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?
\$\begingroup\$
\$\endgroup\$
8
-
\$\begingroup\$ You can simulate the circuit yourself by using this quick online simulator \$\endgroup\$– Long PhamJul 16, 2021 at 9:33
-
\$\begingroup\$ Are you familiar with astable multivibrators...? \$\endgroup\$– Mitu RajJul 16, 2021 at 9:36
-
1\$\begingroup\$ @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. \$\endgroup\$– JREJul 16, 2021 at 9:37
-
1\$\begingroup\$ @JRE Without the capacitor it is just an amplifier You could say that but I'd still call it a Schmitt trigger. \$\endgroup\$– BimpelrekkieJul 16, 2021 at 9:42
-
\$\begingroup\$ @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. \$\endgroup\$– DatJul 16, 2021 at 9:44
|
Show 3 more comments
1 Answer
\$\begingroup\$
\$\endgroup\$
4
It's called a relaxation oscillator: -
Wiki has a full explanation but, if in doubt, read this from TI entitled Relaxation oscillator circuit.
There is also this page from Circuit digest should you need it. And also this page from Analogzoo.
-
\$\begingroup\$ 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"? \$\endgroup\$– DatJul 19, 2021 at 4:04
-
\$\begingroup\$ 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 \$\endgroup\$– Andy akaJul 19, 2021 at 7:06
-
\$\begingroup\$ 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. \$\endgroup\$– DatJul 19, 2021 at 8:21
-
1\$\begingroup\$ 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. \$\endgroup\$– Andy akaJul 19, 2021 at 8:40