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.
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.
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??
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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.
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.
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?
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"What is now proved was once only imagin'd." - William Blake, 1793.
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
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Fat, sugar, salt, beer: the four essentials for a healthy diet.
A capacitor has a very low impedance to high-frequency (i.e., 145MHz)
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,
Two little problems here:
You should have posted more of the schematic so a conlusive evaluation could be made.
You aren't bright enough to understand the offered insight.
Are you wondering why the base of the transistor is at DC ground, or wondering what the RF choke does to ensure that the base of the transistor is not at ground for RF?
People have answered what they think you are asking, which isn't really clear. Clarify that, and you might get an answer you want.
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.
Thanks Michael, I'm sorry I can't post the full diagram as my scanner's bust. I'll try to clarify. I know why the base of the transistor (it's a 2n5771, in answer to Paul's question) is at DC ground. I'm not so dumb as to realise that a choke passes dc but not the high frequncy RF. I guess it boils down to this: how did the designer arrive at the given value of 0.4uH for this inductor? Why not 4nH? Why not 40uH? Why not 100mH?? What's the deal with this value and since I've only got a 1uH in my junk box, will that be okay instead?
Thanks, all,
Steve.
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Fat, sugar, salt, beer: the four essentials for a healthy diet.
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...
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"What is now proved was once only imagin'd." - William Blake, 1793.
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.
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
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Fat, sugar, salt, beer: the four essentials for a healthy diet.
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.
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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.
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