Emitter follower bjt.....

Thread Starter

Zeeus

Joined Apr 17, 2019
600
Hi...Made this circuit earlier...Used 3k for both R1 and R2... I don't remember the capacitor values..sine wave input

Output impedance is around little Re (24 ohms- quiescent point : Re changes yeah?)

So using the 47 ohms, was expecting about 2/3 of 5.3V but there was clipping (output was like half wave rectifier output : I did not read the max value :( )

Please why would it clip? I replaced the 47 ohms with 470 and there was no clipping : output almost same amplitude as Vout (5.3) if I remember correctly
 

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DickCappels

Joined Aug 21, 2008
6,642
In order to drive the load resistor through the capacitor the emitter follower must be able to source and sink current into the capacitor. It can source a "decent" amount of current from the emitter but your emitter resistor cannot sink much current. The result is the driver runs out of current when driving that very low resistance load. Decrease the emitter resistor but be careful to not pull the base voltage too low or cook your transistor :)
 

DickCappels

Joined Aug 21, 2008
6,642
The capacitor changes the time varying positive voltage to AC voltage.

If you were to get 2/3 of 5.3 volts peak-to-peak, for example, you would need 1.76 amps from the positive supply (+12V through the emitter) and 1.76 amps from the negative supply (OV from ground ground through the resistor).

With 5.1 k for the emitter load the maximum negative-going current you could get is 12V/5.1k = 2.4 ma, so the signal clips. Since your signal does not get to +12V the current is even lower.

Slightly different view:
In your circuit with those values the capacitor just charges up and little current passes through it because it is already charged with the + end toward the positive supply and only a high resitance discharge path for the signal, so not much current is going to flow.

Try making your emitter resistor lower and see what happens.
 

Jony130

Joined Feb 17, 2009
5,186
In the emitter follower, the maximum negative voltage swing across the load is limited by RE resistance value and DC voltage at emitter.



The RE and RL form a voltage divider so that the maximum negative voltage across is equal to:

VL = Veq * RL/(RL + RE) = Ieq*(RL||RE)

Where

Veq is a DC voltage across the emitter resistor.
Ieq - quiescent emitter current.

More can be fiand here
https://forum.allaboutcircuits.com/...single-ended-follower-amplifier-pg-91.120993/
 

Thread Starter

Zeeus

Joined Apr 17, 2019
600
In the emitter follower, the maximum negative voltage swing across the load is limited by RE resistance value and DC voltage at emitter.



The RE and RL form a voltage divider so that the maximum negative voltage across is equal to:

VL = Veq * RL/(RL + RE) = Ieq*(RL||RE)

Where

Veq is a DC voltage across the emitter resistor.
Ieq - quiescent emitter current.

More can be fiand here
https://forum.allaboutcircuits.com/...single-ended-follower-amplifier-pg-91.120993/
Thanks..
https://forum.allaboutcircuits.com/...tter-amplifier-using-bjt.152628/#post-1310586

in that post, why did you choose output capacitor that way?
Rout = Rc = 940 ohms is bigger than 600 ohms... load voltage should be less than 1/2 Vc i'm guessing(the capacitor avoids this?)


oh and thanks for the clipping link..

The capacitor changes the time varying positive voltage to AC voltage.

If you were to get 2/3 of 5.3 volts peak-to-peak, for example, you would need 1.76 amps from the positive supply (+12V through the emitter) and 1.76 amps from the negative supply (OV from ground ground through the resistor).

With 5.1 k for the emitter load the maximum negative-going current you could get is 12V/5.1k = 2.4 ma, so the signal clips. Since your signal does not get to +12V the current is even lower.

Slightly different view:
In your circuit with those values the capacitor just charges up and little current passes through it because it is already charged with the + end toward the positive supply and only a high resitance discharge path for the signal, so not much current is going to flow.

Try making your emitter resistor lower and see what happens.
Never really thought of the capacitors that way (charging and discharging), only thought of it as to keep the dc bias....Nice way to think of it

Although I do not understand your math :( but suppose there is a collector resistor(Rc) and Rc + Rl = Re : Capacitor charges and discharges same rate?

Change R2 to 10K, R3 to 560 ohms..
lol...later this morning..

Thinking you chose R2 as 10k to centre the base then assumed beta as 100 and chose R3 = (5k * 10)/100 (keeping it small so output cap can discharge during negative voltage across load)

Correct?
 
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Jony130

Joined Feb 17, 2009
5,186
in that post, why did you choose output capacitor that way?
Bewcouse I can. Normally Cout capacitance forms a high pass filter together with Rc and RL.
And the corner frequency is Fc = 1/(2*pi*Cout*(RC+RL)) ≈ 0.16/( Cout * (Rc+RL) ) and normally we choose Cout using this equation:
Cout ≈ 0.16/(Fc * (Rc + RL)) ≈ 0.16/(Fc*RL)
Where Fc is a corner frequency, the lowest frequency we want to amplifier.
http://www.learnabout-electronics.org/Amplifiers/amplifiers22.php

Never really thought of the capacitors that way (charging and discharging),
But you should.
Try read this
http://forum.allaboutcircuits.com/threads/voltage-divider-bias.71104/ (especially carefully read the panic mode post #3)
 

DickCappels

Joined Aug 21, 2008
6,642
This is easier to picture if you consider that because of the low value of the load resister the negative end of the capacitor "nearly" grounded.

The reason I referred to the capacitor charging is because the emitterl can charge the positive end up from ground but the emitter resistor cannot discharge it very much during the negative half cycle.

In real operation when frequencies below the cutoff frequency Jony130 refered to the positive end of the capacitor will charge up to approximately the peak voltage of the audio signal. With the high value emitter resistor your amplifier turns into a peak detector.
 
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