in the art of electronics chapter 2, section 2.03 H&H go on to explain emitter followers i pretty mnuch understood all the initial stuff but havnt gone beyound the topic- Important points about followers.
here a 10 V sine input is provided and the output is also graphed.. clippin is shown to occur at 5 V.. the clippin occurs due to cutoff i presume.. but i cant seem to understand how the transistor is in cutoff.. the base emitter is still forward biased..

i was hoping someone here can provide me with an explaination.
thank you
Aditya

 

Originally posted by haditya@Dec 18 2005, 06:44 AM
in the art of electronics chapter 2, section 2.03 H&H go on to explain emitter followers i pretty mnuch understood all the initial stuff but havnt gone beyound the topic- Important points about followers.
here a 10 V sine input is provided and the output is also graphed.. clippin is shown to occur at 5 V.. the clippin occurs due to cutoff i presume.. but i cant seem to understand how the transistor is in cutoff.. the base emitter is still forward biased..

i was hoping someone here can provide me with an explaination.
thank you
Aditya

[post=12537]Quoted post[/post]​

I assume you are refering to Figure 2.8 located on page 66 of the book.

If you look closely at the circuit in Figure 2.8, you will see that the emitter of the NPN transistor is tied to the -10V power source through a 1K ohm resistor. The clipping is due to the existence of the 1K ohm load resistor that is tied to ground. The two 1K ohm resistors form a voltage divider. Once the transistor's base voltage reaches minus 4.4 volts, the base-emtter junction begins to enter the reverse bias condition. At this point any further negative travel of the input voltage can no longer influence the voltage at the emitter. Since the two reistors are equal, the resulting voltage divider sets the negative most limit of the emitter voltage at minus 5 volts.

hgmjr

 

Same thing ... different words ...

The emitter resistor and load form a voltage dividing network splitting the negative 10 volts in half.

For the transistor to be forward biased, the base must be more positive than the emitter ... by .6 volts in this case ... the base must be Ee + .6 or neg 5 +.6 = neg 4.4

So, once the base is less than Ee + .6, the transistor cuts off.

 

the voltage divider analysis is wat i missed..thx a lot guys