2
\$\begingroup\$

I simulated and built current circuit. relaxation oscillator

What I got was a rectangular wave generator with a duty cycle significantly lower than 50%.

Than I made R3 = R4 and got 50% duty cycle.

I managed to think that RC constant is defined only by C1 and R1. It appears I was wrong.

How can I define RC constant for this circuit? Can anybody shed some light on the problem?

\$\endgroup\$
  • \$\begingroup\$ I find being able to simulate with realistic timing always helps me to understand how the circuit works. Here ya go: bit.ly/Jl3Y23 \$\endgroup\$ – krb686 Dec 14 '13 at 22:43
  • \$\begingroup\$ Thanks, useful source @krb686 However the circuit provided is meant for bi-polar power supply. \$\endgroup\$ – ivan Dec 14 '13 at 22:59
  • \$\begingroup\$ That simulator can do that also. Right click the op-amp, edit, set minimum voltage to 0, and then add the DC supply and resistor, and change all the other values. I used your values and it works great. I would link it to you, but Paul Falstad left a serious bug in his code where frequently the export window freezes and you lose the circuit. @ivan \$\endgroup\$ – krb686 Dec 14 '13 at 23:33
0
\$\begingroup\$

Yes, you are wrong - it's more complex than that.

The signal on pin 3 is important to analyse in order to determine operating frequency. When the output is low, the voltage on pin 3 is determined by R3, R4 and R2 - this is the voltage the capacitor has to discharge to in order to cause the op-amp to flip from being low to being high. Then, when the op-amp output is high, the capacitor starts to re-charge up to a new high voltage. Pin 3 is now a higher voltage because the output is now high and, via R2, it sets a higher threshold (on pin 3) than what it did when the output was low. This is called hysterisis.

The capacitor charges and discharges at a rate determined mainly by C1 and R1 but, if R2 is relatively big it's going to take longer to attain the "new" threshold in order to flip the op-amp output.

This is how they work - calculate the two voltages that could appear at pin 3 and these two voltages are what the cap charges to and then discharges to during one period of oscillation. Then, use what you know about exponential charging of a capacitor via a resistor to fill in the gaps.

\$\endgroup\$
  • \$\begingroup\$ I calculated pin3 voltages through KCL. Then ΔV_pin3 = ΔV_out*(1-e^(-T/2R1C1)). The role of R2 is not clear. Can you give more details? \$\endgroup\$ – ivan Dec 14 '13 at 22:50
  • \$\begingroup\$ Pin 3 is the non-inverting input according to your diagram - it looks more like you've calculated the voltages on pin 2. Pin 3 (without R2 and with R3 and R4 equal) is half the supply voltage, V3. If you assume the op-amp output swings rail-to-rail R2 is in parallel with R4 when the output is low and in parallel with R3 when the output is high. This means, due to the circuit oscillating, the voltage on pin 3 is a low amplitude square wave. By "low" I mean somewhat less than the power rails. \$\endgroup\$ – Andy aka Dec 14 '13 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.