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I made this circuit. I calculated everything, but all I get is THE LED JUST WONT TURN ON.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ What is the voltage across the LED? What is the voltage across the resistor? \$\endgroup\$ – analogsystemsrf Apr 16 at 4:20
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    \$\begingroup\$ What exactly did you calculate and how? \$\endgroup\$ – jonk Apr 16 at 4:25
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    \$\begingroup\$ Also explain what you want to achieve with this circuit. In my opinion, it will just destroy the LED, the transistor and possibly both. \$\endgroup\$ – Bimpelrekkie Apr 16 at 6:32
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    \$\begingroup\$ Hm..... I smell smokes bro... \$\endgroup\$ – Mitu Raj Apr 16 at 7:28
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Possibility 1: You mixed up the pins on your transistor and didn't hook it up right. Not every transistor has its emitter, base, and collector pins in the same order. Even different packages of the 2N2222 might have different pin order.

Possibility 2: you blew up your LED. If this circuit works, the base of the transistor is at about 2.5 V. Then the current through the base resistor is about 0.5 mA. Multiply this by the \$\beta\$ of the transistor (anywhere from 100 to 300) and there could be as much as 150 mA flowing through the LED. But the LED is only rated for 30 mA continuous. You now have a DED (dark emitting diode).

Generally, you should design BJT circuits with the assumption that the current gain could be arbitrarily high. Use some other mechanism, like an emitter resistor, to limit the collector current.

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  • \$\begingroup\$ Ummmm.... I tested the LED after my 'EXPERIMENT' and found out that it was working just fine......... Must be something with the package then..... M really sorry I am quite new to electronics........ \$\endgroup\$ – Rahul Maurya Apr 17 at 1:15
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    \$\begingroup\$ @RahulMaurya, a photo of your actual circuit might help. Another thing beginners mess up is understanding which parts of a breadboard are connected and which aren't. \$\endgroup\$ – The Photon Apr 17 at 1:17
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I've no idea how you calculated your circuit. If you'd taken a moment to explain your thinking, I'd have been happy to help clarify things. But without that, all I can do is offer a couple of alternate suggestions.

First off, you could just use a resistor with the LED and not use a BJT, at all. That's probably the obvious first answer. The resistor's value would be \$R=\frac{5\:\text{V}-V_\text{LED}}{I_\text{LED}}\$, where \$V_\text{LED}\$ is the expected voltage across the LED when the LED experiences the desired \$I_\text{LED}\$. You can get the approximate values for these from the LED datasheet, by experiment, or by educated guesses. It's not critical. The main thing is to over-estimate the resistor value to start out (just to be safer.) You can always reduce its value to get the current higher, if you want.

But if you want a circuit that will provide a relatively fixed current (you decide what you want) regardless of the LED type, or if you want to be able to turn the LED on and off under some kind of control line to an MCU, then you will need a BJT or two. Since I've no clue where you are headed with this (since you haven't said, at this time), I can't assume anything about the LED except that it probably works on a voltage below \$3.5\:\text{V}\$ (this value was chosen by me because the circuits below won't work if the LED requires more than that) and that you are fine with something around \$10-12\:\text{mA}\$ for the LED (easily adjusted within a small range of reason from there by changing \$R_1\$'s value a bit.)

The circuit on the left is pretty much a standard approach. It uses two standard small signal BJTs. \$Q_2\$ pulls down on the base voltage of \$Q_1\$ such that a set current is measured by the voltage drop across \$R_1\$ and applied to \$Q_2\$'s \$V_\text{BE}\$. If the current in \$R_1\$ increases, then \$Q_2\$ pulls down harder on \$Q_1\$'s base by forcing a greater voltage drop across \$R_2\$. And the reverse also is true. So it works pretty well. However, it is somewhat temperature-sensitive. (Hold onto \$Q_2\$ and let your body heat it up and watch.)

schematic

simulate this circuit – Schematic created using CircuitLab

The circuit on the right also uses two BJTs, but one of them is a PNP. (That may be an advantage if you don't have two NPNs but do have a spare PNP.) If you thermally couple both BJTs together (for example, NPN and PNP transistor arrays) then this circuit will be a little less sensitive to temperature at the expense of two more resistors.

The lead on both left and right circuits, marked as \$+5\:\text{V} = \text{ON}\$, can be used with an I/O pin of an MCU to enable and disable the LED.

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  • \$\begingroup\$ Thank you for the help.... I am quite new to electronics..... and I have this bad habit of raging over stuff....... But hey I will try it again, now that I know what to do :) \$\endgroup\$ – Rahul Maurya Apr 17 at 1:21
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    \$\begingroup\$ @RahulMaurya If you need further explanations, so that you can work out values on your own in different circumstances, just say so. It would also be nice to know what LED you were planning to use (what current it requires and what voltage, too.) But aside from a simple resistor, the circuits shown above are two of many other possibilities. I didn't show it, but you could also use a current mirror for the job, too. But I didn't want to present too many different ideas at once without knowing where you were headed with all this. \$\endgroup\$ – jonk Apr 17 at 1:27
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possibly you have already burned the LED. using the gain of BJT to control the current is not so reliable here. the simplest way is just to connect a resistor in series with the LED for CW output. if you want to modulate the LED, connect like this: power --- resistor --- LED --- npn BJT C --- npn BJT E to ground.

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  • \$\begingroup\$ You can use schematic to add more clarity. \$\endgroup\$ – Mitu Raj Apr 17 at 18:37
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You appear to have your LED on the wrong side of the transistor. Try connecting the LED and a current limiting resistor between the transistor and the + rail.

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    \$\begingroup\$ That may or may not be a good thing for the LED. If the BJT saturates, almost 5 V will be across the LED. But of course that would be too much collector current for the base current. But the OP might get near 100 mA through the LED anyway. So.. not so good. But you are right about the OP's circuit being bonkers, too. \$\endgroup\$ – jonk Apr 16 at 4:25
  • \$\begingroup\$ @jonk well more basic and flawed than bonkers I think. I still have to use a datasheet or reference just to breadboard an LED or fet/transistor =). Good point though. I just googled 2n2222 LED circuit to check if it was N or P channel and didn't even notice the missing current limiting resistor =P. \$\endgroup\$ – K H Apr 16 at 4:33
  • \$\begingroup\$ Ummmmm no.. the LED does not blow up as I can still use it but... Yup... it was on the wrong side...... when I connect it to the 'Right Side' it does give me a faint glow... Thank yalll = ) ....... I am really new to electronics so yea, I am kind of a NOOB. \$\endgroup\$ – Rahul Maurya Apr 17 at 1:24

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