What's this inductor doin'?

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.
Reply to
Steve Nosko
Loading thread data ...

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

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

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

Oh, OK - you've fixed it here.

Cheers! Rich

Reply to
Rich Grise

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

You need to look through these until something starts to make sense to you:

formatting link

Welcome to the zoo! Rich

Reply to
Rich Grise

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

Paul's

be

will

terminology."

is

near

the

the

and

some

Ccb

This

the

value

capacitor

board

about

circuit

Reply to
Joe Rocci

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.
Reply to
Steve Evans

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.
Reply to
Steve Evans

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

can

Reply to
Joe Rocci

can

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.
Reply to
Steve Nosko

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.
Reply to
Robert Monsen

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

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

Roy Lewallen, W7EL

Reply to
Roy Lewallen
[snip]

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.
Reply to
Steve Evans

a
a

must

discharged -

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

of

base

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

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

more

next

not

the

around

voltage

and

Reply to
Ratch

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.
Reply to
Steve Evans

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

Reply to
Joe Rocci

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?

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.

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.

capacitor...

Then you say:

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...

close

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.
Reply to
Steve Nosko

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.
Reply to
Steve Evans

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.
Reply to
Steve Evans

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

Reply to
Joe Rocci

base

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.
Reply to
Steve Nosko

voltage

and

base

Steve Evans responds:

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.

Continues a wiser Steve Evans:

YUP !

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
Reply to
Steve Nosko

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.