2
\$\begingroup\$

The emitter follower bellow cannot follow the base voltage Vb after some point:

enter image description here

Vin input is DC sweeping from +10V down to -10V.

The output voltage Vout however stops following Vb at some particular voltage and settles as in the below plot:

enter image description here

Why is that happening?

edit:

Here is my latest understanding step by step fashion(let me know if it is wrong):

Starting the scanario while the transistor is on active region...

As long as the current flows from Vcc to both "Vee and GND" the emitter voltage follows the base voltage.

But decreasing the base voltage gradually will decrease the base current.

This will in turn will gradually decrease the "Ic = beta*Ib" (current passing through the tarnsistor down to -12V terminal and GND).

There is a moment where the base current approaches to zero so Ic approaches to zero.

At this point the transistor starts going to "cut-off region" i.e. stops passing any current.

At that moment the transistor starts shutting-off leaving almost no current through collector to emitter.

Now the current starts to flow from GND to -12 where Rload and Re forms a voltage divider.

The emitter voltage starts to settle to around -2V.

After this point lowering the base voltage even more negative than zero to -1.2V, will start to reverse bias the base-emitter junction.

This is beacuse the emitter voltage is settled to -2V and the transitor is off.

The emitter follower rules doesnt hold anymore.

If the load wouldn't be there the base voltage wouldnt be limited to -2V.

\$\endgroup\$
5
  • \$\begingroup\$ Could it be that V3 is the wrong way around? \$\endgroup\$
    – JIm Dearden
    Commented Nov 27, 2016 at 19:32
  • \$\begingroup\$ No the transistor is under a a dual supply. Vee = -12V Vcc = +12V. \$\endgroup\$
    – user16307
    Commented Nov 27, 2016 at 19:34
  • 1
    \$\begingroup\$ Looking at V3 it shows the positive terminal connected to Re and negative terminal to ground. Whereas V2 shows the positive terminal to Q1 collector and its negative terminal to ground. In my book that means V3 has been connected the wrong way around. \$\endgroup\$
    – JIm Dearden
    Commented Nov 27, 2016 at 19:54
  • 1
    \$\begingroup\$ @JimDearden: He's declared V3 as -12V, rather than placing it with the negative terminal up, and declaring it as +12V. Apparently an attempt to confuse the initiated. \$\endgroup\$
    – Peter Bennett
    Commented Nov 27, 2016 at 20:26
  • 1
    \$\begingroup\$ "At this point the transistor saturates i.e. stops passing any current." This is wrong. Transistor enters into Cut-off region not a saturation. Saturation is when base voltage is larger than collector voltage. So the Ic current reach the maximum value. \$\endgroup\$
    – G36
    Commented Nov 28, 2016 at 19:25

4 Answers 4

Reset to default
4
\$\begingroup\$

The current in the transistor cannot go below zero although it can go to a very high positive current.

When the input is going in the negative direction the current in the transistor reduces until it reaches zero. If the input goes further negative the base of the transistor is reverse biased the transistor current remains at zero - it cannot drive a negative current into the load.

When the current in the transistor reaches zero the only current driving the load is that provided by Re and the -12V supply. This will provide about 2mA into the load.

In the positive direction the current in the transistor will keep increasing until the emitter of the transistor almost reaches the positive 12V rail (the input will be slightly above 12V at that point).

You can improve the output in the negative direction by reducing the value of Re - it may need to be significantly lower than the load.

Practical circuits with emitter follower outputs will usually have a complementary output stage with an NPN to provide the positive side and a PNP the negative output. That can then provide a symmetrical output.

For example this is the output stage of an LM358 opamp: enter image description here

\$\endgroup\$
5
  • \$\begingroup\$ But when I remove Rload the follower can follow Vb. Why? \$\endgroup\$
    – user16307
    Commented Nov 27, 2016 at 19:35
  • 1
    \$\begingroup\$ When you remove Rload, you've removed the voltage divider. The emitter would only be connected to the 4K7 resistor to -12V, so can be pulled all the way to -12V by that resistor. \$\endgroup\$
    – Peter Bennett
    Commented Nov 27, 2016 at 20:28
  • \$\begingroup\$ @PeterBennett I still dont understand why the voltage at the emitter cannot go below some certain voltage. \$\endgroup\$
    – user16307
    Commented Nov 28, 2016 at 14:33
  • \$\begingroup\$ is there a way to explain this step by step? emitter voltage goes to a certain point and then something prevents it to go further negative. could you explain it step by step in order of things happening? im stuck with it \$\endgroup\$
    – user16307
    Commented Nov 28, 2016 at 15:29
  • \$\begingroup\$ please see my edit and see if it is ok. i re-read what you wrote and tried to describe in a more explicit way. \$\endgroup\$
    – user16307
    Commented Nov 28, 2016 at 19:20
11
\$\begingroup\$

The emitter follower can only pull the output voltage up (positive), it can't push the output down.

Rload and Re form a voltage divider between Ground and -12V. If the transistor wasn't there, the output voltage produced by the Re/Rload divider would be about -2.1 volts, and that is what your graph shows. When the base voltage goes below about -1.5 volts, the transitor will be cut off, and will have no effect on the output voltage.

\$\endgroup\$
5
\$\begingroup\$

It might be easier for you to understand why output voltage is clipping at -2.1V if we redraw the circuit and change the reference point (GND) position.

enter image description here

As you can see I moved the GND and I hope that now the the situation is much more clearer. And now we need to ask yourself on simple question: What voltage at transistor base (VB) is needed to open transistor? To turn-ON a BJT the voltage at the base (VB) must be about 0.6 volts larger than emitter voltage.

What is the emitter voltage in this case ?

Well the Rload and RE form a voltage divider therefore

$$V_E = V_{EE} *\frac{R_E}{R_E + R_L} = 12V * \frac{4.7k\Omega}{4.7k\Omega+ 1K\Omega} = 9.894V $$

So, any voltage at the base larger than 0.6V + 9.894V will Turn-ON the transistor.

All of this means that the minimum voltage at the load resistance is equal to:

VEE - VE = 12V - 9.894V = 2.1V

Because the BJT is in Cut-off region. So, BJT is no longer able to provide any current to the load resistance.

\$\endgroup\$
0
3
\$\begingroup\$

Only a small addition to the excellent G36 response.

If you want the output swing on the entire voltage range, the load should be placed appropriately:

Your Circuit
Your circuit.

Better load placement
Better load placement.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.