# Impedance matching concerns equivalent impedance of circuit?

Hello.

Please look at the attached image first. There are two circuits A and B which have individual electrical components a and b. Components a and b are electrically and directly connected by the red lines. Impedances of individual components a and b are not matched, however, equivalent impedances of circuits A and B are matched.

Now a signal travels from A to B. Then, is there a signal reflection due to impedance mismatching? Which impedances are important for the signal? Impedances of individual components at the ends of the connection or equivalent circuit impedances?

Paul Colby
Gold Member
Not certain I understand the "sub" impedances ##a## and ##b##? The circuit ##A## emits a signal then ##A## may be replaced by an ideal voltage source ##V_s## (with zero internal impedance) in series with impedance ##A##. The loop then formed is the impedance ##A## in series with ##B##. This forms a voltage divider. The voltage across ##B## is then ##V_B = \frac{B}{A+B}V_s##. The best match occurs when ##A=B## and ##V_B = \frac{1}{2}V_s## which is the maximum power transfer.

• goodphy
Svein
There are several answers to this.

The most common type of mismatch entails a transmitter impedance that is lower than the line impedance and a receiver impedance that is higher. With no impedance matching, you will have a positive reflection at the receiver end and a negative reflection on the driver end. We usually call this type of mismatch for ringing. Now there are several ways of reducing those reflections.
1. The parallel load version: Insert an impedance matching network in the receiver end. Since a wire (or a pair of wires) have a characteristic impedance in the 100 - 120Ω range, this will put a rather heavy load on the transmitter.
2. The serial load version: Insert an impedance matching network in the transmitter end. Since a low impedance transmitter usually has an output impedance om some tens of ohms, a serial resistor in the 22 - 47Ω range will usually work well. This type does not load the transmitter but cannot be used on multidrop lines.
3. The active load version: Imagine a circuit in the receiver end that helps terminate the line but also helps driving it (switching the load impedance between "high" and "low" as appropriate).

• goodphy and Paul Colby
Not certain I understand the "sub" impedances ##a## and ##b##? The circuit ##A## emits a signal then ##A## may be replaced by an ideal voltage source ##V_s## (with zero internal impedance) in series with impedance ##A##. The loop then formed is the impedance ##A## in series with ##B##. This forms a voltage divider. The voltage across ##B## is then ##V_B = \frac{B}{A+B}V_s##. The best match occurs when ##A=B## and ##V_B = \frac{1}{2}V_s## which is the maximum power transfer.

Yes, the circuit A can be seen as the ideal voltage source with some series impedance ZA, according to the Thevenin's theorem. Maybe I'm questioning what it is obvious for others. Here, the impedance matching problem may only concern ZA and ZB which is an equivalent impedance of the circuit B.

a and b are, for example, resistors in parallel connection with the red lines. So..the signal transmitted from A to B (A and B are assumed directly connected so we don't care of a characteristic impedance of the line of connection) doesn't care of specific impedances of a and b, but ZA and ZB right?

If this is true, then I have to say this: in a complex circuit, a signal at any point sees an equivalent or a total impedance of whole part of the circuit in front of the signal and a Thevenin equivalent impedance of whole part of the circuit in back of the signal, so the signal behaves according to a difference between these "total" impedances. It's somewhat amazing for me.

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Paul Colby
Gold Member
If this is true, then I have to say this: in a complex circuit, a signal at any point sees an equivalent or a total impedance of whole part of the circuit in front of the signal and a Thevenin equivalent impedance of whole part of the circuit in back of the signal, so the signal behaves according to a difference between these "total" impedances. It's somewhat amazing for me.

Yes, it's actually true provided one recognizes that it's true for each frequency and the impedances change as a function of the frequency. This was kind of an "ah-ha" moment for me when I realized that the thermal noise generated in very large resistors is often not as much of a problem as I thought. The impedance mismatch (voltage divisions) often causes most of the noise to drop across the offending resistor and not across the amplifier front end.

Svein's post on ringing is quit understandable from this perspective. In this case the connection between A and B is a transmission line with it's own impedance at each frequency. The net effect is best understood as a reflection at each impedance interface in the time domain. What you find is if the ##Z_A## and ##Z_B## is equal to the line impedance, ##Z_L##, at each frequency then the max power transfer is realized and there are no reflections.

• goodphy
Svein's post on ringing is quit understandable from this perspective. In this case the connection between A and B is a transmission line with it's own impedance at each frequency. The net effect is best understood as a reflection at each impedance interface in the time domain. What you find is if the ##Z_A## and ##Z_B## is equal to the line impedance, ##Z_L##, at each frequency then the max power transfer is realized and there are no reflections.

I see. Now I almost accept the fact that total or equivalent impedances are important, rather than impedances of individual electrical components in a circuit when we concern an impedance matching problem.

However, for circuits connected with a transmission line, I have one confusion. Let's say circuits A (with impedance ZA ) and B (with impedance ZB ) are connected by the transmission line or cable with a characteristic impedance of Z0 . ZAZ0 = ZB. I simply thought that there is no signal reflection at the interface between the line and B as impedances are matched. However, At this position, when we think about an Thevenin equivalent series impedance of a circuit consisted of A and the line, it is

$${Z_{in}} = {Z_0}\left[ {\frac{{{Z_A} + i{Z_0}\tan \left( {kl} \right)}}{{{Z_0} + i{Z_A}\tan \left( {kl} \right)}}} \right]$$

where k and l are a wavenumber of a signal and a cable length. Then Zin ≠ ZB generally. It means mismatched impedance?

Maybe I should treat the transmission line specially than when I treat a lamped electrical network, but I don't know why.

Paul Colby
Gold Member
Then Zin ≠ ZB generally. It means mismatched impedance?
yes

Maybe I should treat the transmission line specially than when I treat a lamped electrical network, but I don't know why.

Depends on the size of ##kl## for the problem in question. For low frequencies and short lengths, ##kl\approx 0##. In this limit, ##Z_{in}=Z_A##.

yes

Depends on the size of ##kl## for the problem in question. For low frequencies and short lengths, ##kl\approx 0##. In this limit, ##Z_{in}=Z_A##.

Hello. Thanks for giving very quick answer!

So equivalent impedances are always important in impedance matching problem, even for a circuit with a transmission line, right? Let's assume rather high frequency and long length of the transmission line between A and B. So, in the case of Zin ≠ ZB , there is a signal reflection at the interface of the line and the circuit B, even if Z0 = ZB

I would like to confirm this one last time:)

Paul Colby
Gold Member
So equivalent impedances are always important in impedance matching problem, even for a circuit with a transmission line, right?
All answers depend on the questions asked. Reflections typically occur at impedance steps in some problem domains. Their importance depends on the engineering goals. That would be the question or concern being addressed.