The circuit is built using two op-amp chips LM741. First LM741 is used to generate square wave and second one will generate triangular.

In 1^{st} LM741 on inverting terminal, the feedback of output is given through resistor R1 and capacitor C2 while on non-inverting terminal feedback is given by voltage divider of R2 and R5. The output of this is directly given to inverting terminal of 2^{nd} LM741 through R4. 2^{nd} LM741 along with R4 and C1 forms an integrator circuit. Its non-inverting terminal is grounded. Both op-amps are given +Vcc and –Vee supplies to respective pins 7 and 4 as shown.

Here is the photograph of the circuit built on bread board

**Circuit Operation**

· Suppose initially output of U1 is +Vcc

· So capacitor C2 will start charging through R1

· When the capacitor voltage increase to more than voltage at non-inverting terminal given by equation

V1 = (R2 / (R2+R5))× Vo = (10 / 20)× Vo = 0.5× Vo

· The output of U1 becomes –Vee

· So capacitor starts discharging through R1

· Again when capacitor voltage decrease to less than voltage given by same above equation, the output becomes +Vcc again

· Thus the cycle repeats and it produces a square wave output with time period determined by RC components R1 and C2

· This square wave output is given as input to an integrator circuit

· The integrator converts square wave into triangular wave

· When output of U1 is +Vcc, the capacitor C1 charges to max value and gives positive ramp in the output of U2

· And when U1 output is –Vee, capacitor C1 discharges and generates negative ramp in the output of U2

· Thus we get triangular wave as an output from U2

Here is the photograph of final output of the circuit on DSO

Note: - the circuit will produce perfect triangular wave output for certain frequency values only. Because integrator works as low pass filter also. So for very low frequency the output will be clipped (because of high gain) and at high frequencies the output will be very low (because of low gain).