Timeline for Automatic dark detector: base voltage
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 21, 2022 at 14:50 | comment | added | ElectroNewbie | Thank you for your time and answer jp312. | |
| May 21, 2022 at 14:02 | comment | added | ElectroNewbie | Ok, I think it must be calculated simply using KVL. The voltage drop across the BE diode is 0.7V, so that’s the voltage at the base of the transistor. | |
| May 21, 2022 at 13:29 | comment | added | ElectroNewbie | I think the key word in your answer is “clamp” the voltage. What does it mean that the base-emitter clamps the voltage? Does it get “fixed” at 0.7V above earth? I’m sure this is all trivial stuff. Thank you all for your patience. Will try to understand better the case in which we only have a resistor and a diode in series. | |
| May 21, 2022 at 13:10 | comment | added | ElectroNewbie | Thank you for your answer. Yes, intuitively I see that when the LDR is very high R2 dominates the circuit and current gets aplied to the base. But what I don’t get is why the base is at 0.7V. Why this holds? Is it a property of the BE “diode” or something like that? | |
| May 21, 2022 at 13:03 | history | answered | jp314 | CC BY-SA 4.0 |