Beginner question

[Slavko Vorkapitch]

Hi - I am reading up on small signal models where for example a shunt
capacitor is placed across the resistor to ground at the emitter end
of a basic class A amplifier - so I understand the desireability of
eliminating unwanted AC signals but wouldn't the connection to ground
eliminate ALL signals - both DC and AC without a shunt capacitor ?
Maybe this is too mathematically simple ..

Thanks in advance,
El Squid

 

[John Fields]

On 4 Oct 2003 09:13:33 -0700, el_squid_2000@yahoo.com (Slavko
Vorkapitch) wrote:

Hi - I am reading up on small signal models where for example a shunt
capacitor is placed across the resistor to ground at the emitter end
of a basic class A amplifier - so I understand the desireability of
eliminating unwanted AC signals but wouldn't the connection to ground
eliminate ALL signals - both DC and AC without a shunt capacitor ?
Maybe this is too mathematically simple ..

---
When current flows through the unbypassed emitter resistor, a voltage is
developed across it which reduces the base-to-emitter voltage
differential, thereby reducing (but not eliminating) the collector
current. When the emitter resistor is bypassed with a capacitor, the
reactance of the capacitor decreases as frequency increases, the effect
being to lower the value of the emitter resistance (impedance) as
frequency increases.

--
John Fields

 

[John Fortier]

"Slavko Vorkapitch" <el_squid_2000@yahoo.com> wrote in message
news:e0a5b916.0310040813.45143488@posting.google.com...

Hi - I am reading up on small signal models where for example a shunt
capacitor is placed across the resistor to ground at the emitter end
of a basic class A amplifier - so I understand the desireability of
eliminating unwanted AC signals but wouldn't the connection to ground
eliminate ALL signals - both DC and AC without a shunt capacitor ?
Maybe this is too mathematically simple ..

Thanks in advance,
El Squid

Capacitors have an impedance (read resistance in this case) which is
inversely proportional to the frequency in question. All capacitors have a
theoretically infinite resistance at DC and at DC the emitter resistance
will simply be the value of the emitter resistor plus the intrinsic
resistance of the emitter itself.

The impedance of a capacitor is determined by the formula Xc = 1/2*Pi*F*C,
so it can be seen that with F at zero, the impedance will be infinite.

As the frequency increases, the impedance of the bypass capacitor decreases,
following the above formula, until it reaches a stage where it is
insignificant compared to the intrinsic resistance of the emitter. At this
frequency the AC gain of the amplifier is greatest.

The DC conditions, however set the bias on the transistor, which, together
with collector resister and the biasing on the base, will determine the DC
voltage of the collector. Without the emitter resistor the biasing on the
amplifier would be dependent on the emitter intrinsic resistance, which is
very variable.

Try http://ourworld.compuserve.com/homepages/g_knott/. for basics which
will explain all of this, probably a lot better than I can.

Regards

John

 

[Robert Monsen]

"Slavko Vorkapitch" <el_squid_2000@yahoo.com> wrote in message
news:e0a5b916.0310040813.45143488@posting.google.com...

Hi - I am reading up on small signal models where for example a shunt
capacitor is placed across the resistor to ground at the emitter end
of a basic class A amplifier - so I understand the desireability of
eliminating unwanted AC signals but wouldn't the connection to ground
eliminate ALL signals - both DC and AC without a shunt capacitor ?
Maybe this is too mathematically simple ..

Thanks in advance,
El Squid

I'm having trouble understanding what you are asking.

Are you asking us to explain why a common emitter amplifier without an
emitter resistor still amplifies?

If you want the reason, based on some rules of thumb (which are based on the
physics), the answer is easy:

let Zc = 1/(2.PI.F.C.j), where F is the signal frequency.
this is the impedance of the cap.

then if Re is parallel to Zc, then Rt = Re||Zc = (Re.Zc)/(Re + Zc)

You can see that if Zc gets small, (which it will when the frequency goes
up, since its inversely proportional to F) then Rt goes down.

Since the gain of the amp is Rc/(re + Rt), that higher frequency makes the
gain go up. re is just the intrinsic emitter resistance, which is 25/Ic mA.

If there isn't any Rc, then just set that to zero in the above equations,
and note that the gain goes up, since then Rt = 0, so G = Rc/re.

This is all based on a transistor model called the ebers-moll equation,
which you can search for if you need more information.

Regards,
Bob Monsen

 

[Brian]

Slavko Vorkapitch wrote:

Hi - I am reading up on small signal models where for example a shunt
capacitor is placed across the resistor to ground at the emitter end
of a basic class A amplifier - so I understand the desireability of
eliminating unwanted AC signals but wouldn't the connection to ground
eliminate ALL signals - both DC and AC without a shunt capacitor ?
Maybe this is too mathematically simple ..

Thanks in advance,
El Squid

Roughly, the current gain of a transistor class A amplifier, is RB /
RE+h. With a capacitor across RE, the gain of the stage goes up because
now the emitter circuit effectively no longer has RE (the current gain is
now only limited by the current gain of the transistor that you are
using). While this increases the current gain of the stage, it also
decreased the linearity of the stage.
Roughly, the voltage gain of the stage is RC/RE+h. So as RE goes down,
voltage gain goes up.
RB = Base resistor
RE = Emitter resistor
RC = Collector resistor
h = 26 / emitter current in milliampers

With RE not having a capacitor across it, the gain of the stage is very
much effected by the current gain of the transistor. Without a capacitor
across the emitter resistor, the stage gain is controlled primarily by
the resistors.
Hope this helps,
Brian


--
Building user friendly software for electronics.
http://www.fncwired.com

 

[Brian]

Brian wrote:

Slavko Vorkapitch wrote:

Hi - I am reading up on small signal models where for example a shunt
capacitor is placed across the resistor to ground at the emitter end
of a basic class A amplifier - so I understand the desireability of
eliminating unwanted AC signals but wouldn't the connection to ground
eliminate ALL signals - both DC and AC without a shunt capacitor ?
Maybe this is too mathematically simple ..

Thanks in advance,
El Squid

Roughly, the current gain of a transistor class A amplifier, is RB /
RE+h. With a capacitor across RE, the gain of the stage goes up because
now the emitter circuit effectively no longer has RE (the current gain is
now only limited by the current gain of the transistor that you are
using). While this increases the current gain of the stage, it also
decreased the linearity of the stage.
Roughly, the voltage gain of the stage is RC/RE+h. So as RE goes down,
voltage gain goes up.
RB = Base resistor
RE = Emitter resistor
RC = Collector resistor
h = 26 / emitter current in milliampers

With RE not having a capacitor across it, the gain of the stage is very
much effected by the current gain of the transistor. Without a capacitor
across the emitter resistor, the stage gain is controlled primarily by
the resistors.
Hope this helps,
Brian

--
Building user friendly software for electronics.
http://www.fncwired.com

The last paragraph should have read:

With RE having a capacitor across it, the gain of the stage is very much
effected by the current gain of the transistor. Without a capacitor across
the emitter resistor, the stage gain is controlled primarily by the
resistors.


--
Building user friendly software for electronics. http://www.fncwired.com