Author Topic: Emitter Follower  (Read 1469 times)

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Offline AztlanpzTopic starter

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Emitter Follower
« on: October 15, 2015, 01:54:40 am »
I am having some problem with the circuit below (Fig A) and a output of (Fig B); that output is corrected right .  But when I make the emitter negative (see Fig C) I get this output see Fig D.  That is incorrect Right?


Aztlan
 
« Last Edit: October 15, 2015, 01:56:16 am by Aztlanpz »
 

Offline vk6zgo

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Re: Emitter Follower
« Reply #1 on: October 15, 2015, 03:13:43 am »
In Fig (a) your transistor base is not positively biased with respect to the emitter,so you get no output,except during the most positive part of the input sinewave,so Fig (b) is correct.

In Fig (c) you have connected the emitter to a negative supply.
Your transistor base is positively biased with respect to the emitter,so you get an output,which is of similar amplitude & in phase with the input sinewave,so Fig (e) is correct.

 

Offline alsetalokin4017

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Re: Emitter Follower
« Reply #2 on: October 15, 2015, 04:42:08 am »
It is not even necessary to have a power supply connected at all to obtain the "Fig. B" trace. See the screenshot below, which is the Fig. A circuit with _nothing_ connected to the Collector of the 2n3904.
« Last Edit: October 15, 2015, 04:46:58 am by alsetalokin4017 »
The easiest person to fool is yourself. -- Richard Feynman
 

Offline SteveLy

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Re: Emitter Follower
« Reply #3 on: October 15, 2015, 05:03:16 am »
That all looks good.

In the first circuit your transistor is switched off when the base voltage is less than the B-E diode voltage of the transistor. So you only get an output signal for positive voltages above that.

In the second circuit the transistor is already switched on when there is no signal (0V) on the input because current can flow from 0V at the base to -15V past the emitter and R1. Here your input signal is no longer switching the transistor on and off, but it is always switched on. The input signal only increases and decreases the base current, which increases/decreases the emitter current and hence voltage drop across R1 in proportion.
 


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