Thanks for the explanation, Joe, but no. I never said anythink about
multiplicaiton

Steve,
Can you send me a copy of the original post? When these things happen, I
like to see where my crazy notions came from.

Joe
W3JDR

Steve Evans wrote:

Sure did, Steve! I ran it through a spice program and you're right in
every detail. So there's obviously some basic flaw in my understanding
of caps. The textbooks say to treat a cap as a short circuit at AC
(assuiming its reactance isn't too high at the frequency of interest).
That would appear to be grossly misleading as there's a huge
difference between the mean voltage levels on each side. . .

The mean voltage on each side is the DC component. The fact that it's
different on the two sides illustrates the fact that the capacitor is an
open circuit to DC. The shape of the waveforms on the two sides of the
capacitor are the same. That illustrates that the AC component is the
same on both sides -- the capacitor is a short circuit to AC.

It's gonna
take me a while to get this trough my thick skull. :-(
So what happens when you have a small ac ripple riding on a DC bias?

The bias (DC) is removed, and the ripple (AC) is passed through.

I'll spice it but won't understand it, I guess. :-(

Roy Lewallen, W7EL

On Wed, 13 Oct 2004 22:58:47 GMT, "Joe Rocci" wrote:

Thanks for the explanation, Joe, but no. I never said anythink about
multiplicaiton

Steve,
Can you send me a copy of the original post? When these things happen, I
like to see where my crazy notions came from.

Here it is in its entirety....

Hi everyone,

Below you will find my attempt to show in text-form, a circuit
fragment from a 145Mhz amplifier:


--------------capacitor-------------------------------transistor base
|
|
I
|
coil
|
|
|
|
------------------------------------------------------------GND

The cap's value is 1nF; the inductor's is 0.4uH.
The cap (I assume) is to couple one amplifier stage into the next
(50ohm source/load) with minimal attenuation of the desired VHF
signal. But like what's the purpose of this inductor to ground??
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.

On Tue, 12 Oct 2004 17:10:41 -0500, "Steve Nosko"
wrote:

This could be viewed as what we call a "clamper circuit" The AC voltage
AT THE BASE will have its positive peak "clamped" or moved to +0.7 volts and
the waveform will extend from there as negative as the waveform is tall.

IF we had 5 volts peak (10 volts peak-to-peak) the cap left side, the base
voltage will swing from +0.7 volts to -9.3 volts.

Okay, upon further thought about this there's still something amiss in
my understanding. I take what you say about the cap blocking out the
DC component of the waveform to leave the AC largely unaffected.
However, the term "clamping" AIUI means the diode lops off anything
over about 0.7 volts from the input waveform (ie, it conducts it away
to ground) so around half of it is lost (half wave rectification). Now
you state (and the spice progs agree) that what *actually* happens in
this case is that the whole AC waveform gets shifted south into
negative territory. It's still a full wave, but it's way down into
the negative and only the highest peaks just creep above zero volts.
Is this effect *solely* attributable to the steady build-up of
negative charge on the cap's RHS? I think what's really freaking me
out here is the fact that the signal source is grounded on the neg.
side and yet we have that same signal that after going through a cap
can end up going fully negative *below* ground. It just seems like any
such voltage beneath zero/ground potential is breaking the laws of
physics. Ground should be the 'absolute zero' of the potentials in any
circuit and here it is being violated. I need some help to get my
thick head around the concept! :-(
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.

Thanks for re-posting that Steve. I don't know where I got the crazy idea
that this was a multiplier stage....maybe I'm confusing this with a
different thread.

Thanks
Joe
W3JDR

Steve Evans wrote in message
...
On Wed, 13 Oct 2004 22:58:47 GMT, "Joe Rocci" wrote:

Thanks for the explanation, Joe, but no. I never said anythink about
multiplicaiton

Steve,
Can you send me a copy of the original post? When these things happen, I
like to see where my crazy notions came from.

Here it is in its entirety....

Hi everyone,

Below you will find my attempt to show in text-form, a circuit
fragment from a 145Mhz amplifier:


--------------capacitor-------------------------------transistor base
|
|
I
|
coil
|
|
|
|
------------------------------------------------------------GND

The cap's value is 1nF; the inductor's is 0.4uH.
The cap (I assume) is to couple one amplifier stage into the next
(50ohm source/load) with minimal attenuation of the desired VHF
signal. But like what's the purpose of this inductor to ground??
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.

"Steve Evans" wrote in message
...
On Tue, 12 Oct 2004 17:10:41 -0500, "Steve Nosko"
wrote:
[snip]
IF we had 5 volts peak (10 volts peak-to-peak) the cap left side, the
base
voltage will swing from +0.7 volts to -9.3 volts.
WHEW !
Did this work

Sure did, Steve! I ran it through a spice program and you're right in
every detail. So there's obviously some basic flaw in my understanding
of caps. The textbooks say to treat a cap as a short circuit at AC

C O O L !!! So my explanation did it...sorry for the gloat folks. Had
me scared there for a minute...30+ years doing and teaching electronice not
for naught after all.

This is super, Steve Evans, Now you're learning.
Now, look at the Spice circuit _after a bunch of input cycles_ and
monitor the voltage on BOTH sides of the cap and you will see that they are
both the same **AC wise **, but there will be a constant, or DC difference.
You can use the difference probe and you will only see a constant (DC)
difference in potential across the cap. If you look at each side
individually, you will see that the AC signal (yes I know "AC Signal" is
redundant, but I think it helps clarify) is identical on both sides, though
shifted by that DC amount. You can even change the waveform. Put *TWO* AC
sources in series, one less voltage than the other, and different
frequencies. The two cap sides will still have the same waveform.

The concept you quote of "cap as a short circuit at AC" is true, but
more easily seen in the steady-state. This is what we call it when all of
the transient effects have died out. These transient effects are usually
capacitors getting charged up (sorry Ratch, but I gotta say it). The AC
component will "pass through" the cap un-impeeded, BUT you can also have
this DC offset from one side of the cap to the other.
In the case we have focused on (coupling cap & NO inductor) you have a
little more complexity throwing in a monkey wrench. The Base-Emitter
junction/diode is rectifying the AC and charging up the cap _just like it
does_ in the common power supply rectifier circuit--but in this case it
makes trouble for us by reverse biasing the base.
I am glad to help, but as you see, you have sort of jumped into the
middle of circuit theory armed only with the "AC short" part of the
capacitor's characteristics. My bet is that you'll remember this very well.

They say: "we learn from our mistakes". Well, BOY! ARE WE LEARNIN' NOW!

73,
--
Steve N, K,9;d, c. i My email has no u's.

"Steve Evans" wrote in message
...
On Tue, 12 Oct 2004 17:10:41 -0500, "Steve Nosko"
wrote:

This could be viewed as what we call a "clamper circuit" The AC
voltage
AT THE BASE will have its positive peak "clamped" or moved to +0.7 volts
and
the waveform will extend from there as negative as the waveform is tall.

IF we had 5 volts peak (10 volts peak-to-peak) the cap left side, the
base
voltage will swing from +0.7 volts to -9.3 volts.

Steve Evans responds:
Okay, upon further thought about this there's still something amiss in
my understanding. I take what you say about the cap blocking out the
DC component of the waveform to leave the AC largely unaffected.
However, the term "clamping" AIUI means the diode lops off anything
over about 0.7 volts from the input waveform (ie, it conducts it away
to ground) so around half of it is lost (half wave rectification).

DAMN! And I sort of thought of this as I was writing it... STUPID ME! I
did what I (in my mind, usually) will normally criticize others for doing
soften here in an attempt to "help" others "fully understand". Namely,
adding some additional explanation or extreme detail which only serves to
further confuse the OP. Please excuse me.



While not necessary for understanding this circuit, I'll fill-in this bit
here. A "clamper" is a diode and cap circuit which will clamp a particular
point on a waveform to a specific voltage (kinda like being clamped in a
vice on the bench), but NOT change the wave's wiggling *shape*. Sometimes
this is needed. In the coupling circuit in question, it is the POSITIVE
PEAK of the signal that gets clamped to +0.7 volts. The wave's SHAPE is
un-changed, but the whole thing is shifted in its DC component.

What you are thinking of is called a "CLIPPER" because it CLIPS *off*
part of the waveform like barber's scissors.





Back to the actual subject.



In general, any waveform can have an AC component and a DC component.
The DC component only serves to "shift" the position of the signal up or
down; which can also be called a DC offset.



Now

Continues a wiser Steve Evans:
you state (and the spice progs agree) that what *actually* happens in
this case is that the whole AC waveform gets shifted south into
negative territory. It's still a full wave, but it's way down into
the negative and only the highest peaks just creep above zero volts.
Is this effect *solely* attributable to the steady build-up of
negative charge on the cap's RHS?



YUP !




I think what's really freaking me
out here is the fact that the signal source is grounded on the neg.
side and yet we have that same signal that after going through a cap
can end up going fully negative *below* ground. It just seems like any
such voltage beneath zero/ground potential is breaking the laws of
physics. Ground should be the 'absolute zero' of the potentials in any
circuit and here it is being violated. I need some help to get my
thick head around the concept! :-(



Oooooo. BEWARE! GROUND IS NOT ABSOLUTE ! ! ! Nope, nope, nope. This is
going to take some time to explain and more experience/study will be needed
for you to really _get it_.

The bad news for you may be that ground, I must sadly inform you, is
relative. There is no one, solid, never varying, absolute thing which is
ground, except in our imaginations. Many hams believe there is, but there
isn't.



That being said, let's start out simply and build.



Here is a very applicable analogy:

Voltage, also called "potential difference", is a lot like altitude --
height. We can talk about the height above the street level. We might
consider the street level to be "ground". In Physics, moving some object to
a higher level gives it the "Potential" to do damage if it falls on your
head, so the "potential energy" of it is greater. Holding it three feet
above your head gives it a certain potential, right? Voltage is just like
this.

HOWEVER, what about standing on the roof of a building THEN moving the same
object three feet above your head. You must agree that it has the SAME
potential to do damage TO YOUR HEAD that it did in the first example, right?
In this case, the roof of the building is our ground. So ground is relative
and *we* get to pick it. It is usually a known point in our circuit and we
use special symbols to show it. Note that this is why voltage is also is
called potential *DIFFERENCE*.

Unfortunately, this analogy will fall apart when trying to use it for
negative voltage, if we put this negative voltage "Below" our ground, but
for the "relative" concept, I hope it worked.



Now I'll try *negative*.



Ground is a REFERNCE to which we relate all the other voltages in the
circuit we happen to be talking about.

Lets put a screw into a wooden board and call it our ground. Work with
me here, Steve. Connect a wire and run it over to several other screws in
the board. They are all now our "ground".

Take your basic Ray-O-Vac flashlight battery. [[ actually, for Ratch,
this is a "cell", but common usage dies hard. A "battery" is a collection
of cells usually connected in series to get a greater voltage ]]]. Run a
wire from the negative terminal of the battery...oops, cell, and connect it
to our ground.

Now, with the negative lead (rhymes with seed) of our volt meter
connected to our ground, measure the voltage on the Ray-O-Vac positive
terminal with the positive lead of the voltmeter. We get + 1.5 volts.

In electronics (when we want to study some phenomenon carefully) we
have a convention of ALWAYS putting the NEGATIVE lead of the voltmeter on
our ground. That way we can easily see if the voltage it is positive or
negative.



NOW... reverse the connection on the Ray-O-Vac. You will now measure a
NEGATIVE 1.5 volts. So what's the way to interpret this.?



Voltage (potential difference) is also called Electro Motive Force
because it is the "force" that "pushes" the electrons around the circuit to
form Ratch's (and everyone's) current flow (sorry, Ratch, couldn't resist
pulling your chain a bit in some friendly ribbing (:-).

When we reverse the battery/cell, we are now "pushing" in the other
direction. It is still a potential difference, but negative.





In MY model (the one in my head), I visualize this as being "physically"
below ground (and we often refer to it as being "below ground") in the sense
that on the oscilloscope and Pspice display it is below on the graph.

If you REALLY want to make the altitude analogy work, you can imagine
that gravity always "pulls" things toward the "ground" - but this is YOUR
local ground. Then the negative potentials "pull you up".



END.hope this helps, Steve.

73

Sure did, Steve! I ran it through a spice program and you're right in
every detail.
=======================

What makes you think Spice is correct? Its only a buggy computer program.
Rubbish in - rubbish out!

On Fri, 15 Oct 2004 23:06:38 +0000 (UTC), "Reg Edwards"
wrote:

Sure did, Steve! I ran it through a spice program and you're right in
every detail.
=======================

What makes you think Spice is correct? Its only a buggy computer program.
Rubbish in - rubbish out!

Because it agreed exactly with Steve's prediction. I think therefore
I'm entitled to rely on it in this instance at least.

--

"What is now proved was once only imagin'd." - William Blake, 1793.

On Fri, 15 Oct 2004 23:06:38 +0000 (UTC), "Reg Edwards"
wrote:

What makes you think Spice is correct? Its only a buggy computer program.
Rubbish in - rubbish out!

BTW, Reg. Is there a program on your site that can handle stripline
calculations for UHF circuit board inductors?
--

"What is now proved was once only imagin'd." - William Blake, 1793.