# NPN transistor wired correctly?

i'm experimenting with transistors, and wired one to my circuitboard. Though I got the LED to turn on with a NPN transistor, I'm not sure it's setup properly.

I read that current moves against the direction the arrow is pointing for the transistor symbol (emitter to collector), so I wired the emitter to a higher voltage (7.5V) than the base (4.5V). The problem is, when I remove the voltage source to the base, the LED remains lit. Shouldn't it turn off if the base truly acts like a switch for current flowing from emitter to collector?

I included two schematics below, one a crude, most likely incorrect representation of my circuit, and another from a book I'm learning from. My attempt at making a circuit follows my failure to get the schematic in the book to work - please let me know if you spot any errors in either schematic.

simulate this circuit – Schematic created using CircuitLab

• Um. Your wiring, if I'm reading it correctly, puts $7.5\:\textrm{V}$ across the LED. The red wire is coming from ground, but then the yellow wire goes from there to the LED cathode. The blue wire from $7.5\:\textrm{V}$ is patched up to the anode of the LED by the other blue wire! You could pull that BJT out and the LED would still be on, I think. (Not that I think the BJT survived -- I don't.) I hope that trainer is bullet proof (resistor in series with ALL LEDs just for this circumstance, etc.)
– jonk
Jul 12, 2017 at 7:49

## 3 Answers

That is truly a very poor diagram in your book, throw it away. You have also made mistakes in transcribing it.

This is much better. It is drawn as a conventional circuit, with GND at the bottom, and increasing voltage generally up the page, with conventional current flowing downwards (hint, in the direction of the BJT emitter arrow), which makes it much easier to read and interpret.

simulate this circuit – Schematic created using CircuitLab

• Better, but not there yet. The 10 volts needs to be in series with the 5. Jul 12, 2017 at 10:20
• @WhatRoughBeast if we must Jul 12, 2017 at 13:36

Current flow in a NPN is always from collector to emmitter and in PNP its from emitter to collector.

Your circuit will not turn on the transistor.

Just for understanding, assume the npn transistor in form of two diodes as shown below.

simulate this circuit – Schematic created using CircuitLab

In order to turn ON a NPN Transistor you have to forward bias the E-B diode(junction), in your circuit you have reverse biased that junction so the transistor will not turn on.

edit:

The simplified form of the circuit you're trying to replicate will look something like this

simulate this circuit

It seems like you're not using any of these resistors.

Replace the transistor (it may be damaged) and try using the above schematic with the current limiting resistors.

• thanks, I tried switching the voltage sources for base (white) and emitter (blue), is this what is meant by forward biasing the junction? If so, I still couldn't get the LED to light in that configuration Jul 12, 2017 at 14:52
• The picture shows that you're not using any resistors to limit the base current, which may caused a damage to the transistor Jul 12, 2017 at 15:47

The orientation of the diode in your schematic does not match your book's schematic and neither do the Batteries. The Base Must be around .7 Volts higher than the Emitter in an NPN BJT for current to flow between the collector and emitter.

I couldn't see how you got the LED to glow unless you somehow made the Collector Base Junction Forward Biased. So I Tried it, your schematic that is.

And the Led had it's negative lead at GND and it's positive lead at the collector and the Positive voltage was at the collector and I connected the Emitter to the Base (I basically just made your schematic) and the LED lit up...Wild. But when I disconnected the base the LED stopped.

I imagine that the diode became forward biased and the electrons went to the collector and repelled the collector electrons into the Base making the Collector Base Junction Forward Biased. Or the Positive charge simply forward biased the Base-Emitter Junction and the LED became forward biased (assuming the top schematic is correct.