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
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

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
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...arch+th e+Web

Welcome to the zoo!
Rich

"Steve Evans" wrote in message
...
On Sun, 10 Oct 2004 14:01:44 +0100, Paul Burridge
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,
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 wrote in message
...

"Steve Evans" wrote in message
...
On Sun, 10 Oct 2004 14:01:44 +0100, Paul Burridge
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.

On Mon, 11 Oct 2004 21:01:46 GMT, "Joe Rocci" wrote:

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.

No Joe! The frequency is 145Mhz throughout.

An aside to Steve... I'm pouring over your expansive explanation.
It'll take a while to sink in, though!

Thanks,

Steve.
--

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

On Mon, 11 Oct 2004 14:59:11 -0500, "Steve Nosko"
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.

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 wrote in message
...
On Mon, 11 Oct 2004 14:59:11 -0500, "Steve Nosko"
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.

"Steve Evans" wrote in message
...
On Mon, 11 Oct 2004 14:59:11 -0500, "Steve Nosko"
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
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
will start to "fill up" or charge the capacitor. Each time another positive
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.

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
(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.

On Mon, 11 Oct 2004 23:54:00 GMT, "Joe Rocci" wrote:

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...


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

steve
--

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