Connect with us

What's this inductor doin'?

Discussion in 'Electronic Basics' started by Steve Evans, Oct 3, 2004.

Scroll to continue with content
  1. Steve Evans

    Steve Evans Guest

    Hi everyone,

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

    --------------capacitor-------------------------------transistor base

    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??
  2. Joe Rocci

    Joe Rocci Guest

    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.

  3. Joe Rocci

    Joe Rocci Guest

    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.

  4. 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
  5. Right. I didn't pay any attention to the given values. They produce
    a resonance around 8 megahertz.
  6. John Fields

    John Fields Guest


    At 145 MHz, the reactance of the cap is:

    Xc = -------

    = ------------------------------ ~ 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:

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

    it's tuned to

    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.
  7. Reg Edwards

    Reg Edwards Guest

    The input impedance of the transistor is capacitive. So the inductor very
    likely resonates with it at the working frequency.
  8. 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?
  9. 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.
  10. Steve Evans

    Steve Evans Guest

    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

  11. Dbowey

    Dbowey Guest

    Steve (the OP) posted:

    << On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI
    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

    Two little problems here:
    1. You should have posted more of the schematic so a conlusive evaluation
    could be made.
    2. You aren't bright enough to understand the offered insight.

  12. Then rephrase the question.

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

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

    Michael VE2BVW
  13. Perhaps if you posted the proper schematic it would help. What's the
    transistor in question?
  14. Steve Evans

    Steve Evans Guest

    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,

  15. Roy Lewallen

    Roy Lewallen Guest

    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

    Roy Lewallen, W7EL
  16. 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...
  17. Jim Pennell

    Jim Pennell Guest

    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

  18. Steve Evans

    Steve Evans Guest

    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!

  19. Steve Evans

    Steve Evans Guest

    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. :-(
  20. 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.

    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.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day