What's this inductor doin'?

[Steve Evans]

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.

 

[Joe Rocci]

The inductor keeps the transistor base at DC ground potential (probably the
same potential as the not-shown emitter). This makes the transistor only
conduct on positive half-cycles of the drive signal, which is a very
non-linear condition that generates lot's of harmonic content. It's also
common to put a little resistance in series with the inductor, which
slightly reverse-biases the transistor because the RF waveform can then
swing more toward the negative than the positive. A little reverse bias
causes the transistor to conduct over a smaller portion of the input cycle,
which enhances higher-order harmonic generation.

Joe
W3JDR

Steve Evans <smevans@jif-lemon.co.mars> wrote in message
news:ag50m01ki8c5hiaqb80enlsfi62lbv8k2g@4ax.com...

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.

 

[Joe Rocci]

With a 1nf coupling cap, there's no impedance matching happening because the
capacitive reactance is so low that the impedances on both sides of the cap
are essentially connected together.

Joe
W3JDR

John Popelish <jpopelish@rica.net> wrote in message
news:416020A3.F44AFB0E@rica.net...

Steve Evans wrote:

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??

The inductor provides a bias path to ground, to hold the average
transistor base voltage at zero volts, while passing the base
current. It also forms a resonant circuit with the capacitor (and
base capacitance) that has a peak response at some frequency,
hopefully in the middle of the band being amplified. This resonance
lowers the impedance at the input side of the capacitor and raises it
at the base node, stepping the input voltage up and the input current
down.

--
John Popelish

 

[John Popelish]

Steve Evans wrote:

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??

The inductor provides a bias path to ground, to hold the average
transistor base voltage at zero volts, while passing the base
current. It also forms a resonant circuit with the capacitor (and
base capacitance) that has a peak response at some frequency,
hopefully in the middle of the band being amplified. This resonance
lowers the impedance at the input side of the capacitor and raises it
at the base node, stepping the input voltage up and the input current
down.

--
John Popelish

 

[John Popelish]

Joe Rocci wrote:

John Popelish <jpopelish@rica.net> wrote in message
news:416020A3.F44AFB0E@rica.net...
Steve Evans wrote:

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??

The inductor provides a bias path to ground, to hold the average
transistor base voltage at zero volts, while passing the base
current. It also forms a resonant circuit with the capacitor (and
base capacitance) that has a peak response at some frequency,
hopefully in the middle of the band being amplified. This resonance
lowers the impedance at the input side of the capacitor and raises it
at the base node, stepping the input voltage up and the input current
down.

With a 1nf coupling cap, there's no impedance matching happening because the
capacitive reactance is so low that the impedances on both sides of the cap
are essentially connected together.

Right. I didn't pay any attention to the given values. They produce
a resonance around 8 megahertz.

--
John Popelish

 

[John Fields]

On Sun, 03 Oct 2004 15:08:22 GMT, Steve Evans
<smevans@jif-lemon.co.mars> wrote:

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??

---

At 145 MHz, the reactance of the cap is:

1
Xc = -------
2pifC


1
= ------------------------------ ~ 1.1 ohms
6.28 * 1.45E8 Hz * 1.0E-9 F

So it's likely not effecting a match to 50 ohms.


The reactance of the inductor is:


Xl = 2pifL



= 6.28 * 1.45E8 Hz * 4.0E-7 H ~ 364 ohms

so they're not resonant at 145MHz.



Since the resonant frequency of the LC is:


1
f = --------------
2pi(sqrt LC)

it's tuned to

1
f = ----------------------------- ~ 7.96MHz
6.28 * sqrt (4E-7H * 1E-9F)

which is nowhere near 145MHz.


If that's all there is to the circuit, my guess is that it's a
highpass filter with the coil doing double duty as a DC return for the
base as well as a fairly high reactance load for the driver. Also,
(WAG) since the transistor's input resistance and capacitance will
appear effectively in parallel with the coil, it may wind up looking
like something closer to 50 ohms than 364 ohms to the driver.

--
John Fields

 

[Doug Smith W9WI]

Steve Evans wrote:

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

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??

A capacitor has a very low impedance to high-frequency (i.e., 145MHz)
signals and a very high impedance to low-frequency (i.e., DC) signals.

An inductor is the other way around - very low impedance to low
frequency (DC) signals and very high impedance to high frequency (2m).

The inductor allows DC bias currents to flow while not shunting the
desired 2m RF to ground.

The capacitor passes the 2m drive signal from the previous stage without
attenuation, while keeping the DC from the previous stage out of this one.
--
Doug Smith W9WI
Pleasant View (Nashville), TN EM66
http://www.w9wi.com

 

[Steve Evans]

On 10 Oct 2004 00:26:35 GMT, et472@FreeNet.Carleton.CA (Michael Black)
wrote:

Then rephrase the question.

Are you wondering why the base of the transistor is at DC ground, or
wondering what the RF choke does to ensure that the base of the transistor is
not at ground for RF?

People have answered what they think you are asking, which isn't really clear.
Clarify that, and you might get an answer you want.

Thanks Michael,
I'm sorry I can't post the full diagram as my scanner's bust. I'll try
to clarify. I know why the base of the transistor (it's a 2n5771, in
answer to Paul's question) is at DC ground. I'm not so dumb as to
realise that a choke passes dc but not the high frequncy RF. I guess
it boils down to this: how did the designer arrive at the given value
of 0.4uH for this inductor? Why not 4nH? Why not 40uH? Why not 100mH??
What's the deal with this value and since I've only got a 1uH in my
junk box, will that be okay instead?

Thanks, all,

Steve.
--

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

 

[Rich Grise]

On Sunday 10 October 2004 10:51 pm, Robert Monsen did deign to grace us with
the following:

Steve Evans wrote:
On Sun, 10 Oct 2004 14:01:44 +0100, Paul Burridge
pb@notthisbit.osiris1.co.uk> wrote:


Well then it appears Reg's hunch was right. The transistor in question
has an input capacitance of just under 3pF, so the 0.4uH inductor
forms a parallel tuned circuit with it at 145Mhz. This prevents the
input signal from partially shunted to ground via the input
capacitance were the inductor not there. The idea is to allow as much
of the input signal as possible to develop across the BE diode to
maximise input impedance and gain. The input capacitance is in
parallel with this diode and bypasses RF signals around it - which you
*don't* want. Will a 1uH work instead? Do the maths and find out. But
if you don't know the inductor's Q that probably won't help much...


Sorry, but none of this makes sense to me. There's no diode involved
so I don't know where you get that from. And what's "input
capacitance" and "Q"?
Do try to speak in plain English!

Possibly surprisingly, all that jargon actually qualifies as English!
It's just that it's, well, jargon.

Steve


The transistor acts like a tiny capacitor from base to emitter, in
addition to its other jobs. Thus, an inductor in series with the base
will act like a bandpass filter at

f = 1/(2*pi*sqrt(L*C))

An inductor and capacitor with the same values in parallel will act like
a bandstop filter at that frequency.

Q is the "quality or merit factor" of the inductor, and is computed by
dividing the 'reactance' (or AC resistance) of the inductor by the DC
resistance *at the frequency in question*; it will generally be
different at different frequencies. Q is really the ratio of reactive
power in the inductance to the real power dissipated in the resistance,
and can be computed by using the formula:

Q = 2*PI*f*L/ri

where f is frequency, and ri is the resistance of the coil at f. You buy
inductors with a given Q range.

For an LC resonant circuit, the Q of the inductor will affect the 'Q' of
the resulting resonance, which simply means that with a bigger Q, the
passband or stopband will be wider.

You were doing so good up to this point. ;-)

The Q of a resonant circuit is
defined as the resonant frequency divided by the width of the passband.

Oh, OK - you've fixed it here.

Cheers!
Rich

 

[Rich Grise]

On Sunday 10 October 2004 03:29 pm, Steve Evans did deign to grace us with
the following:

On Sun, 10 Oct 2004 05:41:23 -0700, Roy Lewallen <w7el@eznec.com
wrote:

[snip]

(*) Even this explanation is highly simplified. At higher frequencies,
the inductor will exhibit additional series and parallel resonances due
to various parasitic capacitances and inductances and transmission line
effects.

groan
Dur, nope. I still don't get it. Can anyone explain this stuff *in
simple terms*? I'm really struggling here with some of the
terminology. :-(
--

You need to look through these until something starts to make sense to
you:
http://www.google.com/search?q=basic+electronics+tutorial&btnG=Search+the+Web

Welcome to the zoo!
Rich

 

[Steve Nosko]

"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:r2ejm0t0lfstc6pa8edd2o3statd7a4kr7@4ax.com...

On Sun, 10 Oct 2004 14:01:44 +0100, Paul Burridge
pb@notthisbit.osiris1.co.uk> wrote:

Well then it appears Reg's hunch was right. The transistor in question
has an input capacitance of just under 3pF, so the 0.4uH inductor
forms a parallel tuned circuit with it at 145Mhz. This prevents the
input signal from partially shunted to ground via the input
capacitance were the inductor not there. The idea is to allow as much
of the input signal as possible to develop across the BE diode to
maximise input impedance and gain. The input capacitance is in
parallel with this diode and bypasses RF signals around it - which you
*don't* want. Will a 1uH work instead? Do the maths and find out. But
if you don't know the inductor's Q that probably won't help much...

Sorry, but none of this makes sense to me. There's no diode involved
so I don't know where you get that from. And what's "input
capacitance" and "Q"?
Do try to speak in plain English!

Steve

Hi Steve (swell name, by the way),
From this last comment, it appears that you have a lot to learn. Paul's
was a pretty good explanation of a first step at understanding what might be
going on in the circuit shown some time ago (an inductor in shunt with the
base-emitter of the transistor).

The Base-Emitter in a transistor is a semiconductor junction just like a
diode and in the (higher power) RF amplifiers behaves pretty much like a
diode. With RF applied to the base, there will be conduction on the
positive peaks only and this will constitute a DC current flow which must
have a DC path. The inductor provides such a path since the capacitor can
not. If this makes no sense to you then you, indeed are in over your head
in an attempt to understand because it is pretty basic and simple. You will
need to understand diodes and transistors first.

you say you are "really struggling here with with some of the terminology."
Perhaps you can tell us which words are giving you heartburn?



I will respond to Paul's content, however, with this. The BE
capacitance of this device, in this aparent application, I am pretty sure is
not the dominant effect. The Rrverse biased capacitance is the wrong thing
to focus on. While it is interesting that that it and the inductor are near
resonance, this probaly is not what is happening because this would make the
inpedance looking into the base very high and difficult to get power to the
base, contrary to Reg's hunch. The orignal ASCII cricuit simply had a
coupling cap and a base-emitter shunt cap. It looks like class B or C. C
more likely. Therefore the transistor is in conduction part of the time and
not for another part of the time. Therefore we have a nonlinear, large
signal condition. The base impedance under this condition (pulsed
conduction) will be quite low and dominate and therefore, it will need some
impedance matching to get enough into the base (from the preceeding
collector). SO, I say that the inductor is :
1- Providing the obvious DC path and.
2- Impedance matching along with the "coupling" capacitor (did it have a
value??)...BUT!

The one monkey wrench I will throw in, is that the Miller effect will also
have a very significant effect on the input impedance of the stage. The Ccb
is a path providing significant feedback and probably dominating the input
impedance.

If you don't recall, the Miller Effect describes the capacitance looking
into the base which looks like Ccb times the voltage gain (call it A). This
is due to the fact that Ccb connects between the input / base and output /
collector. Because the collector voltage is ~~ 180 degrees out of phase,
with the base voltage, an input voltage change of, say one millivolt, on the
input side of Ccb results in a change in voltage on the output side of Ccb
of one milivolt times the voltage gain, A. This results in a total change
across Ccb of A+1 milivolts and therefore a current change A+1 times a value
that the 1 milivolt input change expected to see. This makes the capacitor
look A+1 times as big as it actually is.

Finally, and possibly the most difficult to quantify (ok two monkey
wrenches--nobody expects the Spanish inquisition), in RF circuits there is
*very frequently* one other confounding factor and this is the circuit board
layout and/or the actual physical construction. All the previous talk about
how inductors and capacitors behave differently at high frequencies (I
believe by Roy Lewallen) is nicely put, but the actual connection methods
also can have a very significant effect on what value components are used.
The "wiring" can add other capacitances and inductances which, very often,
do not show up on the schematic. This can have profound effect on the
components used, completely masking any hope of understanding of the circuit
from the schematic diagram. As the power level in the circuit goes up, the
impedances go down and short wires or PC board runs can become significant
impedances, either to help or hurt the desired matching circuit.
--
Steve N, K,9;d, c. i My email has no u's.

 

[Joe Rocci]

Steve,
I'm not sure, but I think the original post said this stage was a frequency
multiplier with an OUTPUT frequency of about 145 MHz. If that's the case,
then the INPUT frequency would be 72 MHz or less. At that frequency, I don't
think the choke and the input capacitance of the transistor are anywhere
near resonance. Also, the coupling cap was stated as 1nF if I recall.

I think what we're looking at here is a DC -lock coupling cap and a
DC-return RF choke....nothing more.

Joe
W3JDR


Steve Nosko <suteuve.nosukowicuz@moutouroula.com> wrote in message
news:ckeork$sip$1@newshost.mot.com...

"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:r2ejm0t0lfstc6pa8edd2o3statd7a4kr7@4ax.com...
On Sun, 10 Oct 2004 14:01:44 +0100, Paul Burridge
pb@notthisbit.osiris1.co.uk> wrote:

Well then it appears Reg's hunch was right. The transistor in question
has an input capacitance of just under 3pF, so the 0.4uH inductor
forms a parallel tuned circuit with it at 145Mhz. This prevents the
input signal from partially shunted to ground via the input
capacitance were the inductor not there. The idea is to allow as much
of the input signal as possible to develop across the BE diode to
maximise input impedance and gain. The input capacitance is in
parallel with this diode and bypasses RF signals around it - which you
*don't* want. Will a 1uH work instead? Do the maths and find out. But
if you don't know the inductor's Q that probably won't help much...

Sorry, but none of this makes sense to me. There's no diode involved
so I don't know where you get that from. And what's "input
capacitance" and "Q"?
Do try to speak in plain English!

Steve


Hi Steve (swell name, by the way),
From this last comment, it appears that you have a lot to learn.
Paul's
was a pretty good explanation of a first step at understanding what might
be
going on in the circuit shown some time ago (an inductor in shunt with the
base-emitter of the transistor).

The Base-Emitter in a transistor is a semiconductor junction just like a
diode and in the (higher power) RF amplifiers behaves pretty much like a
diode. With RF applied to the base, there will be conduction on the
positive peaks only and this will constitute a DC current flow which must
have a DC path. The inductor provides such a path since the capacitor can
not. If this makes no sense to you then you, indeed are in over your head
in an attempt to understand because it is pretty basic and simple. You
will
need to understand diodes and transistors first.

you say you are "really struggling here with with some of the
terminology."
Perhaps you can tell us which words are giving you heartburn?



I will respond to Paul's content, however, with this. The BE
capacitance of this device, in this aparent application, I am pretty sure
is
not the dominant effect. The Rrverse biased capacitance is the wrong thing
to focus on. While it is interesting that that it and the inductor are
near
resonance, this probaly is not what is happening because this would make
the
inpedance looking into the base very high and difficult to get power to
the
base, contrary to Reg's hunch. The orignal ASCII cricuit simply had a
coupling cap and a base-emitter shunt cap. It looks like class B or C. C
more likely. Therefore the transistor is in conduction part of the time
and
not for another part of the time. Therefore we have a nonlinear, large
signal condition. The base impedance under this condition (pulsed
conduction) will be quite low and dominate and therefore, it will need
some
impedance matching to get enough into the base (from the preceeding
collector). SO, I say that the inductor is :
1- Providing the obvious DC path and.
2- Impedance matching along with the "coupling" capacitor (did it have a
value??)...BUT!

The one monkey wrench I will throw in, is that the Miller effect will also
have a very significant effect on the input impedance of the stage. The
Ccb
is a path providing significant feedback and probably dominating the input
impedance.

If you don't recall, the Miller Effect describes the capacitance looking
into the base which looks like Ccb times the voltage gain (call it A).
This
is due to the fact that Ccb connects between the input / base and output /
collector. Because the collector voltage is ~~ 180 degrees out of phase,
with the base voltage, an input voltage change of, say one millivolt, on
the
input side of Ccb results in a change in voltage on the output side of Ccb
of one milivolt times the voltage gain, A. This results in a total change
across Ccb of A+1 milivolts and therefore a current change A+1 times a
value
that the 1 milivolt input change expected to see. This makes the
capacitor
look A+1 times as big as it actually is.

Finally, and possibly the most difficult to quantify (ok two monkey
wrenches--nobody expects the Spanish inquisition), in RF circuits there is
*very frequently* one other confounding factor and this is the circuit
board
layout and/or the actual physical construction. All the previous talk
about
how inductors and capacitors behave differently at high frequencies (I
believe by Roy Lewallen) is nicely put, but the actual connection methods
also can have a very significant effect on what value components are used.
The "wiring" can add other capacitances and inductances which, very often,
do not show up on the schematic. This can have profound effect on the
components used, completely masking any hope of understanding of the
circuit
from the schematic diagram. As the power level in the circuit goes up, the
impedances go down and short wires or PC board runs can become significant
impedances, either to help or hurt the desired matching circuit.
--
Steve N, K,9;d, c. i My email has no u's.

 

[Steve Evans]

On Mon, 11 Oct 2004 14:59:11 -0500, "Steve Nosko"
<suteuve.nosukowicuz@moutouroula.com> wrote:

The Base-Emitter in a transistor is a semiconductor junction just like a
diode and in the (higher power) RF amplifiers behaves pretty much like a
diode. With RF applied to the base, there will be conduction on the
positive peaks only and this will constitute a DC current flow which must
have a DC path. The inductor provides such a path since the capacitor can
not.

Okay, Steve, I'm gonna have to take your explanation in bite-size
chunks. Kindly indulge me...

I don't see that the inductor is necessary to provide such a path to
ground for the signal peaks, since they (the input signal pos. peaks)
turn on the transistor and complete the circuit to ground via the
base/emitter junction, which will be a low resistance path with
sufficient base drive level on the peaks. Can you tell me why this
path alone isn't good enough and there has to be an inductor across
B/E as well?

Thanks!

Steve
--

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

 

[Joe Rocci]

Steve,

If the inductor was not there to hold the zero-crossings of the input sine
wave at zero volts, then the whole waveform would sink toward a lower DC
voltage because it is capacitively coupled. You can prove this to yourself
by taking a large capacitor and driving a diode that is connected to ground.
The test can easily be done with audio frequencies if you don't have RF
equipment. You could also simulate it on a program like SPICE.

If the choke were removed from the circuit, this input DC shift would
reverse bias the BE junction, preventing the abiltiy of the waveform to
drive current into the BE junction. No base current = no collector current =
no gain.

BTW, I was almost sure your original post said this was a multiplier
circuit. Did the word "multiplier" not appear in it anywhere? Hmmm...

Joe
W3JDR


Steve Evans <smevans@jif-lemon.co.mars> wrote in message
news:7u3mm0lun8fk658o0ibl23neqbf9vmlnt2@4ax.com...

On Mon, 11 Oct 2004 14:59:11 -0500, "Steve Nosko"
suteuve.nosukowicuz@moutouroula.com> wrote:

The Base-Emitter in a transistor is a semiconductor junction just like a
diode and in the (higher power) RF amplifiers behaves pretty much like a
diode. With RF applied to the base, there will be conduction on the
positive peaks only and this will constitute a DC current flow which must
have a DC path. The inductor provides such a path since the capacitor
can
not.

Okay, Steve, I'm gonna have to take your explanation in bite-size
chunks. Kindly indulge me...

I don't see that the inductor is necessary to provide such a path to
ground for the signal peaks, since they (the input signal pos. peaks)
turn on the transistor and complete the circuit to ground via the
base/emitter junction, which will be a low resistance path with
sufficient base drive level on the peaks. Can you tell me why this
path alone isn't good enough and there has to be an inductor across
B/E as well?

Thanks!

Steve
--

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

 

[Robert Monsen]

Rich Grise wrote:

On Sunday 10 October 2004 10:51 pm, Robert Monsen did deign to grace us with
the following:
For an LC resonant circuit, the Q of the inductor will affect the 'Q' of
the resulting resonance, which simply means that with a bigger Q, the
passband or stopband will be wider.


You were doing so good up to this point. ;-)

You are right, higher Q => narrower passband.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Steve Evans]

On Tue, 12 Oct 2004 17:10:41 -0500, "Steve Nosko"
<suteuve.nosukowicuz@moutouroula.com> 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
(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. 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?
I'll spice it but won't understand it, I guess. :-(

--

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

 

[Ratch]

"Steve Nosko" <suteuve.nosukowicuz@moutouroula.com> wrote in message
news:ckhku1$e66$1@newshost.mot.com...

"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:7u3mm0lun8fk658o0ibl23neqbf9vmlnt2@4ax.com...
On Mon, 11 Oct 2004 14:59:11 -0500, "Steve Nosko"
suteuve.nosukowicuz@moutouroula.com> wrote:

The Base-Emitter in a transistor is a semiconductor junction just like
a
diode and in the (higher power) RF amplifiers behaves pretty much like
a
diode. With RF applied to the base, there will be conduction on the
positive peaks only and this will constitute a DC current flow which
must
have a DC path. The inductor provides such a path since the capacitor
can
not.

Okay, Steve, I'm gonna have to take your explanation in bite-size
chunks. Kindly indulge me...

I don't see that the inductor is necessary to provide such a path to
ground for the signal peaks, since they (the input signal pos. peaks)
turn on the transistor and complete the circuit to ground via the
base/emitter junction, which will be a low resistance path with
sufficient base drive level on the peaks. Can you tell me why this
path alone isn't good enough and there has to be an inductor across
B/E as well?
Thanks! Steve

Most certainly, Steve.... Indulge, I will.
This can get confusing... Steve Noskowicz here (K9DCI)

Joe Rocci tried, but I don't think he went through the proper step-by-step
explanation.

OK Here's what happens... Start with the capacitor completely
discharged -
zero volts across it - right end same voltage as left side (whatever that
may be). On the first positive peak, some current flows through the base

Current does not flow. Current IS flow. "Current flow" means "charge
flow flow" which is redundant and ridiculous. One should just say charge
flow.

emitter junction because the voltage on the right side of the cap is the
same as that on the left side. (this assumes there is at least 0.7 volts
of
signal coming from the left. NPN transistors conduct current when the
base
is about 0.7 volts positive.
The current is some quantity of electrons, right. Well these electrons

Er, no. Current is quantity of charge per unit time.

will start to "fill up" or charge the capacitor. Each time another
positive

A capacitor does not charge. The net change of charge is zero for a
capacitor going from zero volts to its breakdown voltage. That is because
for each amount of charge stored on one plate, the same amount of charge
leaves the opposite plate. Now it takes energy to accumulate or deplete
like charges (electrons in this case) because they do not like to get close
together. This energy is stored in a electrostatic field proportional to
the square of the voltage. Therefore a capacitor should be described as
being "energized", not "charged". Ratch

pulse happens, the right side of the capacitor collects more and more
electrons. This charges the cap more and more and makes the right side
more
and more negative. There is no way to drain off this charge before the
next
pulse comes along. The base-emitter junction will be reverse biased and
not
conduct any current. I hope you see this because that's the key.
As the caps gets more charge from each pulse, the right side becomes more
negative. After each pulse, it will take more and more voltage on the caps
left side to get enough voltage on the right side (0.7 volts) to get to
the
base-emitter conduction voltage and get any current to flow.
The end result is that the capacitor will charge to the peak input
voltage (less about 0.7 volts) and the base will never get to the 0.7 volt
level. At this point, the base voltage will be swinging (with the input
signal causing it) from about 0.7 volts plus to a value equal to
*negative* the peak-to-peak input voltage (less the 0.7 volts). That's
minus volts. The AC signal at the bacse will NOT be swinting equally
around
zero volts.

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.

WHEW !
Did this work ???
--
Steve N, K,9;d, c. i My email has no u's.

 

[Steve Nosko]

long post warning directed at Ratch...

Hi there Ratch. Comments inserted below, but I am using terminology that
has been used for a long time and is commonly accepted. If the OP has
problems, I prefer him to pose (sp) the questions for clarification.

Much of what you point out is right along the lines of the following
examples.

ROM stands for "Read Only Memory", right?
RAM stands for "Random Access Memory", right?

Yet ROM is also "Random Access". Therefore, ROM IS RAM. These are
conventions which came about by an imperfect system of terminology. Namely,
whatever sticks gets used.

Example closer to this:

Does a "river flow"? Do you know what this means? I think you do, yet by
your comment a river is a flow of watter and "a river flows" would be
"incorrect". We could only say that "Water Flows".

While I appreciate youre desire to be correct, I do not believe the OP would
be confused by my use of very common terminilogy and if he/she is, then the
questions will come from them/he/she.

I would, however, ask that you comment on the concepts in question. Is it
an explanation of *WHY* the coupling cap causes the Veb to go negative
causing no conduction in the part?



"Ratch" <Watchit@Comcast.net> wrote in message
news:Hkhbd.120265$He1.113453@attbi_s01...

"Steve Nosko" <suteuve.nosukowicuz@moutouroula.com> wrote in message
news:ckhku1$e66$1@newshost.mot.com...
[...snippity snip...]
On the first positive peak, some current flows through the base

Current does not flow. Current IS flow.

I agree (in a manner of speaking) that changing this to:

" some charge flows through the base emitter junction"
is "correct", but I maintain that it is _just as
understandable_.

However, if I really wanted to use the word, current, how would you word
it? What does current do when it does its thing? Since "flow" is included
in the term "current", would I say:
" On the first pulse, some current happens through base emitter junction." ?
We don't talk that way.

The current is some quantity of electrons, right.
Er, no. Current is quantity of charge per unit time.

Certainly it is. However, the point is that the capacitor develops a
potential difference across its two terminals because of the difference of
charge between the two sides. I maintain that my use is common usage in the
field. I agree that I could have said:
"The current is some quantity of electrons per time, right."
I believe that either way, the following sentence would provide the
necessary train of thought.

Well, these electrons will start to "fill up" or charge the
capacitor...

Then you say:

A capacitor does not charge.

It is completely common to refer to "charging a capacitor". You will
probably be very surprised that we also refer to "charging an incuctor".
This use of the word "charge" no longer refers to electron charge (in both
cases), though that is where the term originated. This "charge" is a
generic term meaning, perhaps, to "impart come analog quantity", or "fill
with some desired substance" which can be electrons, magnetic field, or
Halon. Just like we may say to "charge a battery" or "charge" one of those
glow-in-the-dark things by holding it near a strong light, then turning off
the light to watch it glow brightly. Although we commonly "load" a spring,
we could also use this convention ( though unconventional for a spring) and
charge it in some situation. We "charge" fire extinguishers, but no current
flows there eigher and we "charge" gasoline at the pump. Still no
electrons. It's common usage which may be imperfect, but accepted and
understood as a result.

Going in to depth on a capacitor, you say...

The net change of charge is zero for a
capacitor going from zero volts to its breakdown voltage. That is because
for each amount of charge stored on one plate, the same amount of charge
leaves the opposite plate. Now it takes energy to accumulate or deplete
like charges (electrons in this case) because they do not like to get
close
together. This energy is stored in a electrostatic field proportional to
the square of the voltage. Therefore a capacitor should be described as
being "energized", not "charged". Ratch

While this is a detailed description of what may be some of the
properties of electric charge and the capacitor, I did not believe that
going down to this level of information helps understand how the Base bias
goes negative in the target circuit. The operation of the capacitor is a
lower level concept which I was hoping the OP had some understanding of. If
not, then more detail of capacitors would be evident from the OP's
follow-up.

I'll do what I accuse you of doing and see how you respond.
I have to give the statement "net change of charge is zero for a
capacitor going from zero volts to its breakdown voltage" some analysis.
First, I don't think capacitors "go" anywhere when we drive electrons
onto one side. (do they sit in the back seat or front?)
Second, I could argue that there is a *change* of charge because there
are more electrons on one of the capacitor foils than the other...but that's
another thread. I'm just trying to reenforce my point that our use of
terminology is imperfect, but in some circles it is accepted.
Oh yes, I don't believe electrons "like" or "dislike" anything... I
don't think they can. They do, however, appear to have a repelling force
when brought near each other.


Whether or not the net charge changes, we certainly have rearranged that
charge such that there is one hell of a potential differennce between the
capacitor plates. That is the point.

Gee, do modern capacitors have "plates" any more... so...what do I call
these things, eh?



We "energize" circuits and light bulbs. Does that mean that there must be
some stored charge differential, as in a capacitor, in order for us to use
that term?


....Didn't see any more of your comments...

I maintain that it is common and acceptable for:
current to flow.
Capacitors to charge and discharge.

By the way, even though *you* also use some of this imperfect terminology, I
know precisely what you **mean** when you say:
"for a capacitor going from zero volts to its breakdown voltage" - don't
go anywhere
"charge stored on one plate" - no plates in many a common cap these
days
"they do not like to get close together" - what other emotions do
electrons have...

The point of the thread is:
If I replace all of my terms with yours, will the explanation then tell the
OP why the inductor is needed ??


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

 

[Steve Nosko]

"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:f58rm0loi3e0r6ki4tp6pdnrfcgg2qia46@4ax.com...

On Tue, 12 Oct 2004 17:10:41 -0500, "Steve Nosko"
suteuve.nosukowicuz@moutouroula.com> 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 Nosko]

"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:kmutm0pljptt1onk6a01urao5uam3cupo6@4ax.com...

On Tue, 12 Oct 2004 17:10:41 -0500, "Steve Nosko"
suteuve.nosukowicuz@moutouroula.com> 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