Expert: cleggsan - 7/27/2006

Question
Thanks for your help.
" The purpose of the resistor in the emitter leg is to swamp out the small voltage changes at the base and have large voltages at the emitter."
I do not quite understand what you mean here....

As far as I can see, the combination of the voltage divider at the base and the emitter resistor set the current through the base-emitter section so that under quiescent conditions the voltage across the base-emitter is 0.6V.
Also, due to emitter-follower action, the emitter resistor seems to provide negative feedback to allow the amplifier to operate with larger swings of the base voltage?

If we have a basic emitter follower (transistor with emitter resistor) and apply a voltage to the base, the voltage at the emitter will be 0.6V less than the voltage at the base? But if the value of the emitter resistor is changed... does this change the value of the emitter voltage (so for example, if a higher resistor were placed in the emitter leg, the voltage across it may rise to 0.5V below the base? assuming constant base voltage)
Thanks :)


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Followup To

Question -
Is the transistor in an emitter follower always in its active region unless the voltage on the base swings above the supply voltage (+Vcc or below 0V?

with a common-emitter amplifier with a grounded emitter, the voltage on the base can only swing + or - 0.1V or the transistor will saturate (base-emitter voltage will rise above +0.6V) but it appears that with a common-emitter amplifier with an emitter resistor (emitter degredation) the voltage on the base can swing between the supply voltage (+Vcc) and ground giving a wider range of base voltage variations.
Is this correct?

Answer -
Well, sort of. I think you are misunderstanding the operation of a transistor linear (class A amplifier configuration) circuit.  

Transistors are current devices, not voltage devices.  True; where there is current flowing there will be potential difference.  However, the operation of a transistor circuit in linear mode is to bias it such that at no signal condition the transistor is conducing half way between cut-off and saturation.  The trick to do that is to have it pass enough current from base to emitter to accomplish that status.

It turns out that the voltage between the base and emitter (it is just a diode junction, after all) is typically 0.6 volts (in silicon transistors and 0.3 in germanium).

If you look at the IV characteristics you find that the current in the emitter/collector will rise extremely rapidly once that base to emitter voltage reach and surpasses 0.6 volts.  It is extremely non-linear.  The purpose of the resistor in the emitter leg is to swamp out the small voltage changes at the base and have large voltages at the emitter.

Hope thismakes sense to you.  If you study real life emitter folower circuits you will find biasing resistors down from the power supply voltage that are selected to put the transistor in the center of its operation!

(For switching, the transistor is usually biased OFF and when a pulse comes along that is over the 0.6 volts the transistor turns on and goes into saturation or full current.)

Finally, the transistor is a current amplifier. If the transistor you are using in the circuit has a current gain or beta of 100 it means that a current of 1 ma into the base will cause 100 ma to flow in the emitter/collector circuit.

Let me know if you have any questions whatsoever.

Cleggsan

http://www.mathcad.com/Library/LibraryContent/MathML/emitampl.htm

is a good example and explanation of a common emitter design. It looks at both the ac operation and dc conditions for proper bias.  

Answer
Let's look at the emitter follower in two ways: DC for biasing the transistor up into its center point and AC to look at the way the small signal ac characteristics behave.

DC:  The resistor dividers on the base circuit bring the voltage on the base up to .6v dc above the dc voltage at the emitter.  This level is set such that under dc conditions the current in the emitter/collector of the transistor is half way between cutoff and saturation.  Got that part?

AC:  Now, assuming we are biased - as above - into the center of operation, we can apply an ac signal to the input or base either directly if it is dc coupled, but usually via a capacitor that allows ac signals to pass through but does not alter the dc bias settings on the base.

As the ac signal swings positive and then negative it causes the current to increase and decrease above and below the center point of where it was established by the dc bias settings.  When the signal swings positive it increases the current in a positive way and the voltage increase on the emitter to ground resistor increases in exact proportion (because, remember there is a near fixed relationship on voltage between the base and emitter junction due to it being biased in the forward or ON position).  Same is true with a negative going signal; the emitter resistor current changes in exact proporting to the input current  and the instantenous voltage falls at the same rate.

Since the ac voltage variations are in exact lockstep to the input voltage (ac) the gain at the emitter resistor is unity; that is the voltage at the output of the emitter is the same as the voltage at the input.

The larger the resistor in the emitter, the higher the input impedance at the intput of the base; this is because the currents are proportionately lower.

And so on.

Hope this is helpful.

Cleggsan