# impedance matching and dynamic impedance

Discussion in 'Electronic Design' started by rob, May 17, 2004.

1. ### robGuest

Hi to all.
I am trying to match a 50 ohm source to a parallel tuned cct at
13.56Mhz.The parallel
cct consists of 140nH // 984pF . The actual cap would be a fixed value
ie 820p in parallel with a variable cap.
The problem is that I don't know the resistance of the parallel tank
at resonance.The theoretical resistance (perfect inductor , cap) is
infinite. The DC resistance of the coil is less than 0.1 ohms. This
would give a dynamic impedance of 2.8K ohms if R is 0.05 ohms. ( Rd =
L/R.C)
This does not take into effect skin effect of the wire and other
parasitics.

I then thought I would try and measure the impedance of the cct at
resonance.
Would the following method give acceptable results or am I barking up
the wrong tree!!

1. Connect the tuned cct to a signal generator via a 1K resistor.
2. Set the signal gen to 1V out at 13.56Mhz
3. Tune the resonant cct to resonance using a scope (watch for peak)
4. calculate the dynamic impedance using basic voltage divide rule.
Peak voltage across
tank being Vx. Vx = (1v * Rx) / ( Rx + 1K)

Using this method I get a value of 315 ohms.
Does this sound at all correct? I have not taken into account the
capacitance added by the scope probe. I assume this should not make to
much difference to the result , because once removed the cct can be
re-tuned.

I would then match 315 ohms to 50 with a l-match.

As I said before , would this method give acceptable results.

Cheers
Rob

2. ### Tony WilliamsGuest

Q's the other day (related to another thread). The
sums below might give you some ideas.

+---------+--------+
| | |
[Rg] [Rl] |
| | |
Vin----+ [L] ===C
| |
0v----+---------+--------+

The sketch above is a parallel tuned circuit.
L+Rl and C being the primary elements, which are
driven/loaded by an external generator, of source
resistance Rg.

You want to know the unloaded Q of just the tuned
circuit. Call this Q. When driven/loaded by Rg
then the circuit has a lower Q, call this Qe.

Q
Qe = -------------- is the important sum.
1 + wo*L*Q/Rg

Measure Qe, (by using the 0.707 bandwidth points when
Rg is in circuit). You already know wo, L and Rg, so
the unloaded Q can be estimated.

If Q >10 then Z(unloaded) = Q * sqrt(L/C) approximately.

All sums pinched out of "Electron Tuned Circuits", by
Samuel Seely.

3. ### Joe McElvenneyGuest

Hi,

I take that the ||-tuned tank is not working into an open circuit
as there wouldn't be any point in it otherwise. If so, then its load
resistance will likely elbow out any series 'R' as the main contender
for circuit losses.

Cheers - Joe

4. ### John FieldsGuest

---
See below...
---
---
To find out, I tried a couple of experiments using this circuit:

1.25MHz>--[50R]--+------------>Ein
|
[Rs] <--100k, 10k, 1k, 100R
|
+-------+---->Eout
| |
[100pF] [150µH]
| |
GND--------------+-------+---->0V

with the different series resistors shown, just to see what would
happen, and here's how it went:

Considering that the reactances of the L and C in the tank (at

Xc = Xl = 2pi f L

= 6.28*1.25E6Hz*1.5E-4H ~ 1200 ohms,

One might think that after redrawing the circuit to look like this:

Ein
|
[Rs]
|
+---->Eout
|
[Rp]
|
GND

Then setting Rs to 100k ohms, Rp to 1200 ohms (the reactances of the
tank components) and ignoring the 50 ohm generator resistance that
Eout would be:

Eout = (Ein Rp)/Rs + Rp

= (1*1200)/100k + 1200

= .012V

However, because of the Q of the circuit at resonance, the actual
voltage measured at Eout was 0.15V, so since the tank is purely
resistive at resonance if we rewrite Eout = (Ein Rp)/Rs + Rp to solve
for Rp, we get:

Rp = (Eout Rs)/Ein - Eout

and solving for Rp gets us:

Rp = (Eout Rs)/Ein - Eout

= (0.15*100k)/1 - 0.15

= 17,647 ohms

Doing the experiment again with Rs = 10k yields a measured Eout of
0.63VPP, which corresponds to a tank impedance of 17,027 ohms. Pretty
close, (about a 3.5% error) so it looks like we're on the right track.

Dropping Rs down to 1000 ohms and then down to 100 ohms yields just
about 1V out for 1V in which is consistent with the tank being a much
higher impedance than Rs, so I'd have to say that if that's the way
you're doing your measurements, you're doing them right.

5. ### Tom BruhnsGuest

Can you make an estimate of the coil's unloaded Q from its
construction? If it's an air-core single-layer solenoid coil, you
should be able to get quite close. Same if it's wound on a powdered
iron toroid core of known characteristics. If nothing else, you can
put an upper bound on it by considering the RF resistance of the wire
its wound with, assuming you can determine that. This would be a
double-check on the measurement you make. You can make it per Tony's
diagram, but you can also do it by coupling lightly to the tank
circuit (with a tiny capacitor, say 1pF, from a signal generator) and
measuring the 3dB bandwidth.

Another way to do it is just make an educated guess what L-network
values you'll need and tune them until you get a match to 50 ohms,
assuming you have a way to measure that match. Then you have the
values you need for the L-network, and if you can measure them, you
can calculate the load you've matched to 50 ohms. (But read the down
to the last suggestion in this message for a much easier way!)

By the way, unless your coil is quite small, I'd expect an unloaded Q
of 100 or above at 13MHz. It should only take a coil less than 1cm
(less than 3/8 of an inch) diameter to get Q above 100, unless you
have lossy material around its field. So I'd expect a tank resistance
at resonance of more like 100*12 = 1200 ohms or above. Your 315 ohms
seems very low. Is it a very small inductor?

An L network will have its own losses, too, of course. You may find
that you can match to the tank more easily some different way. One
way that used to be quite common in radio transmitters, and is still a
fine way to couple, is a link of a couple turns near the "cold" end of
the tank coil. If you can make it adjustable coupling (by varying the
distance and/or angle) you should be able to adjust for a match. You
can also do it by putting a variable capacitor in series with the

Yet another way is to realize that your L matching network will have a
reactance shunt across your tank. It can be either an inductance or a
capacitance. Then to complete the match you put a series capacitance
or inductance back to the source. But if you say the L network will
have an inductance shunt across your tank, that's JUST the same as
decreasing the capacitance that's already in the tank. So you save a
coil, and just use a capacitor to the top of the tank. Assuming that
your tank is 1000 ohms at resonance, it takes a 54pF series cap back
to the source, and reducing your tank capacitance by about the same
amount, to about 933pF. Beware: tuning will be quite sharp! If
you're right about the 315 ohms at resonance, then it would be more
like 102pF series, and reduce the tank capacitance to 898pF. If you
have a high-Q coil and R=3000 ohms, the series cap is 30.5pF and the
tank shunt is reduced to 954pF. So you can see, with variables for
both the tank and the series cap, you can adjust over a rather large
range of coil Qs. Just be sure the frame of the series variable is
connected to the low-impedance point...the generator...and isolate it
from ground as best you reasonably can. (It will add a small amount
of capacitance to ground at that generator output node.)

Hope that helps...

Cheers,
Tom

6. ### Tam/WB2TTGuest

Why don't you measure the real part when the reactance has been tuned out,
and the loop is *coupled to the load*. BTW, I don't think you have a 50 Ohm
source; you have a source that is meant to drive a 50 Ohm load.

Tam

7. ### Tom BruhnsGuest

As one who is surrounded by network analyzers and signal generators
that really do have 50 ohm sources, I'd never assume that someone
doesn't actually have a 50 ohm source at hand, but in any event, does
it matter in this case? (I didn't get any sense from the OP whether
the excitation was going to be a few milliwatts or a few kilowatts.)

Cheers,
Tom

8. ### Tam/WB2TTGuest

Tom,

I think 13.56 is the industrial heating frequency (among ather things). So,
there is probably a class C or class D amp putting out kilowatts. He doesn't
say what his loop is coupled to. Is he melting aluminum, treating patients,
or what...

Tam

9. ### robGuest

Thanks for all the tips guys. The application is for a wireless tag
reader system. Only low power.Few watts at most.
Have got a working system with a wound coil. Put in a 47 ohm resistor
to bring down the Q and make matching easier. Works fine , but I want
to experiment a bit.
The "Load" is the inductivly coupled Tag.

chers
Rob

10. ### Reinier GerritsenGuest

Just a tip: read the application notes of Texas Instruments (Tag-it).
It has some (practical) info on antenna matching.

Reinier