Understanding Series Termination in High Speed Digital Circuits

[likephysics]

I was reading about series termination in high speed digital circuits -
A series termination comprises a resistor between the driver's output and the line . The sum of the output impedance of the driver, RD, and the resistor value, R, must equal Z0. With this type of termination, only one-half the signal value appears on the line because of the voltage division between the line and the combination of the series resistor and the driver's impedance.

I can't understand how the voltage divides between the transmission line and the series resistor.

 

[schip666!]

Think of the transmission line and the input impedance of whatever it connects to as one resistor, and the output impedance plus series resistor as a second resistor. Then you have a voltage divider, non? If the two impedances are equal you get 1/2 the drive voltage on the transmission line.

 

[likephysics]

Think of the transmission line and the input impedance of whatever it connects to as one resistor

Ok, transmission line is 50ohms and input impedance is another gate, so high impedance like 10Kohms.

and the output impedance plus series resistor as a second resistor.

This combo is 50ohms.

Then you have a voltage divider, non? If the two impedances are equal you get 1/2 the drive voltage on the transmission line.

50 ohms is in series with transmission line(also 50 ohms), which is connected to high input impedance.
I don't see a voltage divider?

 

[schip666!]

I admit to not being an expert on this, but...

The transmission line impedance itself is a "nominal" value so I wouldn't include it in the calculations. If you indeed have a regular (say TTL) gate at the input then it's impedance is probably higher but 10k is fine, and you would be right conceptually. It may be that it's not a regular gate. Do we have part numbers or is this a theory thing?

 

[likephysics]

Don't have part numbers. Just trying to understand the concept.

 

[dlgoff]

I can't understand how the voltage divides between the transmission line and the series resistor.

I know you are asking about a high speed digital circuits but I think it would be similar to a http://hyperphysics.phy-astr.gsu.edu/hbase/electric/vdivac.html" .

 

[schip666!]

Dunno if this helps:
https://www.physicsforums.com/archive/index.php/t-224750.html

I suspect that a "50 ohm line receiver" is exactly that: 50 ohms input impedance, and not a regular TTL load. That said, I'm probably talking out of my a:zzz:

 

[berkeman]

50 ohms is in series with transmission line(also 50 ohms), which is connected to high input impedance.
I don't see a voltage divider?

There is a reason that the full transmitted voltage shows up at the receiver input, and it is related to the high input impedance of the receiver input. Think about what happens with the reflection from the "open circuit" end of the transmission line at the receiver end of the TL...

 

[likephysics]

There is a reason that the full transmitted voltage shows up at the receiver input, and it is related to the high input impedance of the receiver input. Think about what happens with the reflection from the "open circuit" end of the transmission line at the receiver end of the TL...

I got it. It's just a transmission line that is open. So the reflection coefficient for a open termination transmission line is

ZL-Z0/ZL+Z0

ZL is infinity.

So reflection is 1.
Initially the source sees Zsource and Z0, so the voltage is divided between the two.
The reflection adds to this initial voltage.

 

[berkeman]

I got it. It's just a transmission line that is open. So the reflection coefficient for a open termination transmission line is

ZL-Z0/ZL+Z0

ZL is infinity.

So reflection is 1.
Initially the source sees Zsource and Z0, so the voltage is divided between the two.
The reflection adds to this initial voltage.

Correct-a-mundo. Good job.