Timeline for Use of transistor for switching the led
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 4, 2020 at 3:43 | vote | accept | G-aura-V | ||
| Apr 26, 2018 at 9:51 | comment | added | po.pe | It's not "some" portion... it will always be V-0.7V. This means if your transistor has a B of 1000 and you have 1.5V with a 66k resistor the collector current will be (1.5V-0.7V)/66k*1000=12mA | |
| Apr 26, 2018 at 9:45 | comment | added | G-aura-V | some portion of the voltage applied is dropped across the resistor. Doesnt this happen in case of LDR? | |
| Apr 26, 2018 at 9:40 | comment | added | po.pe | Dividing what voltage? | |
| Apr 26, 2018 at 9:27 | comment | added | G-aura-V | Thats the thing I was missing. So if I keep the source voltage below 0.7, there will be no any current through the transistor and I can get my led lit up. Right? Wait, doesnt the LDR function as the voltage divider like the resistor does? | |
| Apr 26, 2018 at 9:16 | comment | added | po.pe | Not exactly... you have a diode in the base emitter path of your transistor which has a forward voltage of 0.7V - so it's not the breakdown voltage. What your LDR does is controlling the current going into the base ((1.5V-0.7V)/RLDR) which is then amplified by factor B... so IC = B * IB. So there will always be some current (probably negligable with high base resistor) trough the transistor. | |
| Apr 26, 2018 at 9:08 | comment | added | G-aura-V | As per my understanding, the current doesnt flow until the circuit is closed. For closing the circuit, the base voltage must exceed the breakdown voltage (0.7V). Here, the LDR(5k & 66k resistor in the schematic) is used to limit the voltage. When the LDR offers 5k resistance, base voltage exceeds the breakdown voltage due to which the current flows through the transistor (less resistive path). But when the resistance offered by LDR is 66k, the current cant flow through the transistor which leaves only the path through the LED to complete the circuit.Is anything wrong with my understanding? | |
| Apr 26, 2018 at 8:33 | comment | added | pipe | I'm neither an upvoter nor a downvoter, but looking at the edit history it looks like the original schematic didn't make much sense unless you read the fine-print. The right way to show that the values doesn't mean anything is to leave them out, not just write anything random and hope that people won't try to build it. | |
| Apr 26, 2018 at 8:16 | history | edited | Russell McMahon♦ | CC BY-SA 3.0 |
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| Apr 26, 2018 at 8:06 | comment | added | Russell McMahon♦ | The downvoter does not understand how to vote properly. If downvoting an answer like this one, which addresses the main fault with the original circuit, a downvoter MUST say why - as otherwise they make it seem like the answer is fundamentally wrong - which is not the case. Downvoting in this manner is about equivalent to trolling. | |
| Apr 26, 2018 at 6:53 | history | edited | po.pe | CC BY-SA 3.0 |
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| Apr 26, 2018 at 6:18 | history | answered | po.pe | CC BY-SA 3.0 |