Can anybody explain Emitter-Follower easily?

I know basic electronics but am totally unable to understand the Emitter-Follower configuration of transistor. Can anybody explain it in easy to understand terms? Consider the following scenario:

-I give a NPN transistor a base voltage of +5V. Voltage at collector is +12V and Emitter is connected to common ground. There is NO LOAD. (I know that if there is load on the collector side the transistor will simply act as a switch.) But if I'm not wrong, when there is no load, the voltage between Emitter and Ground is 4.2V (5V - Voltage Drop), and not 12V (or 11.2V accounting for drop)! Why is this happening? When the transistor is saturated, shouldn't current flow from Collector to Emitter and voltage should be 12/11.2 V?

-Also, the primary application of this config is to change high input impedance to low output impedance. However I am unable to understand impedance at all!

Any help will be highly appreciated. Sorry if the questions are too silly.

One thing to ponder on is that if there is no ‘NO LOAD’ there is no current flow.

srnet:
One thing to ponder on is that if there is no 'NO LOAD' there is no current flow.

In that case, why does current flow when I short the terminals of a battery?

Because a short is a MASSIVE load.

Because you have connected a load.

Do you know anything about 'transistor biasing'?

The way you have fed voltages to the transistor should have killed it in the mean time.

A transistor has three zones of operation: active zone (amplification zone), cut off zone, and saturation zone.

A transistor has to be biased first to set the quiescent point and then apply excitation (input signals) to shift it from cut off zone to saturation zone and vice versa or let is make an excursion in the active zone.

This is an example of an emitter-follower circuit:
ef.png

ef.png

Voltage at collector is +12V and Emitter is connected to common ground.

If the emitter is connected to ground then the voltage at the emitter will be 0V, how can it ever be anything else?

You need a resistor between the emitter and ground, then you will get your expected 4v2 on the emitter. Try 1k ohms or thereabouts.

I don't know what you mean by 'no load', the way you describe it makes no sense to me.

If the Emitter is connected to "Ground" [i.e. the *Negative Supply*], then it's NOT the Emitter Follower configuration.

Have a look at this:


Notice how, in the Emitter Follower configuration, the Emitter voltage [the voltage across R2], is always one VBE less than the input voltage [typically somewhere around 700mV]. In other words, the Emitter closely follows the voltage applied at the Base of the transistor [i.e. across the Base-Emitter junction]. The fact that the output is Vin - VBE, and that VBE hardly changes, is key, to understanding this circuit.

An interesting feature of this is: because of the current gain of the transistor, signified by the parameter "hFE", the current at the Base [the input current] is far lower than the current at the Emitter [the current that flows through R2]. This is what makes this configuration a "unity-gain [nearly] buffer" [we're talkin' AC gain, here :wink: ]. In other words, using this transistor arrangement, you can mirror an AC voltage, at the output [with only a minor, usually insignificant loss], with little current impact, on the input!

Here's the math:

[tt][b][tt]I[sub]B[/sub] = [(Vin - V[sub]BE[/sub])/R[sub]2[/sub]]/h[sub]FE[/sub] = (Vin - V[sub]BE[/sub])/(R[sub]2[/sub]*h[sub]FE[/sub])[/b][/tt][/tt]

So, if R2 is 500Ω and Vin is 5V, and VBE is 700mV, and hFE = 70, then:

** **I[sub]E[/sub] = (Vin - V[sub]BE[/sub])/R[sub]2 = 8.6mA[/sub]** **

** **I[sub]B[/sub] = (5V - 0.7V)/(500Ω*70) = 123µA** **

-- Another way:
** **I[sub]E[/sub]/h[sub]FE[/sub] = I[sub]B[/sub] = 8.6mA/70 = 123µA** **

Notice how a mere 123µA at the base, controls a whopping [by comparison] 8.3mA at the Emitter!

BTW: another name for the Emitter Follower is the Common Collector.

For this sort of learning, I recommend that you download LTSpice. It is a free software simulator where you can build small circuits and test them with voltage and current probes. There is a bit of a learning curve, but you'll find some getting started videos such as Getting Started with LTspice - learn.sparkfun.com

Here is an example of an emitter follower circuit which I created recently:

Edit:
Updated with a more realistic example based on the description in the OP. The simulator cannot usefully model the situation with a completely empty load so I've added a 10M resistor.

Take the left hand circuit of reply #8. That is an emitter follower.

The instant you apply a voltage to the base, current starts to flow from the base to the emitter and causes a bigger current to flow from collector to the emitter. The size of this current is the base current times the gain of the transistor. But as more current flows through the emitter resistor, there is more voltage dropped across the resistor.

This in turn leads to a smaller voltage difference between the base and the emitter and so reduces the base current. This continues until an equilibrium is reached where there is just enough base current to sustain a voltage on the emitter and no more. You might think this equilibrium point is reached when the two voltages are equal, but in fact the circuit from base to emitter looks like a diode. So the equilibrium point is reached when the emitter voltage matches the base voltage minus the voltage drop across a diode which is 0.7V for a silicon diode.

The emitter resistor acts as a sort of negative feedback reducing the effective gain of the transistor to one.

However I am unable to understand impedance at all!

Think of it as resistance to an input. With a low resistance input you get lots of current for a small voltage. A speaker is an example of a device with a low input impedance.

When something has a high impedance very little current flows into it for lots of volts. The analogue inputs of an Arduino are a good example of the, the current flowing into the input is known to engineers as “bugger all”.

The difference between an impedance and resistance is that impedance also encompasses the concept of AC signals as well as DC ones. If you were only dealing with DC then they would be the same. But with AC signals you have to consider the phase difference between the voltage and the resulting current.

livelongpranav:
I know basic electronics but am totally unable to understand the Emitter-Follower configuration of transistor. Can anybody explain it in easy to understand terms? Consider the following scenario:

-I give a NPN transistor a base voltage of +5V. Voltage at collector is +12V and Emitter is connected to common ground.

You've just fried the transistor. The base-emitter junction has been massively overloaded and melted rapidly.
The base-emitter voltage can't rise much above 0.7V without large currents flowing from base to emitter.

An emitter follower is also known as common-collector. The collector is at a fixed voltage, the emitter has a load, and the emitter voltage tracks the base voltage less about 0.6 to 0.7V.

There is NO LOAD. (I know that if there is load on the collector side the transistor will simply act as a switch.) But if I'm not wrong, when there is no load, the voltage between Emitter and Ground is 4.2V (5V - Voltage Drop), and not 12V (or 11.2V accounting for drop)! Why is this happening? When the transistor is saturated, shouldn't current flow from Collector to Emitter and voltage should be 12/11.2 V?

-Also, the primary application of this config is to change high input impedance to low output impedance. However I am unable to understand impedance at all!

Any help will be highly appreciated. Sorry if the questions are too silly.

Change your circuit and add a 100 ohm resistor from emitter to ground. Vary the base voltage and see the emitter voltage follow it almost linearly. If the gain of the transistor is 200, then the base looks like 200 times
the load (ie 20k), since 200 times less current flows for "about" the same voltage drop as the emitter.

An emitter follower circuit is what is states. The Emitter will have the same voltage as the base minus the Emitter Base voltage which is about 0.7 volts. The collector is connected to the power source and the emitter is the output. The voltage you put on the base is about +0,7 of what will be on the emitter. It doesn't matter if NPN or PNP. Use a NPN for positive and PNP for negative, that should get you through what you need.

Have Fun and Good Luck,
Gil

jackrobot:
Emitter Follower means that the input signal follows the output signal.

No. The output follows the input, not the way round you have it.