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In the circuit attached. There are two capacitors. One is parallel to "PWR LED". Although I understand it delays the turning on and off of the LED, why would you need that? Also, the other capacitor is connected in parallel to the phototransistor. This connection I completely fail to understand the purpose of. This circuit diagram is of a flame sensor, which generates an output when the phototransistor detects IR.Flame Sensor Circuit Diagram

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    \$\begingroup\$ One is parallel to "DO LED" No it isn't. \$\endgroup\$
    – Andy aka
    Jan 9, 2020 at 12:35
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    \$\begingroup\$ One is parallel to "DO LED" No there isn't. the 100nF on the left is for supply decoupling (also called a bypass capacitor). The 100 nF cap in parallel with the photo transistor is to filter out any noise from the photo transistor. It makes the circuit respond a bit more slowly to light so that it becomes less sensitive to responding randomly but only responds when it should. \$\endgroup\$ Jan 9, 2020 at 12:38
  • \$\begingroup\$ @Andyaka I'm sorry. I mean to the PWR LED. I will edit it right away. \$\endgroup\$
    – Zelreedy
    Jan 9, 2020 at 12:38
  • \$\begingroup\$ @Bimpelrekkie If I apply what I understood of a decoupling capacitor on this circuit, does it mean that since the phototransistor continues to have different voltage drops depending on the IR intensity levels, the capacitor is there to counter those voltage changes, which may affect the current through the PWR LED, thus keeping the LED stable? \$\endgroup\$
    – Zelreedy
    Jan 9, 2020 at 12:56

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The capacitor that is parallel to the photo-transistor is used to extend the time the DO_LED is on after the flame has disappeared or momentarily ceased. The recharging of that capacitor (100 nF) is via the 10 kohm resistor hence the CR time is 1 millisecond.

The only other capacitor is across the power rails and this is a requirement for most op-amps to ensure stability and/or correct operation.

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  • \$\begingroup\$ Thank you. I also have another quick question. Why are the "DO LED" and the 1kΩ resistors connected in parallel to the 10kΩ resistor, both connected to 5V and the output terminal of the op-amp? When I tested the sensor, the LED would always be on and only turn off when it detects IR. \$\endgroup\$
    – Zelreedy
    Jan 9, 2020 at 13:55
  • \$\begingroup\$ @Zelreedy The LM393 has open collector outputs and, my speculation is, that when the circuit was being developed, a 10 k resistor was used as a pull-up resistor for the open collector output and subsequently in the development, the LED and 1 kohm resistor were added and although these don't require the 10 k resistor, it was left in the circuit. \$\endgroup\$
    – Andy aka
    Jan 9, 2020 at 13:58

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