why must the matching impedance be 50 ohm

[fstspec]

I understand that for maximum power the input and output impedance must be match. But I dun understand why most circuits have to be matched to 50 ohm, can anyone care to explain the reason behind it? What is the reason for most circuits to have an input impedance of 50 ohm, why is it other values like 60 or 70 ohms.

 

[Dr.EM]

Circuits definately arn't all matched at 50 ohms. Telephone systems are usually 600 ohm. Which circuits do you find as 50 ohm?

 

[stevez]

As already stated, impedances should match for maximum power transfer. In some circuits the behavior of the connected circuits can be profoundly impacted by a mismatch. The final amplifier of a transmitter can self-destruct because it's output is mismatched. As I understand it, filters may not perform as intended if not matched.

Do a little reading on transmission lines and you'll see how the size of conductors, spacing and other things, affects what is known as characteristic impedance. If you focus on coaxial cables you'll see that coax can be (and is) made in 50 ohm impedances as well as others - 75 ohm, 93 ohm, etc. What the industry did at some point was realize that for various applications the output impedance of an item should match the available coaxial cables. 50 ohms, for many situations, has been found to be an acceptable design point for the cables so the equipment has to match. For many things, like cable tv, 75 ohms or so is the standard - because in this type of application the 75 ohms is a better compromise in terms of behavior.

 

[stevez]

I still don't trust that my posts will make it so I broke this up.

Think about this example - an RF transmitter feeds a 50 ohm coaxial cable - therefore the load must behave like 50 ohms for a proper match. At 200 watts there will be 2 amps current flowing with 100 volts between the coaxial conductors. If we designed our system to use 450 ohm line we'd have 0.66 amps of RF but now we've got 303 volts across the conductors. I use this to illustrate that voltage and current levels do change and that can drive some choices.

 

[stevez]

Too add further confusion (or clarity) the voltage between the conductors in my prior example only holds up if things are matched. In a mismatched condition the voltage at some points on the transmission like can be many times the matched voltage (too much to explain here). That can and does cause failures in components, transmission lines, etc.

 

[Ambient]

I just took an Electromagnetic Theory course for my major and we went over the fine details of matching the impedances. Trust me, matching is important. And I will never take that painful course again lol.

 

[zevon8]

Ambient: I bet you now know exactly what that funny little P-R key does on your calculator.

here is a link to an explanation of Characteristic Impedance:

Nice work stevez.

 

[Optikon]

Geesh!... why people don't use google, I don't know.

Here is one explanation of why.

**broken link removed**

 

[Sig239]

Hello everyone
About 7 or 8 months ago, I did exactly what you said Optikon. I typed a question regarding electronics into Google. To my pleasant surprise, I found the answer to my question (through Google) right on this website, where I also found a large community of friendly, and social, like minded individuals who enjoy sharing their vast knowledge. When it comes to people coming on here and expecting someone else to complete their school projects or other large projects, then I agree with you wholeheartedly. However, I have no problem with people posting to ask general questions about electronics theory, etc. Isn't that why we are here: to learn from each others experiences, and to interact with real people instead of just reading technical papers online? I'm not trying to flame anyone here, I just thought that's what this forum is all about. I have a lot of respect for all of the experienced folks on here who have been doing this stuff for many years now, but continue to educate people post after post, even if it does get somewhat repetative sometimes. Thanks to everyone for a wonderful forum.

 

[Sig239]

Hello fstspec
There are some instances when you don't want maximum power, and therefore don't want the impedences to match. An oscillator is an example of this, since if you draw too much current from the tank circuit, you will sacrifice stability and reliability. In this case, you "loosely" couple to the tank circuit.
Now ask yourself this. Is directly shorting a battery the way to draw the most current? HMMMMM... Have fun!!!

 

[Nigel Goodwin]

Hello fstspec
There are some instances when you don't want maximum power, and therefore don't want the impedences to match. An oscillator is an example of this, since if you draw too much current from the tank circuit, you will sacrifice stability and reliability. In this case, you "loosely" couple to the tank circuit.
Now ask yourself this. Is directly shorting a battery the way to draw the most current? HMMMMM... Have fun!!!

In MOST instances you don't want matched impedances (and it would be a REALLY bad thing!), but you do need to be aware of the small number of cases where it's ABSOLUTELY VITAL to match impedances (such as RF amplifiers).

Incidently, shorting the battery wouldn't produce maximum power, because you're not matching it's impedance (or more correctly, internal resistance).

 

[Hero999]

To get maximum power from the battery the load impedance would have to be the same as the internal impedance of the battery so the efficiency would only be 50%.

Just thinking, why have I seen transmitters with a higer efficiency than 50%? Could it be because the impedance of the output stage isn't resistive so it doesn't disipate any power?

 

[audioguru]

Just thinking, why have I seen transmitters with a higher efficiency than 50%? Could it be because the impedance of the output stage isn't resistive so it doesn't disipate any power?

The output transistor can be operated in class-C so it conducts only part of the RF wave, and it can switch fully-on and fully-off. Then the tuned LC tank turns the rectangular-wave current into a pretty good efficient sine-wave.

 

[Hero999]

I'm aware of that, but you can by an RF amplifier with an output impedance of 50ohms and it's 60% efficient. How is this possible when the when if for maximum power transfer the load must be 50ohms and it'd be only 50% efficeint?

I presume that the output really isn't 50ohms, it's just designed to drive a 50ohm load, is this correct?

 

[Nigel Goodwin]

I'm aware of that, but you can by an RF amplifier with an output impedance of 50ohms and it's 60% efficient. How is this possible when the when if for maximum power transfer the load must be 50ohms and it'd be only 50% efficeint?

I presume that the output really isn't 50ohms, it's just designed to drive a 50ohm load, is this correct?

No, I would suggest it's simply "playing with the figures", and isn't really more than 50% efficient at all!.

A lot depends on exactly what type of transmitter it is, and how it's measured - the DC input to the final output stage is one VERY easily measured figue, and is one way of specifying amateur radio transmissions. For SSB you use PEP - 'Peak Emitted Power' - which is a far more difficult thing to measure.

 

[Analog]

If you do not match impedances at the RF frequencies, you will get reflections due to the mismatch (known as gamma). These reflections can be troublesome, and can destroy equipment. If you get good with a smith chart, you can even determine lengths of coax that when shorted, look like an open circuit, or an open coax look like a dead short! Its really pretty interesting stuff. Google will give you a treasure trove of info on this matter. EDN had a great article on this as well. (see attached).

 

[RadioRon]

I'm aware of that, but you can by an RF amplifier with an output impedance of 50ohms and it's 60% efficient. How is this possible when the when if for maximum power transfer the load must be 50ohms and it'd be only 50% efficeint?

I presume that the output really isn't 50ohms, it's just designed to drive a 50ohm load, is this correct?

Your presumption is quite correct. In fact, most amplifiers are designed to deliver a certain performance when driving a rated load, such as the popular 50 ohms used in RF work. They are not designed to have an output impedance of 50 ohms. They are designed to drive a load of 50 ohms, that's all. There are many other characteristics that must be traded off to achieve optimum performance and the best tradeoff does not occur when the amplifier's output Z is 50 ohms.

Maximum conversion efficiency is usually achieved by setting the amplifier's output impedance low relative to the load impedance. At times like that, you can get DC to RF conversion efficiencies of up to 90% even at RF frequencies (ref: "The Electronics of Radio", David Rutledge, pp 180-196). This is certainely not the point of maximum power transfer. It is common to find power amplifiers for GSM cellphones that do indeed deliver efficiencies of 60%, at frequencies like 900 or 1800 Mhz. These amplifiers do not have 50 ohm output impedance, and they are not linear but instead operate in Class C.

Another point to make is to reiterate the maximum power theorem, which says that with a given thevenin voltage source with fixed output impedance you will deliver the maximum power to a load if the load impedance is the complex conjugate of the source impedance. However, and this is the key point, if you can vary the source impedance as well as the load impedance you can have better efficiency. Consider taking the source impedance to the extreme of 0 ohms. In this case, all the power available from the source is delivered to the load since there are no internal source losses at all. So the most efficient setup is to set the output Z as close to zero as you can.

The reason this isn't done in some RF circuits is because is impossible to get the RF amp down to zero ohms of output Z while making all the other critical things as good as possible (things like gain, stability, distortion, efficiency etc.). But they do try.

Conversely, the efficiency also tends towards 100% when the load R tends towards infinity but this time the total power also tends towards zero.

A good explanation of this with further references is here:
https://en.wikipedia.org/wiki/Maximum_power_theorem

 

[Hero999]

Thanks for the information.

 

[gangadhara t]

why 50 ohms

hi,

the pdf attachment may help u for ur question.


gangadhar