V+ is set by Vout across a resistive voltage divider.
V- is obtained using Ohm's law and the capacitor differential equation:
Rearranging the V- differential equation into standard form results as the following:
Notice there is two solutions to the differential equation, the driven or particular solution and the homogeneous solution. Solving for the driven solution, observe that for this particular form, the solution is a constant. Inother words,
V_ = A
where A is a constant and,
Using the Laplace transform to solve the Homogeneous equation
results in
V- is the sum of the particular and homogeneous solution
Solving for B requires evaluation of the initial conditions. At time 0,
Vout = Vdd and V- = 0
Substituting into our previous equation,
0 = Vdd + B B = - vdd
Frequency of Oscillation
First lets assume that Vdd = - VSS for ease of calculation. Ignoring the initial charge up of the capacitor, which is irrelevant for calculations of the frequency, note that charges and discharges oscillate between Vdd / 2 and VSS / 2. For the circuit above, VSS must be less than 0. Half of the period (T) is the same as time that Vout switches from Vdd. This occurs when V - charges up from - Vdd / 2 to Vdd /2.
When VSS is not the inverse of Vdd we need to worry about asymmetric charge up and discharge times. Taking this into account we end up with a formula of the form:
Which reduces to the above result in the case that, Vdd = - VSS.
Fig.1 Shows Input Signal (Red) and Output Signal (Yellow)
Fig.2 Shows Input Signal (Red) and Output Signal (Yellow)
