A circuit question on emitter amplifiers

[zalzon]

Shown below is a 2 stage emitter amplifier.

I am unable to figure out why that capacitor is placed in the middle
between the two stages. Also how did they get 2250 for the load
resistance? Please help me someone.

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

 

[Jonathan Kirwan]

On Sat, 26 Jul 2003 22:32:52 GMT, zalzon
<zalzon@zalzonishappy.com> wrote:

Shown below is a 2 stage emitter amplifier.

I am unable to figure out why that capacitor is placed in the middle
between the two stages. Also how did they get 2250 for the load
resistance? Please help me someone.

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

It's called "ac coupled." You can DC couple amplifier stages,
too. But it can eat up your supply voltage headroom pretty
fast. You want to recenter the average voltage at the collector
of the stage's transistor to about midway, so what they are
doing with the AC coupling is to allow the signal oscillations
to pass through the capacitor while blocking the steady DC level
of the 1st stage output. This makes it easier to set the input
bias point for the next stage and to better isolate each stage's
design and allow the recentering of this signal average at the
output back to about midway, again.

The load resistance seen by the 1st stage is approximately (not
exactly) looking at the 1st stage collector resistor in parallel
with each of the two input biasing resistors of the 2nd stage.
That works out to 2330 ohms. It's a little lower than that
because the emitter resistor of the 2nd stage, multipled by the
beta, is also in parallel. So assuming a beta of 100, this
means 2330 in parallel with 68k (100*680), which works out to
2250 ohms.

Also, it's typical to use a larger emitter resistor to ground,
in parallel with another series resistor+capacitor, to set the
AC gain while lowering the required DC quiescent current. But
that's not in the diagram you mentioned.

I'm a hobbyist, not a designer. This is *not* something I'm
practiced at, so take all this with a grain of salt.

Jon

 

[zalzon]

On Sat, 26 Jul 2003 23:24:10 GMT, Jonathan Kirwan
<jkirwan@easystreet.com> wrote:

It's called "ac coupled." You can DC couple amplifier stages,
too. But it can eat up your supply voltage headroom pretty
fast.

What does 'eat up your supply voltage headroom' mean? Please I'm a
novice.

You want to recenter the average voltage at the collector
of the stage's transistor to about midway,

I assume this is the Q point (quiet point). I read about it but could
not understand what its significance was. What really does it
represent?

so what they are
doing with the AC coupling is to allow the signal oscillations
to pass through the capacitor while blocking the steady DC level
of the 1st stage output.

Now these are the two most important questions I have :

1) Why do they want to block DC?

2) Would not the first capacitor on the extreme left already be
blocking the DC from the power supply? Why the need to put another
one in the center and at the load end?

This makes it easier to set the input
bias point for the next stage and to better isolate each stage's
design and allow the recentering of this signal average at the
output back to about midway, again.

What happens if the capacitor is not there? I cannot seem to get my
head around why it makes it easier to set the bias point. All I
understand is that :

- a capacitor charges up and discharges
- it passes AC but blocks DC

In the context of the above, why/how does a capacitor help to set the
bias point for the next stage?

Sorry I'm all messed up.

 

[zalzon]

Ah. That's an excellent explaination Mr Black. I just need to
clarify one thing :

If AC voltage goes into a transistor's emitter, is the output both DC
and AC at the collector?

I was under the impression that one capacitor at the start is all
that's needed to get rid of the DC for good. That is why I could not
understand why 3 capacitors were needed !

So in conclusion, the capacitors are there to strip off DC everytime
the current goes through a transistor! Is this correct?





On 27 Jul 2003 02:02:07 GMT, et472@FreeNet.Carleton.CA (Michael Black)
wrote:

zalzon (zalzon@zalzonishappy.com) writes:

What happens if the capacitor is not there? I cannot seem to get my
head around why it makes it easier to set the bias point. All I
understand is that :

If the capacitor is not there, then the second stage gets it's bias
from the output of the first stage. This likely is not optimal voltage
for biasing the base of the second stage and the second stage suffers.
Or, bias the first stage so it's output voltage biases the second stage
properly, and the first stage will suffer.

By putting that coupling capacitor in there, you have two stages
that are no longer dependent on each other for DC voltages. You can
bias the first stage for optimal results and not worry about what
the static DC output is. The capacitor will strip off that constant
DC voltage, whatever it is, and only pass the AC output of the first
stage. Then, you can have a separate resistor network to bias the
second stage for optimal operation. The coupling capacitor ensures
that the two stages are isolated for DC voltages.

Another way to look at it is to imagine shorting out the capacitor in the
diagram. Since you have a DC voltage on the collector of the first stage,
obviously the resistor network on the base of the second stage cannot
set the bias point of the second stage, since the DC voltage from the
first stage will mess things up.

Michael

 

[zalzon]

Thanks for your message. Most of it made sense to me except the first
part about the q point. I suspect I need to read about it a little
more for it all to sink in especially the DC stuff at the collector.

Anyway, I loaded up electronic workbench (a software to simulate
electronic circuits) and rebuilt your schematic of a circuit. I used
the oscilloscope in the software to check out the amplification. It
works. Compared to the source, the signal coming out the collector
was amplified.

I then built on a second stage exactly like the first (with the
capacitor inbetween) and it was amplified further. I checked the
current at various places using the multimeter in the software..etc

I also was able to followed your calculations quite allright.

Thanks once again!