Electronics > RF, Microwave, Ham Radio

AC Coupling Question

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jmsigler:
Hello, If i am ac coupling a signal output from a clock gen ic say, do I need a 50ohm resistor right after the capacitor? My thought is that I need to present a 50ohm source impedance? Same goes for when ac coupling into an input.

voltz:
Assuming the output impedance from the clock generator is low impedance (like less than a few ohms) then yes, you could place a 50 ohm resistor in series with any ac de-coupling capacitor to give you a 50 ohm output (source) impedance. The capacitor can basically be considered a short circuit at AC.

T3sl4co1l:
An average CMOS logic-level clock generator will have an output impedance comparable to 50 ohms. Don't expect precise source termination, even if you add extra resistance.  Better to use an attenuator (if you can tolerate the loss), or load termination (if the source can put out that much power without overheating).

Tim

evb149:
There are a couple of very different questions here.
As Tim said, the source impedance seen looking into the source from the line will be the driver's output impedance in series with whatever series "series termination" resistance you add between the line and the driver.
The driver IC's data sheet should suggest what the value of driver output impedance is and/or what series termination resistance you might use with the driver to driver a line with a particular impedance.  It is common that fairly strong driver outputs will have at least several ohms of series resistance but more commonly "fairly strong" outputs might have more like 20 or 30 ohms of driver resistance.  So for the case of the latter, one might add series termination resistors around 22R to 33R or so  in front of the driver to match a 50R line.  The values would be correspondingly less for stronger drivers and correspondingly more for weaker ones.

"Same goes for when ac coupling into an input."
-- THAT is a VERY different question.  An INPUT of a typical CMOS / TTL logic gate is fairly high impedance, so relative to a 50R line it'll be closer to an "open circuit" in terms of its reflection.  You do NOT match that kind of high impedance logic input to the line that is driving it with a series termination resistor PLACED AT THE LOAD.  When you're using a series termination scheme your driver should be driving a single "not branched, not multi-drop" line and the receiver end of that line should be effectively an open circuit like a high impedance CMOS/TTL logic input.  That will and should produce something close to a 100% reflection at the load, but that reflection will be "swallowed" when it propagates back and hits the series terminated low impedance driver.  So at the received end incident wave switching can occur due to the reflection from the unmatched high impedance.  At the driving end there is no reflection due to the series termination.  So you place nothing at the receiver end of the line but the receiver in this series termination case with a high impedance receiver.

If you DO NOT use series termination at the DRIVER then you can use parallel / Thevenin / AC termination AT the receiver end of a point to point non multi-drop line having a high impedance receiver at the load end.  Then you'd add shunt or Thevenin or AC shunt termination components right at the receiver to match the line impedance so that there will be no reflection at the load end of the line.  AC terminating the load end or using Thevenin termination may be suitable for continuous clock signals with nearly 50% duty cycles in which case you'd somewhat equalize the high and low side drivers output currents but you'd still need enough dynamic power to drive the low impedance load, and many drivers aren't specified to be strong enough to output enough current to drive 50 ohms to 3.3V or 5V levels.
You'd be driving half-cycles of 33mA @ 3.3V levels or 50mA @ 5V levels into a 50R termination from the driver.

You can also choose the PCB tracks to be higher impedance, maybe 65 to 75 ohms to help reduce the drive required if other impedances in the system aren't problematic to achieve when having your narrowest traces be that high impedance.

Of course if the line is less than a fraction of the transition time long then it isn't really a transmission line and you don't have to worry about impedance matching for SI so much.







--- Quote from: jmsigler on August 06, 2016, 05:13:37 pm ---Hello, If i am ac coupling a signal output from a clock gen ic say, do I need a 50ohm resistor right after the capacitor? My thought is that I need to present a 50ohm source impedance? Same goes for when ac coupling into an input.

--- End quote ---

jmsigler:
Okay, that is what I was looking for. The output impedance is listed at 45ohms suggesting I need a 5ohm resistor.


--- Quote from: evb149 on August 06, 2016, 11:25:15 pm ---Of course if the line is less than a fraction of the transition time long then it isn't really a transmission line and you don't have to worry about impedance matching for SI so much.

--- End quote ---
As a rule of thumb, how do you determine when a trace will begin to act as a transmission line for both sin/square waves? I've seen .5/rise time as an approximation for knee frequency for square waves and can find wavelength from that, but I'm not sure how to relate the wavelength to where impedance matching becomes important.

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