In playing with my simple emitter follower circuit trying to figure
out why there's very faint output when connected to my 8 Ohm speakers,
I came upon some important realizations.

My circuit is a simple emitter follower with one transistor. The base
is biased to midpoint through a voltage divider and the input is
capacitively coupled. The output is also capacitively coupled through
a large capacitor to block DC. The emitter resistor is 5K.

I expected that with a large capacitor, even an 8 Ohm speaker would be
suitable to place the corner frequency of the high pass filter low
enough to pass audio frequencies and high enough to block DC. However
I was faced with two anomalies. Whenever I connect my speaker after
the capacitor, I hear a very faint and distorted sound. Whenever I
connect the speaker before the capacitor (the NPN's emitter), I get an
almost exact replica of the input (which is what the emitter follower
is supposed to do). Simulation confirmed this behavior but I was
puzzled with the result. Do I throw out of the window all I know about
high pass filters and frequency response. Also do I throw out the
importance of bias? Here we have a high pass filter that is not acting
like a high pass filter. And we have a low resistance emitter resistor
(after connecting the speaker in parallel with the 5K emitter
resistor) that should throw the bias point way off midpoint.

After playing with simulation for a while, and replacing the
transistor with equivalent circuit, etc, I hit upon one of my biggest
realizations in electronics (mind me if it's too obvious for you). The
models are correct so long as the transistor is ON and operating in
its linear region. In such a case the high pass filter will act nicely
and block DC and pass the audio signal. However, with such low
resistance, If the output to follow the input faithfully, huge current
will have to pass in the capacitor for the slightest negative output
voltage, and since the current in the emitter resistor (5 KOhm) is
very small compared, this current would have to follow in the
transistor in the opposite direction, so the transistor turns off.
That's my reasoning for why the transistor would turn off.

So now we have another mode of operation. Now it's another circuit.
It's not a high pass filter but a capacitor discharging through a
resistor. At a certain point the capacitor charge will be large enough
to have the transistor turn on only very briefly and the output
voltage change so little.

What caused my confusion is that in most electronic circuits, there
are transistors but often the circuit description doesn't tell if that
transistor will be ON all the time or not. Then analysis follows that
replaces caps with shorts at signal frequencies etc. But that's valid
only if the transistor stays on and hence acts as a linear device. I
wonder if circuit designers even with all their experiences and rules
of thumb take the extra step of analysis (or simulation) to make sure
the transistor is always operating in the linear region. If they do,
it's often not mentioned to us, less experienced people. In other
applications, such as rectifiers, the charging and discharging is
explicitly mentioned and used. This dual role of the capacitor always
puzzled me, when to think of it as a charging/discharging device, and
when to think in terms of frequency response and frequency dependent
impedance.

The other question was why connecting the speaker directly to the
output is OK. Why no blocking capacitor is needed? It turned out that
the small impedance shifts the DC bias point very close to ground, and
hence the output is a replica of the input but shifted up in DC level
just a few millivolts above ground, which seems to not alter the
speaker response much. That seems to be a problem if the gain is high
and clipping will start to happen as you move away from midpoint, but
when the gain is 1, no problem since the input signal is already a few
millivolts around 0.

A possibility here is that the speaker also responds to changes in
magnetic field and hence the DC component wouldn't matter, but i'm not
sure about that one.

I apologize for anyone who took the effort and went through my long
post. Am I right in my conclusions?

Thank you.

M. Hamed wrote:
> In playing with my simple emitter follower circuit trying to figure
> out why there's very faint output when connected to my 8 Ohm speakers,
> I came upon some important realizations.
>
> My circuit is a simple emitter follower with one transistor. The base
> is biased to midpoint through a voltage divider and the input is
> capacitively coupled. The output is also capacitively coupled through
> a large capacitor to block DC. The emitter resistor is 5K.
>
> I expected that with a large capacitor, even an 8 Ohm speaker would be
> suitable to place the corner frequency of the high pass filter low
> enough to pass audio frequencies and high enough to block DC. However
> I was faced with two anomalies. Whenever I connect my speaker after
> the capacitor, I hear a very faint and distorted sound. Whenever I
> connect the speaker before the capacitor (the NPN's emitter), I get an
> almost exact replica of the input (which is what the emitter follower
> is supposed to do). Simulation confirmed this behavior but I was
> puzzled with the result. Do I throw out of the window all I know about
> high pass filters and frequency response.

Not at all. You just have to remember that those concepts
apply to linear circuits, and your circuit effectively has a
rectifier in series with the output (the base emitter junction).


> Also do I throw out the
> importance of bias? Here we have a high pass filter that is not acting
> like a high pass filter. And we have a low resistance emitter resistor
> (after connecting the speaker in parallel with the 5K emitter
> resistor) that should throw the bias point way off midpoint.
>
> After playing with simulation for a while, and replacing the
> transistor with equivalent circuit, etc, I hit upon one of my biggest
> realizations in electronics (mind me if it's too obvious for you). The
> models are correct so long as the transistor is ON and operating in
> its linear region. In such a case the high pass filter will act nicely
> and block DC and pass the audio signal. However, with such low
> resistance, If the output to follow the input faithfully, huge current
> will have to pass in the capacitor for the slightest negative output
> voltage, and since the current in the emitter resistor (5 KOhm) is
> very small compared, this current would have to follow in the
> transistor in the opposite direction, so the transistor turns off.
> That's my reasoning for why the transistor would turn off.
>
> So now we have another mode of operation. Now it's another circuit.
> It's not a high pass filter but a capacitor discharging through a
> resistor. At a certain point the capacitor charge will be large enough
> to have the transistor turn on only very briefly and the output
> voltage change so little.

Yes, the rectifier has switched the transistor out pf the
output circuit.

> What caused my confusion is that in most electronic circuits, there
> are transistors but often the circuit description doesn't tell if that
> transistor will be ON all the time or not. Then analysis follows that
> replaces caps with shorts at signal frequencies etc. But that's valid
> only if the transistor stays on and hence acts as a linear device.

That is incremental AC analysis or operation near a bias
point. For many circuits, including this one, you have to
check the extremes of the signal swing, to make sure the
operating bias has not severely shifted.

> I
> wonder if circuit designers even with all their experiences and rules
> of thumb take the extra step of analysis (or simulation) to make sure
> the transistor is always operating in the linear region. If they do,
> it's often not mentioned to us, less experienced people. In other
> applications, such as rectifiers, the charging and discharging is
> explicitly mentioned and used. This dual role of the capacitor always
> puzzled me, when to think of it as a charging/discharging device, and
> when to think in terms of frequency response and frequency dependent
> impedance.

Linear versus nonlinear circuits. Filters involve linear
assumptions. Charge/discharge processes involve nonlinear
processes. This changes the rules and the tools. Put away
the paint brushes and get out the socket wrenches, so to speak.

> The other question was why connecting the speaker directly to the
> output is OK. Why no blocking capacitor is needed?

Connecting the speaker directly to the emitter (as a DC
load) adds DC current to the speaker, in addition to the AC
current you intend. This distorts the position of the
speaker cone, adds greatly to its heat production, and also
heats the transistor or distorts its operating point. If
you can find a tolerable compromise for all this, you might
eliminate the output capacitor.

> It turned out that
> the small impedance shifts the DC bias point very close to ground, and
> hence the output is a replica of the input but shifted up in DC level
> just a few millivolts above ground, which seems to not alter the
> speaker response much. That seems to be a problem if the gain is high
> and clipping will start to happen as you move away from midpoint, but
> when the gain is 1, no problem since the input signal is already a few
> millivolts around 0.
>
> A possibility here is that the speaker also responds to changes in
> magnetic field and hence the DC component wouldn't matter, but i'm not
> sure about that one.
>
> I apologize for anyone who took the effort and went through my long
> post. Am I right in my conclusions?

I think your understanding is much better than it was when
you first built the circuit. You may now be in a mental
state to appreciate why the complementary emitter follower
was invented.
(at bottom of:)
http://mysite.du.edu/~etuttle/electron/elect26.htm

--
Regards,

John Popelish

On Sun, 24 Feb 2008 11:21:59 -0800 (PST), "M. Hamed"
<mhs000@gmail.com> wrote:

>In playing with my simple emitter follower circuit trying to figure
>out why there's very faint output when connected to my 8 Ohm speakers,
>I came upon some important realizations.
>
>My circuit is a simple emitter follower with one transistor. The base
>is biased to midpoint through a voltage divider and the input is
>capacitively coupled. The output is also capacitively coupled through
>a large capacitor to block DC. The emitter resistor is 5K.
>
>I expected that with a large capacitor, even an 8 Ohm speaker would be
>suitable to place the corner frequency of the high pass filter low
>enough to pass audio frequencies and high enough to block DC. However
>I was faced with two anomalies. Whenever I connect my speaker after
>the capacitor, I hear a very faint and distorted sound. Whenever I
>connect the speaker before the capacitor (the NPN's emitter), I get an
>almost exact replica of the input (which is what the emitter follower
>is supposed to do). Simulation confirmed this behavior but I was
>puzzled with the result. Do I throw out of the window all I know about
>high pass filters and frequency response. Also do I throw out the
>importance of bias? Here we have a high pass filter that is not acting
>like a high pass filter. And we have a low resistance emitter resistor
>(after connecting the speaker in parallel with the 5K emitter
>resistor) that should throw the bias point way off midpoint.
>
>After playing with simulation for a while, and replacing the
>transistor with equivalent circuit, etc, I hit upon one of my biggest
>realizations in electronics (mind me if it's too obvious for you). The
>models are correct so long as the transistor is ON and operating in
>its linear region. In such a case the high pass filter will act nicely
>and block DC and pass the audio signal. However, with such low
>resistance, If the output to follow the input faithfully, huge current
>will have to pass in the capacitor for the slightest negative output
>voltage, and since the current in the emitter resistor (5 KOhm) is
>very small compared, this current would have to follow in the
>transistor in the opposite direction, so the transistor turns off.
>That's my reasoning for why the transistor would turn off.
>
>So now we have another mode of operation. Now it's another circuit.
>It's not a high pass filter but a capacitor discharging through a
>resistor. At a certain point the capacitor charge will be large enough
>to have the transistor turn on only very briefly and the output
>voltage change so little.
>
>What caused my confusion is that in most electronic circuits, there
>are transistors but often the circuit description doesn't tell if that
>transistor will be ON all the time or not. Then analysis follows that
>replaces caps with shorts at signal frequencies etc. But that's valid
>only if the transistor stays on and hence acts as a linear device. I
>wonder if circuit designers even with all their experiences and rules
>of thumb take the extra step of analysis (or simulation) to make sure
>the transistor is always operating in the linear region.


Good grief, of course we do. It's second nature.


>If they do,
>it's often not mentioned to us, less experienced people. In other
>applications, such as rectifiers, the charging and discharging is
>explicitly mentioned and used. This dual role of the capacitor always
>puzzled me, when to think of it as a charging/discharging device, and
>when to think in terms of frequency response and frequency dependent
>impedance.
>
>The other question was why connecting the speaker directly to the
>output is OK. Why no blocking capacitor is needed? It turned out that
>the small impedance shifts the DC bias point very close to ground, and
>hence the output is a replica of the input but shifted up in DC level
>just a few millivolts above ground, which seems to not alter the
>speaker response much. That seems to be a problem if the gain is high
>and clipping will start to happen as you move away from midpoint, but
>when the gain is 1, no problem since the input signal is already a few
>millivolts around 0.
>
>A possibility here is that the speaker also responds to changes in
>magnetic field and hence the DC component wouldn't matter, but i'm not
>sure about that one.
>
>I apologize for anyone who took the effort and went through my long
>post. Am I right in my conclusions?
>
>Thank you.

The emitter follower can only source current, not sink it. It can't
conduct less than zero current, so there's nothing but the wimpy 5K
resistor available to drive the speaker in the negative direction.

If you reduce the value of the emitter pulldown resistor, then you can
get a little more current into and out of the speaker in both
directions. But that's a losing game, which is why real power amps
usually have push-pull output stages.

Connecting the speaker directly to the emitter works at low levels,
but is inefficient and won't scale.

John

On Sun, 24 Feb 2008 12:50:56 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

>On Sun, 24 Feb 2008 11:21:59 -0800 (PST), "M. Hamed"
><mhs000@gmail.com> wrote:
>
[snip]
>> I
>>wonder if circuit designers even with all their experiences and rules
>>of thumb take the extra step of analysis (or simulation) to make sure
>>the transistor is always operating in the linear region.
>
>
>Good grief, of course we do. It's second nature.
>
[snip]

Who is this "M. Hamed", some fricking amateur (that I've previously
plonked)?

I even utilize macros built into PSpice to check that CMOS operation
is in the saturation region, since it's often non-obvious.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

America: Land of the Free, Because of the Brave

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in message
news:dnl3s3ha53lco7r65cr037bmjkis5deeml@4ax.com...
> On Sun, 24 Feb 2008 11:21:59 -0800 (PST), "M. Hamed"
> <mhs000@gmail.com> wrote:
>
>>In playing with my simple emitter follower circuit trying to figure
>>out why there's very faint output when connected to my 8 Ohm speakers,
>>I came upon some important realizations.

[snip]

>
> The emitter follower can only source current, not sink it. It can't
> conduct less than zero current, so there's nothing but the wimpy 5K
> resistor available to drive the speaker in the negative direction.
>
> If you reduce the value of the emitter pulldown resistor, then you can
> get a little more current into and out of the speaker in both
> directions. But that's a losing game, which is why real power amps
> usually have push-pull output stages.
>
> Connecting the speaker directly to the emitter works at low levels,
> but is inefficient and won't scale.
>
> John
>

This is the key concept ("the emitter follower can only source current...").

Read those paragraphs over and over until you think you understand them.
Then, read them over and over again until you do understand them.

Bob

Jim Thompson wrote:
> On Sun, 24 Feb 2008 12:50:56 -0800, John Larkin
> <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
>
>> On Sun, 24 Feb 2008 11:21:59 -0800 (PST), "M. Hamed"
>> <mhs000@gmail.com> wrote:
>>
> [snip]
>>> I
>>> wonder if circuit designers even with all their experiences and rules
>>> of thumb take the extra step of analysis (or simulation) to make sure
>>> the transistor is always operating in the linear region.
>>
>> Good grief, of course we do. It's second nature.
>>
> [snip]
>
> Who is this "M. Hamed", some fricking amateur (that I've previously
> plonked)?
>
> I even utilize macros built into PSpice to check that CMOS operation
> is in the saturation region, since it's often non-obvious.
>
> ...Jim Thompson
In 40 years of electronics, I have never simulated anything.
Unless you count breadboards as simulation

On Sun, 24 Feb 2008 22:18:19 +0100, Sjouke Burry
<burrynulnulfour@ppllaanneett.nnlll> wrote:

>Jim Thompson wrote:
>> On Sun, 24 Feb 2008 12:50:56 -0800, John Larkin
>> <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
>>
>>> On Sun, 24 Feb 2008 11:21:59 -0800 (PST), "M. Hamed"
>>> <mhs000@gmail.com> wrote:
>>>
>> [snip]
>>>> I
>>>> wonder if circuit designers even with all their experiences and rules
>>>> of thumb take the extra step of analysis (or simulation) to make sure
>>>> the transistor is always operating in the linear region.
>>>
>>> Good grief, of course we do. It's second nature.
>>>
>> [snip]
>>
>> Who is this "M. Hamed", some fricking amateur (that I've previously
>> plonked)?
>>
>> I even utilize macros built into PSpice to check that CMOS operation
>> is in the saturation region, since it's often non-obvious.
>>
>> ...Jim Thompson
>In 40 years of electronics, I have never simulated anything.
>Unless you count breadboards as simulation

Of course you haven't. You also haven't designed a chip with 10,000
transistors on it either

I cut my teeth on paper (and breadboards)... pretty much why I'm so
successful now... and why young bucks ask, "Why'd you do it that way?"

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

America: Land of the Free, Because of the Brave

"M. Hamed" <mhs000@gmail.com> wrote in message
news:6c69d150-9a24-46fb-ba21-d9276afcf86e@q70g2000hsb.googlegroups.com...
> In playing with my simple emitter follower circuit trying to figure
> out why there's very faint output when connected to my 8 Ohm speakers,
> I came upon some important realizations.
>
> My circuit is a simple emitter follower with one transistor. The base
> is biased to midpoint through a voltage divider and the input is
> capacitively coupled. The output is also capacitively coupled through
> a large capacitor to block DC. The emitter resistor is 5K.
>
> I expected that with a large capacitor, even an 8 Ohm speaker would be
> suitable to place the corner frequency of the high pass filter low
> enough to pass audio frequencies and high enough to block DC. However
> I was faced with two anomalies. Whenever I connect my speaker after
> the capacitor, I hear a very faint and distorted sound. Whenever I
> connect the speaker before the capacitor (the NPN's emitter), I get an
> almost exact replica of the input (which is what the emitter follower
> is supposed to do). Simulation confirmed this behavior but I was
> puzzled with the result. Do I throw out of the window all I know about
> high pass filters and frequency response. Also do I throw out the
> importance of bias? Here we have a high pass filter that is not acting
> like a high pass filter. And we have a low resistance emitter resistor
> (after connecting the speaker in parallel with the 5K emitter
> resistor) that should throw the bias point way off midpoint.

In the first case, the cap isn't discharging properly, so you end up with
the input to the cap pegged at Vin(max)-Vbe. This means that the output
transistor can't actually put any power through the cap, because it's
voltage isn't varying.

The way to understand this is to figure out what the impedance is that the
speaker sees when the transistor is on, vs the impedance it sees when the
transistor is off. In the first case, it sees something like 1/10 the input
impedance. In the second case, it sees 5k (since the transistor is off, so
all you see is the emitter resistor).

As a consequence, the circuit acts kinda like a diode, where the cap can be
charged when the transistor is on, but can't be discharged when the
transistor is off. Current in, no current out means it pumps up to the
maximum voltage.

>
> After playing with simulation for a while, and replacing the
> transistor with equivalent circuit, etc, I hit upon one of my biggest
> realizations in electronics (mind me if it's too obvious for you). The
> models are correct so long as the transistor is ON and operating in
> its linear region. In such a case the high pass filter will act nicely
> and block DC and pass the audio signal. However, with such low
> resistance, If the output to follow the input faithfully, huge current
> will have to pass in the capacitor for the slightest negative output
> voltage, and since the current in the emitter resistor (5 KOhm) is
> very small compared, this current would have to follow in the
> transistor in the opposite direction, so the transistor turns off.
> That's my reasoning for why the transistor would turn off.

No, this will happen with a real circuit too. The problem isn't the
simulation, it is the circuit.

>
> So now we have another mode of operation. Now it's another circuit.
> It's not a high pass filter but a capacitor discharging through a
> resistor. At a certain point the capacitor charge will be large enough
> to have the transistor turn on only very briefly and the output
> voltage change so little.
>

Right.

> What caused my confusion is that in most electronic circuits, there
> are transistors but often the circuit description doesn't tell if that
> transistor will be ON all the time or not. Then analysis follows that
> replaces caps with shorts at signal frequencies etc. But that's valid
> only if the transistor stays on and hence acts as a linear device. I
> wonder if circuit designers even with all their experiences and rules
> of thumb take the extra step of analysis (or simulation) to make sure
> the transistor is always operating in the linear region. If they do,
> it's often not mentioned to us, less experienced people. In other
> applications, such as rectifiers, the charging and discharging is
> explicitly mentioned and used. This dual role of the capacitor always
> puzzled me, when to think of it as a charging/discharging device, and
> when to think in terms of frequency response and frequency dependent
> impedance.

You need another transistor to pull down on the cap. The typical output
driver needs both a way to pull up, and a way to pull down.
>
> The other question was why connecting the speaker directly to the
> output is OK. Why no blocking capacitor is needed? It turned out that
> the small impedance shifts the DC bias point very close to ground, and
> hence the output is a replica of the input but shifted up in DC level
> just a few millivolts above ground, which seems to not alter the
> speaker response much. That seems to be a problem if the gain is high
> and clipping will start to happen as you move away from midpoint, but
> when the gain is 1, no problem since the input signal is already a few
> millivolts around 0.

Connecting the speaker directly means you end up with a DC current through
it all the time of (Vcc/2 - Vbe)/R

Where R is the DC resistance of the speaker. Not good. However, the
transistor gets to vary the current on top of the bias, which causes the
magnetic field to vary, causing the paper to move.

>
> A possibility here is that the speaker also responds to changes in
> magnetic field and hence the DC component wouldn't matter, but i'm not
> sure about that one.
>

You are thinking too hard.

> I apologize for anyone who took the effort and went through my long
> post. Am I right in my conclusions?
>
> Thank you.

Try this circuit instead:



-------------------o------------.
| |
| |
.-. |
470 | | |
| | |
'-' |
| |
| |/
o----------| NPN
| |>
V |
- |
|| | | || __ /|
------||--------o o----||--------| | |
|| | | || .--|__| |
V | | \|
1uF - | 10uF |
| |< |
o----------| PNP |
| |\ ===
| | GND
| |
.-. |
470 | | |
| | |
'-' |
| |
------------------o------------'
(created by AACircuit v1.28.6 beta 04/19/05 http://www.tech-chat.de)

It still sucks, but it sucks far less than your circuit.

Filters are interesting, but that isn't your problem here. Notice that when
the input goes down below the midpoint, that causes the NPN to turn off, and
the PNP to turn on. This is also a 'follower' of a kind, it just has a more
balanced impedance in both directions.

Regards,
Bob Monsen

Bob Monsen wrote:

> "M. Hamed" <mhs000@gmail.com> wrote in message
>
> > In playing with my simple emitter follower circuit trying to figure
> > out why there's very faint output when connected to my 8 Ohm speakers,
> > I came upon some important realizations.
> >
> > My circuit is a simple emitter follower with one transistor. The base
> > is biased to midpoint through a voltage divider and the input is
> > capacitively coupled. The output is also capacitively coupled through
> > a large capacitor to block DC. The emitter resistor is 5K.
> >
> > I expected that with a large capacitor, even an 8 Ohm speaker would be
> > suitable to place the corner frequency of the high pass filter low
> > enough to pass audio frequencies and high enough to block DC. However
> > I was faced with two anomalies. Whenever I connect my speaker after
> > the capacitor, I hear a very faint and distorted sound. Whenever I
> > connect the speaker before the capacitor (the NPN's emitter), I get an
> > almost exact replica of the input (which is what the emitter follower
> > is supposed to do). Simulation confirmed this behavior but I was
> > puzzled with the result. Do I throw out of the window all I know about
> > high pass filters and frequency response. Also do I throw out the
> > importance of bias? Here we have a high pass filter that is not acting
> > like a high pass filter. And we have a low resistance emitter resistor
> > (after connecting the speaker in parallel with the 5K emitter
> > resistor) that should throw the bias point way off midpoint.
>
> In the first case, the cap isn't discharging properly, so you end up with
> the input to the cap pegged at Vin(max)-Vbe. This means that the output
> transistor can't actually put any power through the cap, because it's
> voltage isn't varying.
>
> The way to understand this is to figure out what the impedance is that the
> speaker sees when the transistor is on, vs the impedance it sees when the
> transistor is off. In the first case, it sees something like 1/10 the input
> impedance. In the second case, it sees 5k (since the transistor is off, so
> all you see is the emitter resistor).
>
> As a consequence, the circuit acts kinda like a diode, where the cap can be
> charged when the transistor is on, but can't be discharged when the
> transistor is off. Current in, no current out means it pumps up to the
> maximum voltage.
>
> >
> > After playing with simulation for a while, and replacing the
> > transistor with equivalent circuit, etc, I hit upon one of my biggest
> > realizations in electronics (mind me if it's too obvious for you). The
> > models are correct so long as the transistor is ON and operating in
> > its linear region. In such a case the high pass filter will act nicely
> > and block DC and pass the audio signal. However, with such low
> > resistance, If the output to follow the input faithfully, huge current
> > will have to pass in the capacitor for the slightest negative output
> > voltage, and since the current in the emitter resistor (5 KOhm) is
> > very small compared, this current would have to follow in the
> > transistor in the opposite direction, so the transistor turns off.
> > That's my reasoning for why the transistor would turn off.
>
> No, this will happen with a real circuit too. The problem isn't the
> simulation, it is the circuit.
>
> >
> > So now we have another mode of operation. Now it's another circuit.
> > It's not a high pass filter but a capacitor discharging through a
> > resistor. At a certain point the capacitor charge will be large enough
> > to have the transistor turn on only very briefly and the output
> > voltage change so little.
> >
>
> Right.
>
> > What caused my confusion is that in most electronic circuits, there
> > are transistors but often the circuit description doesn't tell if that
> > transistor will be ON all the time or not. Then analysis follows that
> > replaces caps with shorts at signal frequencies etc. But that's valid
> > only if the transistor stays on and hence acts as a linear device. I
> > wonder if circuit designers even with all their experiences and rules
> > of thumb take the extra step of analysis (or simulation) to make sure
> > the transistor is always operating in the linear region. If they do,
> > it's often not mentioned to us, less experienced people. In other
> > applications, such as rectifiers, the charging and discharging is
> > explicitly mentioned and used. This dual role of the capacitor always
> > puzzled me, when to think of it as a charging/discharging device, and
> > when to think in terms of frequency response and frequency dependent
> > impedance.
>
> You need another transistor to pull down on the cap. The typical output
> driver needs both a way to pull up, and a way to pull down.
> >
> > The other question was why connecting the speaker directly to the
> > output is OK. Why no blocking capacitor is needed? It turned out that
> > the small impedance shifts the DC bias point very close to ground, and
> > hence the output is a replica of the input but shifted up in DC level
> > just a few millivolts above ground, which seems to not alter the
> > speaker response much. That seems to be a problem if the gain is high
> > and clipping will start to happen as you move away from midpoint, but
> > when the gain is 1, no problem since the input signal is already a few
> > millivolts around 0.
>
> Connecting the speaker directly means you end up with a DC current through
> it all the time of (Vcc/2 - Vbe)/R
>
> Where R is the DC resistance of the speaker. Not good. However, the
> transistor gets to vary the current on top of the bias, which causes the
> magnetic field to vary, causing the paper to move.
>
> >
> > A possibility here is that the speaker also responds to changes in
> > magnetic field and hence the DC component wouldn't matter, but i'm not
> > sure about that one.
> >
>
> You are thinking too hard.
>
> > I apologize for anyone who took the effort and went through my long
> > post. Am I right in my conclusions?
> >
> > Thank you.
>
> Try this circuit instead:
>
> -------------------o------------.
> | |
> | |
> .-. |
> 470 | | |
> | | |
> '-' |
> | |
> | |/
> o----------| NPN
> | |>
> V |
> - |
> || | | || __ /|
> ------||--------o o----||--------| | |
> || | | || .--|__| |
> V | | \|
> 1uF - | 10uF |
> | |< |
> o----------| PNP |
> | |\ ===
> | | GND
> | |
> .-. |
> 470 | | |
> | | |
> '-' |
> | |
> ------------------o------------'
> (created by AACircuit v1.28.6 beta 04/19/05 http://www.tech-chat.de)
>
> It still sucks, but it sucks far less than your circuit.
>
> Filters are interesting, but that isn't your problem here. Notice that when
> the input goes down below the midpoint, that causes the NPN to turn off, and
> the PNP to turn on. This is also a 'follower' of a kind, it just has a more
> balanced impedance in both directions.

If you want it to work rather better, Use 2 input caps, commoned at the input
side and connect one to each base.

Stabilising the quiesecent current with some emitter Rs wouldn't go amiss either
(use an extra diode in the biasing network).

Graham

On Sun, 24 Feb 2008 14:04:08 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:

>On Sun, 24 Feb 2008 12:50:56 -0800, John Larkin
><jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
>
>>On Sun, 24 Feb 2008 11:21:59 -0800 (PST), "M. Hamed"
>><mhs000@gmail.com> wrote:
>>
>[snip]
>>> I
>>>wonder if circuit designers even with all their experiences and rules
>>>of thumb take the extra step of analysis (or simulation) to make sure
>>>the transistor is always operating in the linear region.
>>
>>
>>Good grief, of course we do. It's second nature.
>>
>[snip]
>
>Who is this "M. Hamed", some fricking amateur (that I've previously
>plonked)?
>
>I even utilize macros built into PSpice to check that CMOS operation
>is in the saturation region, since it's often non-obvious.
>
> ...Jim Thompson


We have been properly chastized for not mentioning this to less
experienced people. Let's try to be more careful in the future.

John

On Sun, 24 Feb 2008 22:18:19 +0100, Sjouke Burry
<burrynulnulfour@ppllaanneett.nnlll> wrote:

>Jim Thompson wrote:
>> On Sun, 24 Feb 2008 12:50:56 -0800, John Larkin
>> <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
>>
>>> On Sun, 24 Feb 2008 11:21:59 -0800 (PST), "M. Hamed"
>>> <mhs000@gmail.com> wrote:
>>>
>> [snip]
>>>> I
>>>> wonder if circuit designers even with all their experiences and rules
>>>> of thumb take the extra step of analysis (or simulation) to make sure
>>>> the transistor is always operating in the linear region.
>>>
>>> Good grief, of course we do. It's second nature.
>>>
>> [snip]
>>
>> Who is this "M. Hamed", some fricking amateur (that I've previously
>> plonked)?
>>
>> I even utilize macros built into PSpice to check that CMOS operation
>> is in the saturation region, since it's often non-obvious.
>>
>> ...Jim Thompson
>In 40 years of electronics, I have never simulated anything.
>Unless you count breadboards as simulation


Simulation can be neat, when nonlinear behavior exceeds reasonable
math analysis, or to train your instincts as regards causalities.

Breadboarding is good when device models are unavailable or flat
wrong, but can bite you on production tolerances.

John

On Sun, 24 Feb 2008 14:47:48 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

>On Sun, 24 Feb 2008 14:04:08 -0700, Jim Thompson
><To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:
>
>>On Sun, 24 Feb 2008 12:50:56 -0800, John Larkin
>><jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
>>
>>>On Sun, 24 Feb 2008 11:21:59 -0800 (PST), "M. Hamed"
>>><mhs000@gmail.com> wrote:
>>>
>>[snip]
>>>> I
>>>>wonder if circuit designers even with all their experiences and rules
>>>>of thumb take the extra step of analysis (or simulation) to make sure
>>>>the transistor is always operating in the linear region.
>>>
>>>
>>>Good grief, of course we do. It's second nature.
>>>
>>[snip]
>>
>>Who is this "M. Hamed", some fricking amateur (that I've previously
>>plonked)?
>>
>>I even utilize macros built into PSpice to check that CMOS operation
>>is in the saturation region, since it's often non-obvious.
>>
>> ...Jim Thompson
>
>
>We have been properly chastized for not mentioning this to less
>experienced people. Let's try to be more careful in the future.
>
>John
>
>

Sno-o-o-o-ort

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

America: Land of the Free, Because of the Brave

"M. Hamed" wrote:
>
> On Feb 24, 2:04 pm, Jim Thompson <To-Email-Use-The-Envelope-I...@My-
> Web-Site.com> wrote:
> > Who is this "M. Hamed", some fricking amateur (that I've previously
> > plonked)?
> >
>
> We're all amateurs in the school of life, aren't we? one way or
> another.
>
> > I even utilize macros built into PSpice to check that CMOS operation
> > is in the saturation region, since it's often non-obvious.
> >
>
> At this point, I don't give a **** (use your imagination) what you do
> with your PSpice
>
> For an attitude like yours, I wouldn't post my company's name. Not
> good for the business.


You probably couldn't get online without using one of Jim's 'Pspice'
IC designs. What have you done, other than act like a spoiled brat?


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

"Michae A. Terrell" wrote:
>
> "Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message news:47C309D4.6D7681D@earthlink.net...
> > "M. Hamed" wrote:
> >>
> >> On Feb 24, 2:04 pm, Jim Thompson <To-Email-Use-The-Envelope-I...@My-
> >> Web-Site.com> wrote:
> >> > Who is this "M. Hamed", some fricking amateur (that I've previously
> >> > plonked)?
> >> >
> >>
> >> We're all amateurs in the school of life, aren't we? one way or
> >> another.
> >>
> >> > I even utilize macros built into PSpice to check that CMOS operation
> >> > is in the saturation region, since it's often non-obvious.
> >> >
> >>
> >> At this point, I don't give a **** (use your imagination) what you do
> >> with your PSpice
> >>
> >> For an attitude like yours, I wouldn't post my company's name. Not
> >> good for the business.
> >
> >
> > You probably couldn't get online without using one of Jim's 'Pspice'
> > IC designs. What have you done, other than act like a spoiled brat?
> >
> >
> > --
> > Service to my country? Been there, Done that, and I've got my DD214 to
> > prove it.
> > Member of DAV #85.
> >
> > Michael A. Terrell
> > Central Florida
>
> And you DumbFuck Imposter think you know it all? Bullshit what makes you think you know above every one fucker? By the way, my news agency said you faked to be me to block me from posting. So you want to be a dictator hugh? Shitstain?
>
> --
> Service to my country? Been there, Done that, and I've got my DD214 to
> prove it.
> Member of DAV #85.
>
> Michael A. Terrell
> Central Florida


Do you think people can't see you are posting fake messages,
'jackthehammer'? I have never had an aioe.org account, and anyone with
half a brain can see from your headers that you're as phony as you are
stupid.


Xref:
sn-us sci.electronics.design:803324
sci.electronics.basics:267290
Path:

sn-us!sn-feed-sjc-01!sn-xt-sjc-10!sn-xt-sjc-01!sn-xt-sjc-08!sn-xt-sjc-14!supernews.com!news.astraweb.com!border2.newsrouter.astraweb.com!news.glorb.com!news.unit0.net!newsfeed.straub-nv.de!aioe.org!not-for-mail
From:
Michae A. Terrell <mike.temrell@earthlink.net>
Newsgroups:
sci.electronics.design, sci.electronics.basics
Subject:
Re: Realizations with transistors and capacitors in an
emitter follower
Date:
Mon, 25 Feb 2008 03:23:07 -0800
Organization:
http://home.earthlink.net/~mike.terrell/
Lines:
28
Message-ID:
<fq0srg$nga$1@aioe.org>
References:

<6c69d150-9a24-46fb-ba21-d9276afcf86e@q70g2000hsb.googlegroups.com>
<dnl3s3ha53lco7r65cr037bmjkis5deeml@4ax.com>
<alm3s3hljp9e6lubvcrkvkqdpupqmb9pe0@4ax.com>

<2e61833f-e0ac-410f-8a73-2b3444d8b5da@h11g2000prf.googlegroups.com>
<47C309D4.6D7681D@earthlink.net>
NNTP-Posting-Host:
7cHXlYzb5o/n/ZRDbgV2UQ.user.aioe.org
Mime-Version:
1.0
Content-Type:
text/plain; charset="iso-8859-1"
Content-Transfer-Encoding:
base64
X-Complaints-To:
abuse@aioe.org
X-MimeOLE:
Produced By Microsoft MimeOLE V6.00.2900.2180
X-Newsreader:
Microsoft Outlook Express 6.00.2800.1106
X-Priority:
3
X-MSMail-Priority:
Normal


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

"ShamShoon" <mhs000@gmail.com> wrote in message
news:4f451149-a73d-4ace-bad2-5fa4e45d79e7@u72g2000hsf.googlegroups.com...
> Last thing I did in my quest for learning about simple amplifiers was
> the common emitter amplifier. The output is opposite of polarity to
> the input. I thought that would be fun to try to have the voice
> reversed. However after trying it I realized that speakers don't
> really have a polarity (despite the clear + & - signs on the speaker)
> which makes sense. The output of the CE amp was very good. I had a
> blast when I added a potentiometer in place of the collector resistor,
> and used it to amplify my little Mp3 player signal before it gets into
> my car. It used to be not powerful enough to drive the car sound
> system and I had to crank the volume all the way up.
>
> No more! this simple circuit worked and it worked beautifully! Now I
> have an amplifier with volume control and very simple, one transistor,
> one OpAmp, and a few resistors and capacitors, and a battery. Sound
> quality is not bad but certainly much better than I expected with such
> a simple configuration. The only thing I need now is to figure out how
> to reduce the power (actually I need to calculate it), because I don't
> want it to drain the battery quick!

Go learn what class B or class D amplifier is.

Bob