 VDi=on=0.7
Using Luc's method, IS=10-15
Using Luc's method, IS=10-12
With the diode voltage Vanode versus Vcathode (and diode current Id running from anode to cathode) the diode model could be:
(Where Is is the leakage current, generally below 1 pA, k is Boltzmann's constant, e is the electron charge and T is absolute temperature, around 300 K at room.)
The capacitor current Ic depends on the voltage across it Ud, as follows:
now the diode voltage depends on the input voltage Ui as:
Ud=Uc-Ui
And finally Ic+Id=0, which leads to the differential equation:
The solution is:
Which brings in a(n integration) constant C0 (which is a voltage).
This expression can be worked to an expression for Uc:
To find C0 we use the initial condition set: at t=0, Uc=0, but also the integral is 0. So we get:
So the expression for Uc simplifies to:
With Ui known,
(where A=1 V and f=1*kHz)
the integral can be solved, but not symbolically, and then Ui can be subtracted from it to get Ud.
Remember that Ud was defined upside-down. In fact V2 = -Ud, so we can write:
Lets take some values and plot:
We see the voltage across the capacitor Uc slowly growing more negative as time progresses. This moves V2 further above Ui.
Success!
Luc
