# Question about impedance matching

Discussion in 'General Electronics Chat' started by tpny, May 11, 2013.

1. ### tpny Thread Starter Member

May 6, 2012
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0
Hi, I read that we should match impedances between the driving circuit and the load so as to prevent frequency from reflecting back to the driving circuit.

But then I thought, wouldn't that create a voltage divider situation and cut the driving signal in half? I thought the input impedance should be ideally infinite and output impedance ideally zero so as not to create such a loading problem. Could you please shed some light onthis? Thank you so much!

2. ### crutschow Expert

Mar 14, 2008
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You didn't read the complete explanation. Impedance matching is used in high frequency circuits where the wavelength or transition time of the signal is short compared to the electrical length of the cable carrying the signal. In such a situation the signal will reflect back from the load and distort the signal unless the load impedance matches the cable impedance.

Yes, that does cut the signal in half but that's unavoidable if you want to maintain signal integrity.

For situations where the cable length is short compared to the signal wavelength or risetime (such as a audio amp or power line) then you do want the source impedance to be as low as possible compared to the load impedance.

3. ### circuitfella11 Member

May 10, 2013
56
5
it is necessary to keep it low.. ratio of impedance of the line and impedance of the load..
most likely you always want a result close to 1.. its a reduction of reflection apparently, ratios close to 1 mean that the signal won't be distorted as much.

--regards

4. ### WBahn Moderator

Mar 31, 2012
23,201
6,993
There are a number of factors at play. First, it isn't "frequency" that is reflecting back, it is power. If you send a lot of power down a cable to a load, you do NOT want it coming back at you.

The goal is usually to maximize the power that makes it into the load. If the load impedance is infinite, then this fraction is tiny. Think back to when you did maximum power transfer calculations for a non-ideal source driving a load. The maximum power transfer to the load occurs when you adjust the load to match the source's impedance (assuming that the source impedance is fixed and unchangeable).

In the case of a transmission line, you are not dealing with a physical impedance, but rather a characteristic impedance. If the line is properly terminated in a matched load, the source sees the line's characteristic impedance and that is how much power it is sending down the line. All of that power makes it into the load because none of it is reflected back. So you don't really have a voltage divider at play in this case.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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One needs to be careful about the notions of incident & reflected power in transmission lines. The generator end only ever supplies the actual power delivered to the load [plus line losses]- it doesn't send excess power down the line and then get some returned to the generator for re-absorption.

Perhaps incident & reflected energy might be more helpful terminology.

Last edited: May 13, 2013
6. ### chuckey Well-Known Member

Jun 4, 2007
75
10
"create a voltage divider situation and cut the driving signal in half?". The 50 ohm impedances as stated are not resistive, they are reactive, i.e. a mixture of Ls and Cs which resonate at the required frequency, so they are lossless. The 50 ohms is a statement of the fact that the voltage is 50 times that of the current, i.e. V/I = 50. If you took a 50 ohm transmission line and terminated with a 50 ohm resistor (which would dissipate power) at any point on the line, the voltage divided by the current would always be 50, at any frequency at any level. If you terminated the line with a different resistance (or impedance) the ratio V/I would not be 50 and would vary all the way down the line. You can use a transmission line to act as a transformer under these circumstances, changing the ratio V/I.
Frank

7. ### cabraham Active Member

Oct 29, 2011
82
31
But the line characteristic impedance Z0 is given by sqrt(L/C). This characteristic impedance is indeed resistive, i.e. I is in phase with V. The generator "sees" Z0, or for example 50 ohms. That is a resistive 50 ohms just like it was terminated in a 50 ohm resistor.

The signal gets attenuated by half due to the voltage divider formed by the generator 50 ohm internal resistance and the 50 ohm line resistance. Half the power is lost in the generator resistance, the other half is dissipated in the load, assuming a loss-less t-line. It is common when routing video (I've done this) to use an amp with a fixed gain of 2. This compensates for the attenuation inevitable with impedance matching which is one half the signal.

If the line is terminated in 50 ohm, the generator does not see any reflected waves returning, but the 50 ohms is real. As far as resonating at the required frequency goes, please elaborate. At any frequency high enough for transmission line model to be applicable, we can regard the line as 50 ohm resistive over a broad range of frequencies.

Claude

8. ### tpny Thread Starter Member

May 6, 2012
216
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What about connecting signals between two pcb's where cable length < 2 feet.. Do I impedance match the source and load? It doesn't apply in this case right? Why or why not?

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Depends - what signal frequencies are you dealing with?

10. ### studiot AAC Fanatic!

Nov 9, 2007
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Yes and no.

That's all I can answer when I don't know if the signal is measured in Hz, kHz or GHz.

11. ### tpny Thread Starter Member

May 6, 2012
216
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< 2 feet in Hz.

At what distance should Hz signals need impedance matching?

12. ### crutschow Expert

Mar 14, 2008
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You need impedance matching when the length of the line is more then 1/8th wavelength of the highest signal frequency in the cable.

13. ### WBahn Moderator

Mar 31, 2012
23,201
6,993
If the signal has frequencies of interest whose wavelenths are more than about 10% of the signal path length, then you should look at the path as a transmission line. Sometimes this means you can use miles of interconnect and not worry and other times you have to treat the internal bond wires on an IC as a transmission line.

14. ### tpny Thread Starter Member

May 6, 2012
216
0
Seems like there's a lot of leeway.. I used an online wave length calculator and for a Mhz frequency, wavelength = 300 meters. So my cable has to be > 3000 meters before I need to consider impedance matching? For Hz signals the wavelength is huge, so there's even more leeway. So why are there so many impedance matching examples for audio, antennaes even when just connecting from one console to another several feet away..

15. ### t_n_k AAC Fanatic!

Mar 6, 2009
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One might need to be cautious about including audio equipment with antennas - presumably RF antennas.

16. ### crutschow Expert

Mar 14, 2008
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You have it backwards. For 1MHz you need to consider impedance matching when the cable electrical length approaches 1/8 to 1/10 of the wavelength or around 30 meters.

The only audio impedances that are matched are the standard 600 ohms used for professional audio connections and that's just an arbitrary standard that has nothing to do with matching cable impedance or the electrical length of the cable (for example the wavelength for a 20kHz electrical signal is 15,000 meters).

17. ### WBahn Moderator

Mar 31, 2012
23,201
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Impedance matching is also more important as your power goes up. If you are trying to deliver hundreds of watts of power to something, you want to maximize how much power actually makes it into the load. For a given source impedance, this happens with the load impedance matches the source impedance (well, actually when it is the complex conjugate).

18. ### Tesla23 Senior Member

May 10, 2009
365
78
My 2 cents worth:

Impedance matching achieves maximum power transfer, so tends to be important when power is at a premium - this tends to be in low power situations. So it is important to match low power signal sources: antennas, microphones etc, and also between elements in the signal chain so as not to waste power and added gain.

Why transmission lines come into this is that if either the source or load is not matched to the transmission line impedance, then the performance of the system varies (often a lot) as the length of the transmission line varies. For example, a high impedance load connected through a piece of transmission line that happens to be λ/4, appears as a near short, but if the line is λ/2 it appears high impedance again. In the time domain severe pulse distortion will occur of the lines are not matched and the line delay is comparable to or larger than the pulse risetime. As has been said, if the lines are electrically short this effect can be ignored.

In high power situations, the active devices tend to be matched for highest efficiency not maximum power transfer. At low frequencies this is often an output impedance much lower than the load impedance. At higher frequencies it is important for the load to be matched to the transmission line, both to avoid dependence on line length, and to avoid standing waves on the line that waste power (and radiate on open structures).

When I used to teach, whenever I found someone keen to match everything, I used to ask what would happen if their toaster was matched to the power outlet.