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simple question about 90 degree hybrid coupler

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winglj

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90 degree hybrid coupled line

Hi, I have a simple question about the 90 degree (quadrature) hybrid coupler

For this type of coupler, there are two outputs with 90 degree phase differences, one input and one isolation port (terminated by 50 ohm resistor).

But if the isolation port is open, will it affect the performance of the hybrid coupler two output?

Thanks.
 

90 degree hybrid coupler

No this is not true you do not need to terminate the isolated line with Zo because it is "theoretically" isolated, but it is wise to terminate it because in a real life line it will not be totally isolated.

I have attached a picture of a coupler that I quickly put together in HFSS you can see how the fields are transmitted from one line to the other, this is because of the direction of the propagation of the wave. One port will look like nothing is going on, but on that same line the port on the other side will have power flowing out of it due to the direction of prorogation and crosstalk from the line that the wave flowed into.

Hope this helps let me know if you need it better explained.
 
90 degree hybrid couplers

I agree with b4bb4ge, even though his example was for a coupled line rather than a hybrid coupler, that for ideal, matched circumstances the load resistor is not necessary. However, as stated, if the circuit deviates in any way from the ideal case then there will be power exiting the 'isolated' port of the coupler. Once you get this happen, for whatever reason, then if the load resistor in not there the performance of the coupler will be degraded.
 

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    winglj

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coupled line rather than a hybrid coupler
Click to expand...

I have always understood that when referring to couplers, hybrid meant "same power but different phase" as opposed to "hybrid" used to describe construction method.


Rod <-often wrong
 

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    winglj

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yendori,

you are thinking about a 3dB, 90° coupler.

A 3 dB, 90° hybrid coupler is a four-port device that is used to equally split an input source with a result of either a 90° phase shift between output ports or to combine two signals while maintaining high isolation between the ports.

The simple configuration of a hybrid coupler I have attached to this, it illustrates two cross-over transmission lines,corresponding with the center frequency that you want to operate at, with a length of one-quarter wavelength at that operation frequency. When power is introduced at the IN port, 3dB of the power (½ the power) flows to the 0° port and the other half is coupled (in the opposite direction) to the 90° port. Reflections from mismatches sent back to the output ports will flow directly to the ISO port or cancel at the input (this is where you would need a resistor on your isolation line). This is why hybrids are so widely used to split high power signals in applications where unwanted reflections could easily damage the driver device.

for more information I would refer to the e-meca or the mini-circuits website they usually have good explanations.
 

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    winglj

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Yes, b4bb4ge. Your last post is exact what I mention for the hybrid coupler.

From your post, my understanding is :

So ideally the ISO port should not affect the performance with or without loading. But if the frequency diverges from center frequency, without the loading, the performance of the coupler will be degraded.

Is it correct?
 

Yes that is correct winglj!

your performance is going to degrade no matter what you do when you are not at the center frequency but it will be less when you load the iso port.

Hope this helps.
 

What do you mean no... just just can't put no to get some points....

This is correct when you drop below the ¼ wavelength you will not get reflections on your iso port but you will not have as good as coupling and when you go above ¼λ you could have more or less power but it will not be a 90° coupler any more (this is true for high and low case), therefore the performance will degrade!

if your other port is not perfectly matched (port 3 90° port) you will build up a standing wave between the two ports and you will need a load on the end of your iso port to get rid of this standing wave.
 

b4bb4ge said:
yendori,

you are thinking about a 3dB, 90° coupler.

A 3 dB, 90° hybrid coupler is a four-port device that is used to equally split an input source with a result of either a 90° phase shift between output ports or to combine two signals while maintaining high isolation between the ports.

The simple configuration of a hybrid coupler I have attached to this, it illustrates two cross-over transmission lines,corresponding with the center frequency that you want to operate at, with a length of one-quarter wavelength at that operation frequency. When power is introduced at the IN port, 3dB of the power (½ the power) flows to the 0° port and the other half is coupled (in the opposite direction) to the 90° port. Reflections from mismatches sent back to the output ports will flow directly to the ISO port or cancel at the input (this is where you would need a resistor on your isolation line). This is why hybrids are so widely used to split high power signals in applications where unwanted reflections could easily damage the driver device.

for more information I would refer to the e-meca or the mini-circuits website they usually have good explanations.
Click to expand...


Thanks b4bb4ge.
 

Hi

Lets not develop a new theory when an apropriate theory, developed by Montgomery in mid 40ties is still valid and working fine.

According to the wellknown S parameter theory, a hybrid, (either 90deg or 180deg) is ANY FOURPORT CIRCUIT THAT IS SIMULTANEOUSLY MATCH AT ALL PORTS (simulataneously meaning at the same frequency). Therefore, if one port is open, you simply don't have a hybrid any more.

Your circuit is effectively a three port circuit and this cannot be a 90 deg hybrid.

flyhign
 

it is not effectively a threee port circuit because if you send power the other way the ports are switched. it is a 4 port device and theoreticly can be lossless, and matched on all ports but it is not reciprical.

because of this yes because it is isolated at 1 frequecy you will get zero reflections.

it is still a 90° at this frequency.
 

That would be interesting, but it's not non-reciprical.
 

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