# What's this inductor doin'?

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

1. ### Steve EvansGuest

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??

2. ### Joe RocciGuest

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

3. ### Joe RocciGuest

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

4. ### John PopelishGuest

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.

5. ### John PopelishGuest

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

6. ### John FieldsGuest

---

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.

7. ### Reg EdwardsGuest

The input impedance of the transistor is capacitive. So the inductor very
likely resonates with it at the working frequency.

8. ### Paul BurridgeGuest

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. ### Doug Smith W9WIGuest

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 EvansGuest

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

11. ### DboweyGuest

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
up?
Thanks,

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

Don

12. ### Michael BlackGuest

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. ### Paul BurridgeGuest

Perhaps if you posted the proper schematic it would help. What's the
transistor in question?

14. ### Steve EvansGuest

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.

15. ### Roy LewallenGuest

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

16. ### Paul BurridgeGuest

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 PennellGuest

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

18. ### Steve EvansGuest

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

19. ### Steve EvansGuest

[snip]
<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. :-(

20. ### Robert MonsenGuest

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.