Timeline for Understanding simple circuit with two transistors and one led
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 10, 2016 at 2:11 | comment | added | x-ray | Exactly. But with the addition that after the switch is opened, the current that flows from VCC to the switch goes off, too. The switch may still be connected to ground, but not to VCC. That's why I tried to bring power to the switch from the transistor. | |
| Apr 10, 2016 at 2:04 | comment | added | Wesley Lee | Ok I need some clarification. A normally closed switch lights up an LED. When you press and release it you want the LED to stay permanently off? | |
| Apr 10, 2016 at 2:01 | comment | added | x-ray | What I really try to do (apart from understanding why this circuit doesn't work as I thought it would) is the following: I want to close a circuit (e.g. light up an led) until I open a normally closed switch. Of course that would be the most straightforward task, but: Once the switch opens (that's a one time operation), the voltage at the switch should drop to (almost) zero. I thought I could add that switch where I just added D2. | |
| Apr 10, 2016 at 1:52 | comment | added | Wesley Lee | As I commented on your question, adding D2 gives about the same result as removing that connection. You want a circuit that powers the LED after a push button is pressed and released, is that correct? Do you want it to turn off after you press it again or you want it to turn off after a time? | |
| Apr 10, 2016 at 1:49 | comment | added | x-ray | Yes, S1 is normally open. I changed the circuit by adding D2. Now S1 cannot short circuit the transistors but only bring current to Q2's base. But the circuit behaves exactly the same: Pressing S1 -> LED on, releasing S1 -> LED off. Why doesn't the LED stay lit up now? | |
| Apr 10, 2016 at 1:13 | comment | added | Tim Spriggs | yes, the transistors are useless in this scenario. | |
| Apr 10, 2016 at 1:07 | history | answered | Wesley Lee | CC BY-SA 3.0 |