hi..can anybody tell me how to convert a sine wave form into a square waveform by using op-amps(IC-741 specifically)....its my term project and i have just half a week left to submit it so plzzz hurry and reply..plz be elaborative and include diagrams if possible..thanks.regards
Dude, no one in this forum will do someone else's work. We only HELP. It is your work so you have to do it yourself.
However I will help you.
In order to convert a sine wave into a square wave you only have no use an Opamp (like IC-741) using a non-inverting amplifier circuit with a very large gain.
The very large gain willl make the circuit saturate quickly, transforming your sine into a square.
ohk thank you for your feedback . i was actually able to "test run" my circuit in the lab..but you see there's now a tiny miny BIG problem..i am now suppose to convert my square wave output into a trangular wave...Now how i do that?..and also to convert a sine wave into square wave using IC-741 op-amp?...and my project is now worth 50% weightage of my total course..what a misery!!
that was an excellent help sir!!...i really appreciate it..and the slide show well that was so elaborative and easy to understand..thanks once again.you ppl are great help..kkep up the good work
hey!! hey.!!..i made the project and m just a day away from showing it to my teacher..guys ive made the integrator and adjusted the but i didnt have any idea what i was doing actualy..i read it somewhere that if you change the resistance and capacitance you can adjust the wave into traingular wave but what i want to ask is that is there any specific easy to understand "formula" for voltage output of an integrator?..not that one that includes an 'integral' type of stuff but a simple universal formula so i can easily understand it and make my teacher understand it as well..thanks waiting for ur feedback
In simple terms, the output will decrease (because it is inverting) at a rate determined by the resistor and capacitor at dV/dT=-Vin/RC so if you had 100k and 100nF and a 5V signal, for example, the output would change at the rate of 500V/s.
Keith.
Added after 3 minutes:
The simple explanation, by the way, is that the opamp keeps the inverting input at zero volts, so the current in the resistor is Vin/R. This current must go through the capacitor and the equation for a capacitor charge (Q=CV), when differentiated gives I=C.dV/dt. So, dV/dt=I/C = Vin/RC. The minus sign appears because of the dire4ction of the current in the capacitor, resulting in -Vin/RC.
Perhaps TANSAH wants to see an integral explicitely in the formula.
Therefore, read this explanation for an intergator circuit (without signal inversion):
* Voltage across a capacitor: Vc=(1/C)∫idt with i=Vin/R (due to virtual ground at the inverting opamp terminal)
* Thus, Vc=(1/RC)∫(Vin)dt.
* For Vin=const. this leads to Vc=(1/RC)*Vin*t (>>linear ramp prop. to time t).