# How does this circuit generate a high frequency signal?

I am new to electronics. I am watching this video.

At 5:37, you can see the first half of the circuit generate a high frequency signal.

How do an op-amp, 9 VDC, and a few resistors generate a high frequency signal?

• You can simulate the circuit yourself by using this quick online simulator Jul 16, 2021 at 9:33
• Are you familiar with astable multivibrators...? Jul 16, 2021 at 9:36
• @Dat: The critical point is that there's a capacitor in the circuit. Without the capacitor it is just an amplifier. With an appropriately connected capacitor, an amplifier becomes an oscillator.
– JRE
Jul 16, 2021 at 9:37
• @JRE Without the capacitor it is just an amplifier You could say that but I'd still call it a Schmitt trigger. Jul 16, 2021 at 9:42
• @JRE ok, do you mean the 100p capacitor connected to the GND in the photo? I see the capacitor, but I don't know how they work.
– Dat
Jul 16, 2021 at 9:44

It's called a relaxation oscillator: -

Wiki has a full explanation but, if in doubt, read this from TI entitled Relaxation oscillator circuit.

• These sentences from Analogzoo page: "When the op amp’s output is at Vcc (high), the positive input pin is held at 1/2 Vcc by the R2/R3 resistor divider. At this point, C1 begins charging up through resistor R1 until it too crosses the 1/2 Vcc mark. The op amp sees that its inverting input is more positive than its non-inverting input, at which point it drives its output to Vss (low)". I have a question: both input pins of the op apms are at 1/2 Vcc, why do the author say: "The op amp sees that its inverting input is more positive than its non-inverting input"?
– Dat
Jul 19, 2021 at 4:04
• Your quote says "until it too crosses the 1/2 Vcc mark" and that means the inverting input has risen slightly higher than the non-inverting input. I guess the word "too" is an error because the non-inverting input is held at Vcc/2 and doesn't cross it @Dat Jul 19, 2021 at 7:06
• Please help me more, also from Analogzoo page: "The problem is that resistor R1 and R4 (a required pull up resistor for the LM393 comparator) are both are connected to the comparator’s output pin, forming a voltage divider between Vcc and C1." What is pull up resistor? What is its role in the circuit? Where is Vcc? I can't see the voltage divider between Vcc and C1.
– Dat
Jul 19, 2021 at 8:21
• Stop please. I do not offer support to other people's web pages. A pull-up resistor can be googled. This is a Q and A site. You ask a question and someone provides an answer. Tutorials in the comment area are very restricted. Jul 19, 2021 at 8:40