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A simple transistor switch for an LED light.


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Hi all,
I want to make a very simple circuit. I have a few components that I have dug out of storage from when I took some electronics courses way back in college. I want to learn some stuff about transistors.

The description of the circuit is to use a NPN transistor switch to power an LED light. The way I would envision doing this, is by creating an on/off switch at the base of the transistor and powering the LED light through the collector of the transistor. My question is, is this the right way to do this? What conditions do I have to meet in order to make my LED work?

Another question is, where can I find a simple, free program where I can lay out my design graphically? I remember I used PSPice back in college to calculate voltage drops and currents. From what I remember, I had a student version at the time. Are there any free PSPice applications?


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HI!  Well transistor theory is a deep subject. I have one textbook that focuses on just PNP/NPN type transistors.  It’s about 2” thick.

The SWITCHING method is thankfully the easer ways to use a transistor.

The Emitter (the arrow on the schematic symbol) needs to connect to ground.

The collector connects to the load, in this case the LED. 
  Note: Notice that the arrow on the schematic NPN transistor points toward the ground.  The LED should also.

Because the LED and the Transistor have a low resistance, if we connected power to them the current flow right now would blow them both up.  So we need a current limiting resistor.  I’d guess a 2.2k is safe for about 9v.  Place the current limiting resistor between the LED and the positive power.

Now the BASE of the transistor.  All transistors have a value called ‘Beta’ in AC theory it is called hFE.  It means how much raw amplification the device has. Most standard duty NPN’s are above 80. some go as high as 1000. (See Darlington)

Because of this we only need a tiny current to control a larger current.

The current that goes into the base flows through the Emitter to ground.
This current times the Beta is what should be flowing through the collector into the emitter and to ground. 

So if we put 1ma into the base of the transistor and it has a beta of 100 then
We should see 100ma flowing through the collector and to ground.
(Note: the current at the Emitter is the SUM of both Base and collector currents 110ma in this case.)

Because we have a LED and a 2.2k ohm resistor on the collector, if we assume the transistor is a dead short and we are using fully charged 9V, the MOST current we can pull is 4ma.  So what happens if we put 1ma in at the base?

The Beta says that the current through the Collector should be 100ma.  We know we can’t pull 100ma though the diode and resistor we are using because there isn’t enough voltage to create that kind of current flow.

The situation is impossible, as a result the transistor is said to be in ‘Saturation’.
This chiefly means, “It’s as ON as it can get.”

So how do we get 1ma through the base?

The path from the base to the emitter is a diode.  All diodes have a voltage drop, (Called the Knee Voltage)  You will need to check your data sheet for the part you are using but it is somewhere around 1 volt.
So the first step is to subtract 1 volt from our supply voltage.
9v – 1v = 8v

Now we know the current and the voltage, all we need to do is use Ohms to find the resistance.

R = E / I  =  ( 8v / 1ma ) = 8k.

Among the COMMON resistor values the two values closest to that are 8.2k and 7.5k
While 8.2 will result in LESS current into the transistor, we would still be deep into saturation so it would still work.

In fact if we try a 10K ohm resistor.

8v / 10k = 0.8ma
0.8ma * 100beta = 80ma Collector Current (Ic)
Still Deep into saturation. So even a 10k ohm resistor will work.

What is the LARGEST resistor you can use?

Well we are assuming that 4ma is all the LED needs to light up.
So if we work backwards from that,

4ma current flowing through the collector
Divide by the BETA (ex:100) to get the Base current = 0.04ma
So now we have the Base current.  We also know the voltage at
The base as the battery voltage (+9v) subtract the Knee (-1v)  so it’s 8v
Given the voltage and the current we again can use ohms to find the resistance.

R = E / I = (8v / 0.04ma) = 200k ohms.

So given a 200k ohm resistor the led will still light.
But if you use a 270k ohm then what happens?

Well you have just crossed into the Linear realm.
As the resistor goes up, the Base current goes down
And the collector current follows. 

The important fact to remember is that a transistor in full saturation provides very little resistance to the current and as a result it heats up very little.  As you enter the Linear region the transistor begins to behave a little like a resistor. At 400k ohms the LED will only see 2ma of current flowing. 

Here is the magic: 
    The current through the Collector is 2ma now.  That is a fact, so what is the voltage through the current limiting resistor?
Well E = I * R = (2ma * 2.2k) = 4.4volts.  Almost half the supply voltage!
The resistor is heating up. The wattage is the Volts times Current. (4.4v * 2ma = 8.8mw)
The interesting thing is that if we add the 1v lost at the LED and the 4.4v at the current limiting resistor we will find that 3.6v is missing.
If you measure the voltage from the Collector of the transistor to the Emitter of the transistor you will find it.  The transistor is acting like a resistor of 1.8k !
That also means it is heating up Wattage = Volts times Amps  = (3.6v * 2ma = 7.2mw)

At these levels it’s nothing to worry about but at higher currents transistors can heat up and self destruct.  That’s why Power amps have those huge heat sinks on the back.

What about the SMALLEST resistor you can use?

Well we already know that anything under 200K will put the transistor into saturation. However as the base resistor values get smaller, the current through the base goes up.  If you check your data sheet you will see a MAXIMUM Ibase this value is
Usually very low.  Definitely around 270 ohms the currents would be getting dangerously close to blowing most transistors. (Around 20ma)

This is all based on the NPN transistor, similar to the 3904.  Others have higher Betas and different Base to Emitter Knee voltages.

I know it was long but I hope there is useful data in there you can use.

Good Luck!

-Mike (Ward)


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  • 4 weeks later...

Hi Gogo,
Rason is incorrect.
He says to use a base current of only the load current divided by the minimum hFE of the transistor. But a transistor's datasheet shows then the transistor isn't saturated, its collector voltage is from 1V to 4V.
Datasheets show how well a transistor saturates when its base current is its load current divided by 10.

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You are correct. Rason's method will get the job done sufficiently. Try it on the workbench and you will see.

Yeah, for lighting a 2V LED from a 9V battery, a high gain transistor will give 20mA and a low gain transistor will give a 4V loss and 8.6mA but the LED will still give some light.

If you use the correct value of base resistor for a transistor to saturate then a high gain transistor and a low gain transistor will operate the same.
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Very Nice Explaining! Good job Theatronics!!!

About this part:

So if we put 1ma into the base of the transistor and it has a beta of 100 then
We should see 100ma flowing through the collector and to ground.
(Note: the current at the Emitter is the SUM of both Base and collector currents 110ma in this case.)

Shouldn't the Ve be 101ma?
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