Another transistor question

[zalzon]

Please take a look here. Tell me if you can why putting an emitter
bypass capacitor on a common emitter amplifier makes the signal the
way it is. (i.e. amplified and clipped). Exactly why does sending
the AC to ground at the emitter increase the gain?

http://www.anycities.com/user/msc/normal.html

thank u in advance to anyone who helps me understand this.

 

[John Popelish]

zalzon wrote:

Please take a look here. Tell me if you can why putting an emitter
bypass capacitor on a common emitter amplifier makes the signal the
way it is. (i.e. amplified and clipped). Exactly why does sending
the AC to ground at the emitter increase the gain?

http://www.anycities.com/user/msc/normal.html

thank u in advance to anyone who helps me understand this.

The emitter resistor develops an AC voltage component (in addition to
its DC drop) that subtracts from the base to emitter input signal.
This subtraction is a form of negative feedback that reduces the
effective stage gain of the transistor.

Bypassing the AC current that would have passed through the emitter
resistor by providing a very low impedance capacitive path, reduces
the voltage this AC current produces at the emitter, so much more of
the input signal voltage appears across the base emitter junction.
This larger input signal produces a larger output signal, to the
point, in this case, of saturating the transistor on or off (or both)
at the peaks of the output signal.

Put an additional resistor in series with the bypass capacitors, and
you can adjust the effect between no bypass and full bypass, and get
any intermediate effect you want.

--
John Popelish

 

[zalzon]

On Sun, 03 Aug 2003 20:08:09 GMT, John Popelish <jpopelish@rica.net>
wrote:

Bypassing the AC current that would have passed through the emitter
resistor by providing a very low impedance capacitive path, reduces
the voltage this AC current produces at the emitter, so much more of
the input signal voltage appears across the base emitter junction.

Is this because of the phases of the signal waves. i.e. the phase of
the signal wave in the base is out of step with the phase of the
signal wave at the emitter-base (without the bypass capacitor). Do
the waves cancel/muddle each other out thereby reducing the signal
voltage at base?

Does my description below accurately portray what's happening :

A bypass capacitor is putting the AC signal at the emitter at ground
level. So the difference in potential between base and emitter
(ground) makes the signal voltage appear larger.

This greater signal voltage than makes more or less DC current to flow
through from emitter to collector thereby amplifying the signal
voltage even further.

Is that all correct? I think I would be devastated if i saw the
word... NO :0

 

[John Jardine]

zalzon <zalzonishappy@thisisnotzalzon.com> wrote in message
news:44pqiv80lj2sigvi9u1leb9f11hhiaj3kl@4ax.com...

Please take a look here. Tell me if you can why putting an emitter
bypass capacitor on a common emitter amplifier makes the signal the
way it is. (i.e. amplified and clipped). Exactly why does sending
the AC to ground at the emitter increase the gain?

http://www.anycities.com/user/msc/normal.html

thank u in advance to anyone who helps me understand this.

No emitter cap' and the emitter voltage follows sympathetically whatever

voltage is fed to the base.
This emitter voltage is across the emitter resistor,
Therefore the emitter current also follows in sympathy.
As the emitter current is pretty much the collector current, then the
collector voltage must also follow suit.
As the collector resistor is a bigger resistance than the emitter resistor
then the voltage across the collector resistor must be a bigger version of
the emitter voltage. You've already fixed the current (a.c.) that can flow
down the transistor by using the emitter resistor, so a value of collector
resistor can be chosen to give whatever voltage gain you want (or even
voltage loss).

Add an emitter cap' and you lose the ability to develop a voltage across the
external emitter resistor *but* you still have a small resistance inside the
transistor at work. This resistor value is = 25 / emitter ma's.
(In your case, the quiescent transistor emitter current is set about 0.45ma,
so this internal resistor is 25 / 0.45 =56ohm.)
Thus the input signal voltage across this internal emitter resistance sets
the emitter current and hence the collector current and in turn the colector
voltage and overall voltage gain.

So ...

Without an emitter cap' your circuit has a fluctuating emitter current of
100mVac / 1500ohms = 67uA. This turns up across the 10k collector resistor
as a fluctuating voltage of 67uA X 10kohms = 0.67Vac. Thus the stage has a
voltage gain of 0.67Vac / 100mVac =6.7 times.

With an emitter cap' the fluctuating emitter current is 100mVac / 55ohms =
1.8ma. This tries to force a fluctuating voltage across the 10k collector
resistor of 1.8ma X 10k = 18Vac!. Thus the voltage gain is massively higher
(consequently clipped at the rails), as can be seen on the scope display.

There's other knock on effects caused by this high gain and in practice it's
better to leave some undecoupled external emitter resistance.
Essentially, go for quality rather than quantity.

regards
john

 

[zalzon]

Thanks john your reply cleared up some issues. Its starting to make
sense.

Who would have thought learning electronics could be so complex

 

[zalzon]

Thank you all for your reply, I'm going through each message twice
trying to follow the examples & calculations laid out.

You people must all be pros cos you seem to understand it like the
back of your hands.

Thanks once again.

 

[zalzon]

On Mon, 04 Aug 2003 12:13:56 GMT, Tom MacIntyre
<tom__macintyre@hotmail.com> wrote:

Presumably the math was the primary reason, because some IT

No its not the math at all. Its the lack of clear, proper
explainations of what's going on in a circuit.

Sometimes when i read explainations, i get the impression the author
of a textbook/message is talking to himself. He has lost conciousness
of the fact that he's explaining it to a beginner.

If a beginner could understand all that techno bable on electronics,
he would not be asking for an explaination in the first place.

Too many instructors/authors just fly over the essential or difficult
points assuming its all obvious when its obviously not. It may be
obvious after spending 3 yrs in the field but its not to a beginner.

 

[A E]

zalzon wrote:

On Mon, 04 Aug 2003 12:13:56 GMT, Tom MacIntyre
tom__macintyre@hotmail.com> wrote:

Presumably the math was the primary reason, because some IT

No its not the math at all. Its the lack of clear, proper
explainations of what's going on in a circuit.

Sometimes when i read explainations, i get the impression the author
of a textbook/message is talking to himself. He has lost conciousness
of the fact that he's explaining it to a beginner.

If a beginner could understand all that techno bable on electronics,
he would not be asking for an explaination in the first place.

Too many instructors/authors just fly over the essential or difficult
points assuming its all obvious when its obviously not. It may be
obvious after spending 3 yrs in the field but its not to a beginner.

You want clear explanations? Head for the used book store, and find books from
the golden era of electronics hobbyists: the 1960s. Understanding electronics,
for example, the Schaum outline series. I think that most books these days exist
mainly to make money through university sales, but back then, it was for the
passion of the thing.