# Impedance Matching

Discussion in 'General Electronics Chat' started by joshgreaves332, Jun 28, 2014.

1. ### joshgreaves332 Thread Starter New Member

Jun 28, 2014
4
0
Hi

this is a great source of info and a great forum. I wonder if someone could explain Impedance matching to me as they would to somebody who understands the operations of circuits, and roughly how the components behave.

For example in a filtering circuit, using the textbook example of a low pass LC circuit, why does this result in a band pass filter if the impedances are not 'matched'?

The textbook says:
"The capacitance and inductance in this filter circuit are attaining resonance at that point(to create the spike resulting in band pass), creating a large voltage drop around C1, which is seen at the load, regardless of L2's attenuating influence"

Does this mean that C is resonating with both L1 and L2 in parallel? What does it mean that the voltage drop on C is seen at the load? Is this because C and L1 are only resonating, not with L2 aswell? Surely they are the same value so there should be resonance, and in parallel which says to me that it should behave like an open circuit with max impedance like a parallel resonance circuit should?

It continues
". The output voltage to the load at this point actually exceeds the input (source) voltage! A little more reflection reveals that if L1 and C2 are at resonance, they will impose a very heavy (very low impedance) load on the AC source, which might not be good either."

Why is the output voltage to the load higher than the source?
If L1 and C are resonating, are they not in parallel? Why do they not impose a very heavy HIGH impedance on the source? Why might this not be good?

thanks in advance for any assistance; I feel its a simple problem just cant seem to see what I am missing!

2. ### dougc314 Member

Dec 20, 2013
38
11
First of all, assuming your asking about the fifth circuit, labeled "low pass filter" which consists of a L-C-L "tee" with a voltage source driving and a resistor terminating, the comment of changing a lowpass into a bandpass is misleading IMHO. It is a lowpass filter. DC passes, "high" frequencies don't. At some point, in between, due to the ratios of the part values (including the resistances across the input and output) the circuit has some voltage gain at an AC frequency.

We can gain some insight by looking at extremes. Consider R to be arbitrarily high. It starts to look like simply L1 and C in series and the circuit is resonant at about 503 Hz. Now, the resistance in the L1-C-V loop is very low, in fact 0 except for the fact that perfect parts are unobtainable, so, at series resonance where the only impedance is the real part the resistance across V1 is very small, and that means lots of current. The voltage across the capacitor will be very big, and appear across R because we made it arbitrarily big (could even be open).

Now if we make R very small it begins to look like its making L2 be across (in parallel with) C1 and L1. (The impedance of the voltage source is 0). That makes L1 and L2 form a 50 mH inductance and the circuit resonates at about 711 Hz. Because R is small (could be nearly 0) the voltage across can be as small as we care to make it. With values of R in between small and big we will get an AC peak somewhere between 503 and 711 Hz, and there could be AC gain or loss there. To get the exact numbers we have to calculate the impedance's for each frequency and write the loop equations. In between DC and the AC peak the output voltage can droop to be very much less than then the input. Above resonance the impedance of the C gets smaller and smaller, the L's bigger and bigger so the output always falls off, another characteristic of a lowpass filter.

Note that as long as R isn't 0 the DC voltage across R is Vin, and its a low pass filter.

3. ### dougc314 Member

Dec 20, 2013
38
11
The voltage across the resistor as a function of w (2*pi*f) is:
R*(j*R-L*w)/(R+L*w*j)/(C*L^2*w^3-C*R*L*w^2*j-2*L*w*R*j) X Vin

where j is the sqrt(-1). These equations get very algebraic intensive, even for simple networks. It also requires calculating The magnitude of a complex number. I got this by using Mathcad's algebraic solver, I didn't do it by hand.

Computers are a wonder invention for solving and understanding the details of how these circuits work.

There is a solution for the magnitude of the output voltage, but it escapes me. In Mathcad its just |V|.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
The book states ...

Interestingly there is no C2 in the circuit so this is presumably an incorrect annotation.

Frankly, I don't get the point of the discussion about adding source resistance to achieve impedance matching. One effect of adding source resistance is to increase the circuit damping which will flatten the resonance peak which occurs with zero source resistance. This also clearly reduces the source loading effect at the resonant condition. It's a little like comparing apples and oranges.

I have attached a more detailed graphical comparison of the "matched" and "unmatched" cases. This (hopefully) gives a clearer perspective of the two responses. One could alternatively interpret the unmatched case as a low pass filter with a somewhat peaky response near resonance. The matched case is clearly flatter. The differences aren't surprising given the aforementioned reasons. The Q in the matched case will be close to unity - in contrast with the unmatched value of around 3 (when viewed as a band-pass filter).

As a further comment: one often finds the typical LC filter network applied to DC rectifier filtering is the capacitive input LC filter or the PI filter section. This doesn't appear to even rate a mention in the discussion. An omission which seems curious.
In any event, one tends to set the filter resonant frequency well below the ripple frequency present on the rectifier output. With a 120Hz ripple frequency present on a full-wave rectified 60Hz output, one would typically design the filter resonant frequency to be no more than a tenth of the ripple frequency ~ 12Hz.

• ###### Filter Graphical Comparison.jpg
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Last edited: Jun 29, 2014