AP 3302 Pt. 3 Section 2 CHAPTER 2 Square Waves applied to CR circuits Since the same current is flowing through each resistor at any given instant, the voltages across the resistors will be in proportion to their resistances. For example, if 100V is applied to a CR circuit containing two series resistors valued at l0kW and 90kW then at the instant of applying the voltage the 10kW resistor would have 1OV across it and the 90kW resistor 90V. Suppose we have a circuit consisting of a 90kW resistor and a l0kW resistor connected in series with a 2uF capacitor which is initially charged to 20V (Fig 6a). The time constant CR is 2 x 106 x 100 x l03 = 0.2 second. If we now apply a voltage step of +100V to this circuit, Vc remains initially at +20V, because C cannot change its charge immediately. Since V = Vc + VR1 + VR2. we then have 80V divided between R1 and R2 in proportion to their resistances. Across R1 we have 90/100 x 80= 72V and across R2 we have 10/100 x 80= 8V (Fig 6b). C then commences to charge, rising to 20 + (63/100 x 80) = 70.5V in CR seconds (0.2 second) and to 100V in 5 CR seconds (1 second). Both VR1 and VR2 fall to zero as Vc rises to + l00V At all times the relationship V = Vc + VR1 + VR2 is maintained. Comparison of CR Time Constant with Pulse Duration When the voltage input to a CR circuit is a square wave the exact shapes of the waveforms across C and across R depend upon how the CR time constant compares with the pulse duration of each part of one cycle of the square wave. We have already seen the waveforms of VC and VR when the relationship between the CR time constant and the pulse duration is such that the capacitor can just charge and discharge in the time available (Fig 3). If the time constant is very short compared with the pulse duration, C can charge and discharge in a fraction of the available time and the waveforms of VC and VR are then different from those in Fig 3. 

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