Since the capacitor is charging its impedance Xc keeps increasing and the gain Xc2/R4 also keeps increasing. The negative feedback compels the opamp to produce a voltage at its out so that it maintains the virtual ground at the inverting input. The gain of the entire circuit (Xc2/R4) will be very low and the entire voltage gain of the circuit will be close to the zero.Īfter this initial “kick” the capacitor starts charging and it creates an opposition to the input current flowing through the input resistor R4. As a result the inverting input terminal (tagged A) of the opamp behaves like a virtual ground because all the current flowing into it is drained by the capacitor C2. A serious amount of current flows through the input resistor R4 and the capacitor C2 bypasses all these current. The feedback resistor R5 connected in parallel to C2 can be put aside because R5 has almost zero resistance at the moment. By virtue capacitor C2 offers very low resistance to this sudden shoot in the input and C2 behaves something like a short circuit. Let’s assume the positive side of the square wave is first applied to the integrator. The opamp integrator part of the circuit is shown in the figure below. The square wave signal is applied to the inverting input of the opamp through the input resistor R4. Resistor R5 in conjunction with R4 sets the gain of the integrator and resistor R5 in conjunction with C2 sets the bandwidth. The opamp IC used in this stage is also uA741 (IC2). Instead of using a simple passive RC integrator, an active integrator based on opamp is used here.
Bipolar square wave generator#
Next part of the triangular wave generator is the opamp integrator. If the values of R2 and R3 are made equal, then the frequency of the square wave can be expressed using the following equation: This cycle is repeated over time and the result is a square wave swinging between +Vcc and -Vcc at the output of the opamp. Now the capacitor discharges trough R1 and starts charging in positive direction. When the voltage across the capacitor has become so negative that the voltage at the inverting pin is less than the voltage at the non-inverting pin, the output of the opamp swings back to the positive saturation. Now a fraction of the negative high output (-Vcc) is fed back to the non-inverting pin by the feedback network R2, R3. The capacitor quickly discharges through R1 and starts charging in the negative direction again through R1. When the voltage across the charging capacitor is increased to a point the voltage at the inverting pin is higher than the non-inverting pin, the output of the opamp swings to negative saturation (-Vcc). A fraction of this high voltage is fed back to the non- inverting pin by the resistor network R2, R3. When the power supply is switched ON, the C1 starts charging through the resistor R1 and the output of the opamp will be high (+Vcc). Initially, when power is not applied the voltage across the capacitor C1 is 0. Resistor R2 and R3 forms a voltage divider setup which feedbacks a fixed fraction of the output to the non-inverting input of the IC. Resistor R1 and capacitor C1 determines the frequency of the square wave. The square wave generator is based on a uA741 opamp (IC1).
The square wave generator section and the integrator section of the circuit are explained in detail below.
The circuit diagram is shown in the figure below. The circuit uses an opamp based square wave generator for producing the square wave and an opamp based integrator for integrating the square wave. An integrator which converts square waves to triangular waves. Like triangular waves, square waves have equal rise and fall times so they are more convenient to be converted to a triangular waveform. In this project, we are using square waves for input. To generate triangular waves we need an input wave.