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What's this inductor doin'?
Hi everyone,
Below you will find my attempt to show in text-form, a circuit fragment from a 145Mhz amplifier: --------------capacitor-------------------------------transistor base | | I | coil | | | | ------------------------------------------------------------GND The cap's value is 1nF; the inductor's is 0.4uH. The cap (I assume) is to couple one amplifier stage into the next (50ohm source/load) with minimal attenuation of the desired VHF signal. But like what's the purpose of this inductor to ground?? -- Fat, sugar, salt, beer: the four essentials for a healthy diet. |
Steve Evans ) writes: 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?? It's there to bias the transistor. In this case, the base of the transistor (I'm assuming bipolar so I say base) is at DC ground. But if you just shorted the base to ground, the signal could not get into the transistor. So you put that RF choke there, so the base sees ground at DC, but the signal sees a relatively high impedance. MIchael VE2BVW |
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 wrote in message ... 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. |
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 wrote in message ... 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 |
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 |
Joe Rocci wrote:
John Popelish wrote in message ... 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 |
On Sun, 03 Oct 2004 15:08:22 GMT, 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?? --- 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 |
The input impedance of the transistor is capacitive. So the inductor very
likely resonates with it at the working frequency. ---- Reg, G4FGQ |
On Sun, 3 Oct 2004 19:20:22 +0000 (UTC), "Reg Edwards"
wrote: The input impedance of the transistor is capacitive. So the inductor very likely resonates with it at the working frequency. You might be on to something here, Reg. Maybe the inductor's there to 'neutralise' the transistor's input capacitance. The parallel tuned circuit formed by the inductor and the transistor input capacitance would have a maximum impedance at 145Mhz if the transistor's (capacitive) input impedance were about 3pF., which doesn't sound far out for an RF small-signal tranny. Without that inductor, sure there'd be no bias on the base, but additionally, the input capacitance of the transistor will shunt away much of the VHF input signal to ground. Does that make sense? -- "What is now proved was once only imagin'd." - William Blake, 1793. |
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 |
On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI
wrote: 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). Sure, I'm aware of all this basic crap, but every reply I've had to this question has thrown up a different answer and none of them make much sense. Can't you guys come up with something in common that adds up? Thanks, Steve -- Fat, sugar, salt, beer: the four essentials for a healthy diet. |
On Sat, 09 Oct 2004 22:40:20 GMT, Steve Evans
wrote: On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI wrote: 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). Sure, I'm aware of all this basic crap, but every reply I've had to this question has thrown up a different answer and none of them make much sense. Can't you guys come up with something in common that adds up? Perhaps if you posted the proper schematic it would help. What's the transistor in question? -- "What is now proved was once only imagin'd." - William Blake, 1793. |
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On Sat, 09 Oct 2004 22:40:20 GMT, Steve Evans
wrote: On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI wrote: 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). Sure, I'm aware of all this basic crap, but every reply I've had to this question has thrown up a different answer and none of them make much sense. Can't you guys come up with something in common that adds up? it looks /IIRC schematics posted/ that that is a some kind of RF power amp in "C" class .... that inductor keeps shorted transistor base for DC to ground & capacitor keeps away DC from previous stage. Also that CR circuit is an impedance transformer (from previous higher to transistor lower RF Z input & also a high pass filter @ the same time. If enough RF current would pass capacitor to rise the potential over few hundred mV RF, every half-period of the signal will open transistors E - C gate to conduct thru (parallell) LC tank that should be connected there from C transistor terminal to power line. LC tank will recover after one period also other (opposite) semiperiod, so you will have a complete sinusoide (RF) to transmit (or amplify it further).. usefull only for CW, modul.CW or FM (not AM or SSB). Hope I helped & understood the schematics & question. -- Regards, SPAJKY ® & visit my site @ http://www.spajky.vze.com "Tualatin OC-ed / BX-Slot1 / inaudible setup!" E-mail AntiSpam: remove ## |
Steve Evans wrote:
. . . 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? I'm guessing you're not experienced enough to realize that inductors are far from ideal. Among their many imperfections, the most problematic in an application like this is shunt capacitance. This resonates with the inductance to create a parallel resonant circuit. At resonance, the impedance is very high -- higher than that of the inductance alone. But above the self-resonant frequency, the impedance drops, and at some point becomes less, then very much less, than the impedance of just the inductance. Well above self-resonance, all you see is the self capacitance -- it looks like a capacitor, not an inductor. That's why a designer doesn't just use 100 mH for everything. Just about any 100 mH inductor looks like a capacitor at 145 MHz, with a very low impedance, much lower than an inductor with smaller inductance value.(*) The trick is to choose an inductor that's below or at its self resonant point while still having enough impedance so it doesn't disturb the circuit it's across. A good rule of thumb is an inductor whose reactance is about 5 - 10 times the impedance it's across. (In your case, this might not be easy to determine. You might be able to get it either from knowing someting about the previous stage, or from the S parameter specfication of the transistor.) 5 - 10 times is usually enough, and if you try for too much, it won't work any better and you run the risk of being above the self-resonant frequency. Of course, an individual inductor can be measured to make sure its impedance is high enough at the frequency of use, if you have the equipment to make the measurement. So, now, is your 1 uH ok? It depends on its shunt C, which depends on its construction. If you can't measure it, just try it. The worst that's likely to happen is that it'll kill the signal (due to low impedance). At only 2-1/2 times the value of the original, there's a good chance it'll work ok. If it doesn't, and if your 1 uH inductor isn't potted, you can get the value down to 0.4 uH by unwinding about 1/3 of the turns. (*) 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. Roy Lewallen, W7EL |
On Sun, 10 Oct 2004 11:57:58 GMT, Steve Evans
wrote: 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? 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... -- "What is now proved was once only imagin'd." - William Blake, 1793. |
On Sat, 09 Oct 2004 22:40:20 GMT, Steve Evans
wrote: On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI wrote: 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). Sure, I'm aware of all this basic crap, but every reply I've had to this question has thrown up a different answer and none of them make much sense. Can't you guys come up with something in common that adds up? The problem is, there are quite a few possibilities for what the inductor is doing. For example, it might be providing a DC path to ground for the base of the transistor, or it might be supressing a parasitic oscillation in the amplifier. Or, for that matter, it might be part of an impedance transform from one section to another. Without the complete schematic, there is no way to tell. Jim N6BIU |
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 -- Fat, sugar, salt, beer: the four essentials for a healthy diet. |
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. :-( -- Fat, sugar, salt, beer: the four essentials for a healthy diet. |
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! 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. The Q of a resonant circuit is defined as the resonant frequency divided by the width of the 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. |
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. |
Thanks for the explanation, Joe, but no. I never said anythink about
multiplicaiton Steve, Can you send me a copy of the original post? When these things happen, I like to see where my crazy notions came from. Joe W3JDR |
Steve Evans wrote:
Sure did, Steve! I ran it through a spice program and you're right in every detail. So there's obviously some basic flaw in my understanding of caps. The textbooks say to treat a cap as a short circuit at AC (assuiming its reactance isn't too high at the frequency of interest). That would appear to be grossly misleading as there's a huge difference between the mean voltage levels on each side. . . The mean voltage on each side is the DC component. The fact that it's different on the two sides illustrates the fact that the capacitor is an open circuit to DC. The shape of the waveforms on the two sides of the capacitor are the same. That illustrates that the AC component is the same on both sides -- the capacitor is a short circuit to AC. It's gonna take me a while to get this trough my thick skull. :-( So what happens when you have a small ac ripple riding on a DC bias? The bias (DC) is removed, and the ripple (AC) is passed through. I'll spice it but won't understand it, I guess. :-( Roy Lewallen, W7EL |
On Wed, 13 Oct 2004 22:58:47 GMT, "Joe Rocci" wrote:
Thanks for the explanation, Joe, but no. I never said anythink about multiplicaiton Steve, Can you send me a copy of the original post? When these things happen, I like to see where my crazy notions came from. Here it is in its entirety.... Hi everyone, Below you will find my attempt to show in text-form, a circuit fragment from a 145Mhz amplifier: --------------capacitor-------------------------------transistor base | | I | coil | | | | ------------------------------------------------------------GND The cap's value is 1nF; the inductor's is 0.4uH. The cap (I assume) is to couple one amplifier stage into the next (50ohm source/load) with minimal attenuation of the desired VHF signal. But like what's the purpose of this inductor to ground?? -- Fat, sugar, salt, beer: the four essentials for a healthy diet. |
On Tue, 12 Oct 2004 17:10:41 -0500, "Steve Nosko"
wrote: This could be viewed as what we call a "clamper circuit" The AC voltage AT THE BASE will have its positive peak "clamped" or moved to +0.7 volts and the waveform will extend from there as negative as the waveform is tall. IF we had 5 volts peak (10 volts peak-to-peak) the cap left side, the base voltage will swing from +0.7 volts to -9.3 volts. Okay, upon further thought about this there's still something amiss in my understanding. I take what you say about the cap blocking out the DC component of the waveform to leave the AC largely unaffected. However, the term "clamping" AIUI means the diode lops off anything over about 0.7 volts from the input waveform (ie, it conducts it away to ground) so around half of it is lost (half wave rectification). Now you state (and the spice progs agree) that what *actually* happens in this case is that the whole AC waveform gets shifted south into negative territory. It's still a full wave, but it's way down into the negative and only the highest peaks just creep above zero volts. Is this effect *solely* attributable to the steady build-up of negative charge on the cap's RHS? I think what's really freaking me out here is the fact that the signal source is grounded on the neg. side and yet we have that same signal that after going through a cap can end up going fully negative *below* ground. It just seems like any such voltage beneath zero/ground potential is breaking the laws of physics. Ground should be the 'absolute zero' of the potentials in any circuit and here it is being violated. I need some help to get my thick head around the concept! :-( -- Fat, sugar, salt, beer: the four essentials for a healthy diet. |
Thanks for re-posting that Steve. I don't know where I got the crazy idea
that this was a multiplier stage....maybe I'm confusing this with a different thread. Thanks Joe W3JDR Steve Evans wrote in message ... On Wed, 13 Oct 2004 22:58:47 GMT, "Joe Rocci" wrote: Thanks for the explanation, Joe, but no. I never said anythink about multiplicaiton Steve, Can you send me a copy of the original post? When these things happen, I like to see where my crazy notions came from. Here it is in its entirety.... Hi everyone, Below you will find my attempt to show in text-form, a circuit fragment from a 145Mhz amplifier: --------------capacitor-------------------------------transistor base | | I | coil | | | | ------------------------------------------------------------GND The cap's value is 1nF; the inductor's is 0.4uH. The cap (I assume) is to couple one amplifier stage into the next (50ohm source/load) with minimal attenuation of the desired VHF signal. But like what's the purpose of this inductor to ground?? -- Fat, sugar, salt, beer: the four essentials for a healthy diet. |
"Steve Evans" wrote in message
... On Tue, 12 Oct 2004 17:10:41 -0500, "Steve Nosko" wrote: [snip] IF we had 5 volts peak (10 volts peak-to-peak) the cap left side, the base voltage will swing from +0.7 volts to -9.3 volts. WHEW ! Did this work Sure did, Steve! I ran it through a spice program and you're right in every detail. So there's obviously some basic flaw in my understanding of caps. The textbooks say to treat a cap as a short circuit at AC C O O L !!! So my explanation did it...sorry for the gloat folks. Had me scared there for a minute...30+ years doing and teaching electronice not for naught after all. This is super, Steve Evans, Now you're learning. Now, look at the Spice circuit _after a bunch of input cycles_ and monitor the voltage on BOTH sides of the cap and you will see that they are both the same **AC wise **, but there will be a constant, or DC difference. You can use the difference probe and you will only see a constant (DC) difference in potential across the cap. If you look at each side individually, you will see that the AC signal (yes I know "AC Signal" is redundant, but I think it helps clarify) is identical on both sides, though shifted by that DC amount. You can even change the waveform. Put *TWO* AC sources in series, one less voltage than the other, and different frequencies. The two cap sides will still have the same waveform. The concept you quote of "cap as a short circuit at AC" is true, but more easily seen in the steady-state. This is what we call it when all of the transient effects have died out. These transient effects are usually capacitors getting charged up (sorry Ratch, but I gotta say it). The AC component will "pass through" the cap un-impeeded, BUT you can also have this DC offset from one side of the cap to the other. In the case we have focused on (coupling cap & NO inductor) you have a little more complexity throwing in a monkey wrench. The Base-Emitter junction/diode is rectifying the AC and charging up the cap _just like it does_ in the common power supply rectifier circuit--but in this case it makes trouble for us by reverse biasing the base. I am glad to help, but as you see, you have sort of jumped into the middle of circuit theory armed only with the "AC short" part of the capacitor's characteristics. My bet is that you'll remember this very well. They say: "we learn from our mistakes". Well, BOY! ARE WE LEARNIN' NOW! 73, -- Steve N, K,9;d, c. i My email has no u's. |
"Steve Evans" wrote in message ... On Tue, 12 Oct 2004 17:10:41 -0500, "Steve Nosko" wrote: This could be viewed as what we call a "clamper circuit" The AC voltage AT THE BASE will have its positive peak "clamped" or moved to +0.7 volts and the waveform will extend from there as negative as the waveform is tall. IF we had 5 volts peak (10 volts peak-to-peak) the cap left side, the base voltage will swing from +0.7 volts to -9.3 volts. Steve Evans responds: Okay, upon further thought about this there's still something amiss in my understanding. I take what you say about the cap blocking out the DC component of the waveform to leave the AC largely unaffected. However, the term "clamping" AIUI means the diode lops off anything over about 0.7 volts from the input waveform (ie, it conducts it away to ground) so around half of it is lost (half wave rectification). DAMN! And I sort of thought of this as I was writing it... STUPID ME! I did what I (in my mind, usually) will normally criticize others for doing soften here in an attempt to "help" others "fully understand". Namely, adding some additional explanation or extreme detail which only serves to further confuse the OP. Please excuse me. While not necessary for understanding this circuit, I'll fill-in this bit here. A "clamper" is a diode and cap circuit which will clamp a particular point on a waveform to a specific voltage (kinda like being clamped in a vice on the bench), but NOT change the wave's wiggling *shape*. Sometimes this is needed. In the coupling circuit in question, it is the POSITIVE PEAK of the signal that gets clamped to +0.7 volts. The wave's SHAPE is un-changed, but the whole thing is shifted in its DC component. What you are thinking of is called a "CLIPPER" because it CLIPS *off* part of the waveform like barber's scissors. Back to the actual subject. In general, any waveform can have an AC component and a DC component. The DC component only serves to "shift" the position of the signal up or down; which can also be called a DC offset. Now Continues a wiser Steve Evans: you state (and the spice progs agree) that what *actually* happens in this case is that the whole AC waveform gets shifted south into negative territory. It's still a full wave, but it's way down into the negative and only the highest peaks just creep above zero volts. Is this effect *solely* attributable to the steady build-up of negative charge on the cap's RHS? YUP ! I think what's really freaking me out here is the fact that the signal source is grounded on the neg. side and yet we have that same signal that after going through a cap can end up going fully negative *below* ground. It just seems like any such voltage beneath zero/ground potential is breaking the laws of physics. Ground should be the 'absolute zero' of the potentials in any circuit and here it is being violated. I need some help to get my thick head around the concept! :-( Oooooo. BEWARE! GROUND IS NOT ABSOLUTE ! ! ! Nope, nope, nope. This is going to take some time to explain and more experience/study will be needed for you to really _get it_. The bad news for you may be that ground, I must sadly inform you, is relative. There is no one, solid, never varying, absolute thing which is ground, except in our imaginations. Many hams believe there is, but there isn't. That being said, let's start out simply and build. Here is a very applicable analogy: Voltage, also called "potential difference", is a lot like altitude -- height. We can talk about the height above the street level. We might consider the street level to be "ground". In Physics, moving some object to a higher level gives it the "Potential" to do damage if it falls on your head, so the "potential energy" of it is greater. Holding it three feet above your head gives it a certain potential, right? Voltage is just like this. HOWEVER, what about standing on the roof of a building THEN moving the same object three feet above your head. You must agree that it has the SAME potential to do damage TO YOUR HEAD that it did in the first example, right? In this case, the roof of the building is our ground. So ground is relative and *we* get to pick it. It is usually a known point in our circuit and we use special symbols to show it. Note that this is why voltage is also is called potential *DIFFERENCE*. Unfortunately, this analogy will fall apart when trying to use it for negative voltage, if we put this negative voltage "Below" our ground, but for the "relative" concept, I hope it worked. Now I'll try *negative*. Ground is a REFERNCE to which we relate all the other voltages in the circuit we happen to be talking about. Lets put a screw into a wooden board and call it our ground. Work with me here, Steve. Connect a wire and run it over to several other screws in the board. They are all now our "ground". Take your basic Ray-O-Vac flashlight battery. [[ actually, for Ratch, this is a "cell", but common usage dies hard. A "battery" is a collection of cells usually connected in series to get a greater voltage ]]]. Run a wire from the negative terminal of the battery...oops, cell, and connect it to our ground. Now, with the negative lead (rhymes with seed) of our volt meter connected to our ground, measure the voltage on the Ray-O-Vac positive terminal with the positive lead of the voltmeter. We get + 1.5 volts. In electronics (when we want to study some phenomenon carefully) we have a convention of ALWAYS putting the NEGATIVE lead of the voltmeter on our ground. That way we can easily see if the voltage it is positive or negative. NOW... reverse the connection on the Ray-O-Vac. You will now measure a NEGATIVE 1.5 volts. So what's the way to interpret this.? Voltage (potential difference) is also called Electro Motive Force because it is the "force" that "pushes" the electrons around the circuit to form Ratch's (and everyone's) current flow (sorry, Ratch, couldn't resist pulling your chain a bit in some friendly ribbing (:-). When we reverse the battery/cell, we are now "pushing" in the other direction. It is still a potential difference, but negative. In MY model (the one in my head), I visualize this as being "physically" below ground (and we often refer to it as being "below ground") in the sense that on the oscilloscope and Pspice display it is below on the graph. If you REALLY want to make the altitude analogy work, you can imagine that gravity always "pulls" things toward the "ground" - but this is YOUR local ground. Then the negative potentials "pull you up". END.hope this helps, Steve. 73 |
Sure did, Steve! I ran it through a spice program and you're right in
every detail. ======================= What makes you think Spice is correct? Its only a buggy computer program. Rubbish in - rubbish out! |
On Fri, 15 Oct 2004 23:06:38 +0000 (UTC), "Reg Edwards"
wrote: Sure did, Steve! I ran it through a spice program and you're right in every detail. ======================= What makes you think Spice is correct? Its only a buggy computer program. Rubbish in - rubbish out! Because it agreed exactly with Steve's prediction. I think therefore I'm entitled to rely on it in this instance at least. -- "What is now proved was once only imagin'd." - William Blake, 1793. |
On Fri, 15 Oct 2004 23:06:38 +0000 (UTC), "Reg Edwards"
wrote: What makes you think Spice is correct? Its only a buggy computer program. Rubbish in - rubbish out! BTW, Reg. Is there a program on your site that can handle stripline calculations for UHF circuit board inductors? -- "What is now proved was once only imagin'd." - William Blake, 1793. |
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