Question

6. The following operational amplifier RC oscillator generates square waveform whose values change between +Vcc and -Vcc. Consider Ro = Rp = R = 10 K12 in this circuit. (a) Describe the operation of the circuit. First assume Vout is at +Vcc. Explain how the switching to-Vec takes place at the output. Draw the waveform for Vout(t), VREF(t) and Vsig(t). (b) Calculate the value of the capacitor, C, if the period of oscillation is 2 us. That means that Vout(t) is at +Vcc for a period of 1 us and at -Vcc for a period of 1 us. R = 10K +Vcc VSIG VsIGN — Vout(t) -Vec > Ro = 10K VREF RF = 10K

Answer 1

+ usat Ui>Vc VICVC output output =-veat division vo tvsat Vi= VORI (by voltak R2 + R2 Rule) VI = B Vo Operation to = tvsat Vi=Bro Capautor charge through Rf Vc increase gradutty if VC < Busat Vos tvsat Reanin TH Vcsport output change Usar - Vsat NOW capcultor disharge through RF Il capantor voltase below - Breat Vce Brest then noc+ Usat Usat Broat 7 dave Y/RC - Ersat? -Vsat To k 2 to Vc - Voo) + (vin-vo) e Ro vc. Veat + C-Broat - kratle" Vc Vsat CI-CHA) VRC) at T2 T0/2 Vc - Brsat
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-17) Bisat = ksat (1-CHple 1-B2 AltBlettro To carcluche) To= 2 RC CHAD Given Ri=RO = 1016 R2 = Rf = 10k RzR = 10ks B² R2 10 = 1/2. 10+10 R 1 To = 2usec 2= 2x168 xc lu(3/2) 2x109xC lu(3) Tc - 91.0pt