INPUT TERMINAL of the integrator B.The output of the integrator is a triangular wave and it is feedback as input to the comparator A through a voltage divider R2R3. Analysis:Just before switching occurs from +Vsat to −Vsat, the voltage at point P is zero.This means the −Vramp must be developed across R2 and +Vsat must be developed across R3. i.e.\( - \FRAC{{{V_{ramp}}}}{{{R_2}}} = - \frac{{ + {V_{sat}}}}{{{R_3}}} \cdots \left( i \right)\)\( - {V_{ramp}} = \frac{{{R_2}}}{{{R_3}}} + {V_{sat}} \cdots \left( {ii} \right)\) Similarly, +Vramp, the output of integrator at which the output of comparator switches from −Vsat to +Vsat, is given by:\( + {V_{ramp}} = - \frac{{{R_2}}}{{{R_3}}}\left( { - {V_{sat}}} \right) \cdots \left( {iii} \right)\)The peak to peak output amplitude of a triangular wave is:\({V_0} = + {V_{ramp}} - {V_{ramp}} = 2\frac{{{R_2}}}{{{R_3}}}{V_{sat}} \cdots \left( {iv} \right)\)Where Vsat = | + Vsat| = | − Vsat|.Equation (iv) represents the amplitude of triangular wave decreases with an increase in R3.The time is taken by the output to swing from −Vramp to +Vramp (or from +Vramp to −Vramp) is equal to half the period (T/2).This time can be calculated from the integrator output equation as follows:\({V_0} = - \frac{1}{{{R_1}{C_1}}}\mathop \smallint \limits_0^{\frac{T}{2}} \left( { - {V_{sat}}} \right)dt = \frac{{{V_{sat}}}}{{{R_1}{C_1}}}\left( {\frac{T}{2}} \right)\)\(\frac{T}{2} = \frac{{{V_0}}}{{{V_{sat}}}}{R_1}{C_1}\)\(\frac{T}{2} = 2{R_1}{C_1}\left( {\frac{{{V_0}}}{{{V_{sat}}}}} \right)\)Substituting V0 from equation (iv) in the above equation.\(T = \frac{{4{R_1}{C_1}{R_2}}}{{{R_3}}}\)We know that the frequency is inversely proportional to the time.\(f = \frac{{{R_3}}}{{4{R_1}{C_1}{R_2}}}\)Thus the frequency of oscillation increases with an increase in R3.