NPN Transistor as a switch: what's the use of the emitter resistor?

[Moriambar]

Hi,
as I stated elsewhere I'm a noob with BJT transistor, since I almost always used FETs (which I find way easier).
Regardless, I've seen some design of npn transistors used as switches and they almost always feature a 100R resistor from the emitter to ground.

What's the purpose of it?
Thanks

[bd139]

That's not a plain switch then. It's an emitter follower.

[Moriambar]

That's not a plain switch then. It's an emitter follower.

I thought that in the emitter-follower configuration the load should be connected to the emitter. Mine is connected to the collector.
The emitter has just 100R to ground.

I provided an example to clarify what I'm saying

[bd139]

The load should be connected between emitter and ground with the emitter follower. This is used for linear devices and current sources rather than switching though. It may switch but it'll be incredibly inefficient.

To switch with a transistor, the emitter should be grounded, the load connected across the positive rail and the collector. The base current should be appropriate to switch the transistor on fully ((Ic / beta) * 2 approx to jam the transistor on hard)

[Moriambar]

The load should be connected between emitter and ground with the emitter follower. This is used for linear devices and current sources rather than switching though. It may switch but it'll be incredibly inefficient.

To switch with a transistor, the emitter should be grounded, the load connected across the positive rail and the collector. The base current should be appropriate to switch the transistor on fully ((Ic / beta) * 2 approx to jam the transistor on hard)

Ok so either I found wrong examples (I've not them at hand) or I did not understand the purpose of the circuit. 100% they were not emitter followers, though.

[bd139]

Yep. Could be either.

Sometimes it's impossible to work out if the design is rubbish, the designer was drunk or the circuit does what is intended but the reason isn't obvious.

[Moriambar]

Yep. Could be either.

Sometimes it's impossible to work out if the design is rubbish, the designer was drunk or the circuit does what is intended but the reason isn't obvious.

Thanks. Well at least I have reasons to doubt the design, then.

[leeatljs]

Sometimes you put a resistor in the emitter to provide a constant current or limit.

[Peabody]

Yes, I think it's a current limiting circuit.  Current will flow through the transistor and the voltage across the emitter resistor will rise.  But at some point that voltage will be too high for the base current to continue to flow, so the collector current will drop.  When it settles out, you should get a fixed current through the load determined by the voltage driving the base.  Something like that.  Anyway, I think the emitter resistor is there for a reason.

[Benta]

Yep, constant current (if the input voltage is known), this works perfectly for LEDs. Otherwise just a simple overcurrent protection.

[soldar]

Not an emitter follower. The resistor limits the current in the device and prevents the bipolar transistor from going into saturation. Switching out of saturation takes longer time so fast counters and other switches will often prevent going into saturation by using such a resistor. Today we just use FETs, which don't have this problem.

[bionic]

The emitter resistor will decide what Beta or amplification for that specific transistor will be!

Remember that:
IC/IE  = Alfa

Where Alpha gives:
Beta=(alpha/alpha-1)=IC/IB

[Benta]

bionic, I don't think you've quite understood this circuit.
But welcome to the Forum.

[gf]

Per se - as drawn - it just looks like an amp with emitter feedback...
But where/how is the load supposed to be connected?
Or if really anything is supposed to be switched with the transistor, where/how is it connected to the given circuit?
[ If nothing else is connected to the given circuit, then I'd consider it rather useless. ]

EDIT: Or is RL itself supposed to the load? If yes, what kind of load is it?

[Benta]

Per se - as drawn - it just looks like an amp with emitter feedback...
But where/how is the load supposed to be connected?
Or if really anything is supposed to be switched with the transistor, where/how is it connected to the given circuit?
[ If nothing else is connected to the given circuit, then I'd consider it rather useless. ]

EDIT: Or is RL itself supposed to the load? If yes, what kind of load is it?

RL is the load, yes. And it could be anything. Another input. A pullup resistor. An LED.
The collector of the transistor is a current source.
Let's say you have a 3.3 V logic output driving the transistor base. 3.3 - 0.7 = 2.4 V across the emitter resistor, equals 24 mA. Wonderful for driving an LED as load.
Or another input through a cable that might be shorted to plus through a connector short cut: no damage to the transistor. And, and, and...

Forget the emitter follower picture you have in your mind. It's a switch.

[gf]

I did not call it emitter follower. I said "amp with emitter feedback" (which is eventually a voltage controlled current source, driving RL). With proper biasing at the input, the circuit could well act as amplifier with a closed-loop voltage gain of approx. RL*beta/(R1+R2*beta). But agreed, if the input voltage is low enough, then the current through RL can be turned off completely - then it becomes a switch. Contrary to the "traditional notion of a switch" (-> low impedance in the "on" position -> low power dissipation at the switch), the transistor and R2 do well dissipate power, when the "switch" (i.e the current source) is turned on.

[Jwillis]

Emitter resistor is a way of biasing the transistor  to prevent thermal run away.
Some times there will be a base resistors with paralleled transistors to prevent load hogging because not all transistors of like kind will turn on at exactly the same voltage.
Look up transistor biasing for further explanation because the way a transistor is biased depends on application. 

[Audioguru]

The transistor circuit shown is called a "phase splitter". The phase at the emitter output is the same phase as the base input but the phase at the collector output is -180 degrees.

[Benta]

Why do you guys keep going on about this? The headline clearly states is it to be used as a SWITCH.
As an amplifier stage, it's far too simple and has no biasing provisions.

[atmfjstc]

I think @soldar has it right. It is indeed a switch configuration, with Re there to ensure maximum switching speed. I remember reading about this trick in school. There is some detail here:

http://www.engineering-bachelors-degree.com/electronic-components/uncategorized/bipolar-junction-transistor-circuitshigh-speed-bjt-switching/

[magic]

This isn't guaranteed to prevent saturation, Baker clamp should be used for that.

[Benta]

This isn't guaranteed to prevent saturation, Baker clamp should be used for that.

Correct. But it depends on the load, if it has too high resistance, saturation may indeed occur.

[David Hess]

As shown in Moriambar's example, the emitter resistance limits the current to (Vin-Vbe)/R2 where Vbe is about 0.6 volts.

Base series resistor R1 has only a minor effect because the transistors current gain (hfe) should be much greater than the ratio between base series resistor R1 and emitter series resistor R2.  It however should be included because if the load is missing, then the drive signal will see an effective short to ground through the emitter series resistor.

Nothing about emitter series resistor R2 prevents the transistor from going into saturation if the collector voltage falls and again, this is why R1 is included to protect the driver.

[Moriambar]

The transistor circuit shown is called a "phase splitter". The phase at the emitter output is the same phase as the base input but the phase at the collector output is -180 degrees.

The output (shown as RL) is at the collector, not at the emitter

[bson]

The emitter resistor makes it a current switch rather than a voltage switch.  Some loads, particularly semiconductors but also solenoids, are current operated and need a particular current rather than voltage.  With an emitter resistor you don't want a base resistor; base current is low as the resistance looking into the base is RE*hfe = high for high hfe, and a base resistor just makes life more complicated.  You don't use a collector resistor with a current switch.