Why is my transistor accepting power from a base?

Question

I have a transistor (NPN, to be exact it's a SS9014). Whenever I run 4.5 V through the base, connect a resistor and LED to the emitter and connect the LED to ground, the LED has a faint glow. Why is any power being transmitted to the LED when there is nothing connected to the collector? How can I avoid this? Here's the schematic:

I am actually still in high school and want to major in electrical engineering. I know this may seem simple for professionals, so thanks for taking the time to help me out.

Answer 1

The base-emitter junction (BE) of a NPN transistor behaves like a diode:

So you basically have this equivalent circuit:

As you can see, your LED is in series with the resistor and the transistor's diode (the BE junction). Some current (depending on the value of your resistor) is passing through this circuit, making the LED light.

One way to avoid this is, as The Photon said, using a MOSFET. Another way is using the NPN transistor in the common-emitter configuration:

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From this comment of yours:

Actually, I read the specifications on the transistor and the maximum base input is 5V

I can understand why you think you should drive the transistor's base from 4.5V (< 5V). But you can drive the base with a higher voltage as long as you have a resistor in series. The BE junction, acting as a diode, will have a fixed voltage drop (typically 0.7V) across it. The rest of the voltage will go across the resistor which also limits the current through the BE junction.

Answer 2

This is how transistor really works - small current running thru base->emitter makes big current flow thru collector->emiter.

You should connect emiter to ground and put resistor and led between collector and voltage source.

Edit

Schematics added:

Answer 3

What you built is a common collector circuit, and the others already try to persuade you to change that to a common emitter circuit. Common emitter is indeed better for switching, but common collector also works if you keep a couple of things in mind.

While common emitter needs less than a volt to drive the transistor, common collector needs a higher voltage. If the LED's voltage is 2 V you need at least 2.7 V at the base to get the least emitter current. To get 20 mA for the LED you need 20 V extra for R1, and you don't have that, so R1 must be a lower value, like 50 \$\Omega\$. Then 20 mA will drop 1 V across R1, and the base voltage will have to be 3.7 V minimum. Then there will be 0.8 V across R2 and the base current will be 800 \$\mu\$A.

That's not how it works. We would have a calculated base current of 800 \$\mu\$A and a collector (or emitter) current of 20 mA, which would give an \$\mathrm{H_{FE}}\$ of 25. But we don't decide how high \$\mathrm{H_{FE}}\$ is, the transistor does. And that's 280 typical. So our calculation is wrong.

You can leave out R2. Then the base will be at 4.5 V, and the emitter at 3.8 V. With a 2 V drop across the LED we have 1.8 V for R1, and then the current is 36 mA. A bit high, let's increase R1 back to 90 \$\Omega\$ to get our 20 mA back.

But wouldn't there be too much base current without R2? No. To get 20 mA collector current we'll have 71 \$\mu\$A base current, the transistor takes care of that. If the base current would increase because the supply voltage increases then so will the collector current, and hence the voltage drop across R1. The emitter voltage will rise and counteract the increase in base current. A similar automatic regulation occurs when the base current would decrease.

So R1 indirectly takes care of the base current and makes R2 superfluous. But you can't calculate base current as (4.5 V - 0.7 V - 2 V)/R1. The resistance seen from the base is R1 \$\times\$ \$\mathrm{H_{FE}}\$. Why is that? Say you increase base current by 1 \$\mu\$A. Then collector current will increase by 280 \$\mu\$A (\$\mathrm{H_{FE}}\$ = 280), and the voltage drop across R1 will increase by 90 \$\Omega\$ \$\times\$ 280 \$\mu\$A = 25.2 mV. So the resistance seen from the base is 25.2 mV / 1 \$\mu\$A = 25200 \$\Omega\$, or 280 \$\times\$ 90 \$\Omega\$.

And that explains why the LED in your circuit lights so faintly: I = (4.5 V - 0.7 V - 2 V)/(R1 \$\times\$ \$\mathrm{H_{FE}}\$ + R2) = 6 \$\mu\$A! It's a wonder it lights at all.

Answer 4

The base-emitter current is flowing through your diode which causes it to light. I don't know the resistor value, but most likely the resistor is large enough that it is limiting the current. The brightness of an LED depends on the current. More current, more light (until the LED is damaged, then no current and no light!)

The base-emitter terminals of a transistor form an effective diode, so the circuit you've described contains a series connection of two diodes (the transistor and LED) and a resistor.

In terms of how to avoid the LED turning on, I'm not sure what you are trying to do, so I can't offer a meaningful suggestion beyond not connecting the battery? :)

Answer 5

The circuit is doing exactly what it should.

This is not a usual way to use a transistor.

The 'be' (base emitter) junction is a forward biased diode.
When you apply 4.5V to the base current flows
From supply inth the base
Through the forward biased be junction though the resistor (if it is above the LED)
through the LED
then to ground.

A circuit has been made, current flows, the LED lights.

Swapping the resistor and LED will produce an identical result.

If you apply Voltage of 4.5V or more to the collector current will flow via CE path and the LED will light more brightly.

The normal way to use a transistor to control an LED is to connect a resistor from V+ to the LED
Connect LED to collector
Connect emitter to ground
Drive base with a voltage VIA A RESISTOR (say 10k).

When base drive is 0V transistor will be off.
When base drive is more than about 0.6V LED will start to turn on
When transistor base drive via resistor is a few volts transistor will be fully on.

Example:

IF LED is a red LED then Iled will be ~~~= (Vcc-VLED)/R = (4.5 say -2.5)/R = 2V/Rled
If Rled = say 33o ohms then I led will be about 2/330 ~= 0.006A = 6 mA