Proper termination with 75Ohm cable

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Discussion Overview

The discussion revolves around the proper termination resistance for a 75Ohm cable connected between a pulse generator with a 50Ohm internal impedance and an oscilloscope set to 1MOhm. Participants explore the implications of impedance mismatches and reflections on signal integrity, focusing on theoretical calculations and practical considerations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants calculate the reflection amplitude at point A due to the impedance mismatch, suggesting it will be V_r = 1V * (75-50)/(75+50) = 1/5V.
  • There is uncertainty about whether the reflection will increase or decrease the amplitude of the pulse arriving at point B.
  • One participant suggests that the proper termination at B should be 75Ohms, questioning how to make the oscilloscope appear as a 75Ohm load.
  • Another participant proposes a calculation involving the oscilloscope's impedance and the termination resistance needed to achieve a match.
  • Discussion includes the idea of adding a 25Ohm resistor in series with the generator to achieve a match at point A, though this is contested regarding its effectiveness in forward direction matching.
  • Clarifications are made regarding the independence of the oscilloscope's termination from the voltage of the signal applied.

Areas of Agreement / Disagreement

Participants express differing views on the effectiveness of various termination strategies and the implications of reflections, indicating that multiple competing views remain and the discussion is unresolved.

Contextual Notes

Participants note the dependence of calculations on assumptions about signal amplitudes and the nature of impedance matching, highlighting the complexity of achieving proper termination in this scenario.

hadoque
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Homework Statement


A pulse generator is connected to an oscilloscope, as in the image. The pulse generator has an internal impedance of 50Ohm and the oscilloscope is set to 1MOhm. The cable between the devices has a characteristic impedance of 75Ohm. The pulse from the generator has an amplitude of 1V and it's long enough to interfer with the reflecting pulse (if any). What is the proper resistance of the termination at B?


Homework Equations


[itex]V_r = V_0 \frac{Z-Z_0}{Z+Z_0}[/itex]

The Attempt at a Solution


At A there will be a reflection because of the higher impedance. The reflection amplitude will be
[itex]V_r = 1\text{V} \frac{75-50\Omega}{75+50\Omega}=\frac{1}{5}\text{V}[/itex]

Now, would this reflection increase (decrease seems more reasonable...) the amplitude of the pulse arriving at B, with 1/5V?
Would then a proper calculation of the termination be using V0= 1.2V, Z0=75Ohm and Z=-0.2?
 

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hadoque said:
Now, would this reflection increase (decrease seems more reasonable...) the amplitude of the pulse arriving at B, with 1/5V?
You are correct with "decrease is more reasonable".

Would then a proper calculation of the termination be using V0= 1.2V, Z0=75Ohm and Z=-0.2?
No, "-0.2" looks more like the reflection coefficient from the mismatch at A and is not an impedance. The question asks "what is the proper resistance of the termination at B". For a 75-ohm cable, one should be thinking: 75-ohms. So, how do you make the oscilloscope that looks look like a 1M-ohm load look like a 75-ohm load? I'll give you a chance to think about that.
 
lewando said:
You are correct with "decrease is more reasonable".

No, "-0.2" looks more like the reflection coefficient from the mismatch at A and is not an impedance. The question asks "what is the proper resistance of the termination at B".
Yes, that's Vr, which is the amplitude of the reflected pulse at A. I'm thinking that the first step would be to calculate the amplitude of the incident pulse at B.
lewando said:
For a 75-ohm cable, one should be thinking: 75-ohms. So, how do you make the oscilloscope that looks look like a 1M-ohm load look like a 75-ohm load? I'll give you a chance to think about that.
Well, I would say [itex]\frac{1}{1\text{M}\Omega} + \frac{1}{Z}= \frac{1}{75\Omega} \rightarrow Z \approx 75\Omega[/itex], but that assumes that the signal at B is 1V, which I don't think it is since the pulse generator is not matched to the cable...
 
True, the pulse generator is not matched to the cable. That is a shame. As for the oscilloscope, its termination is independent of the voltage of the signal applied. It just needs to match the cable.
 
Ok, but putting a 25Ohm resistor in series between cable and generator would give a match at A aswell, right?
 
hadoque said:

Homework Statement


A pulse generator is connected to an oscilloscope, as in the image. The pulse generator has an internal impedance of 50Ohm and the oscilloscope is set to 1MOhm. The cable between the devices has a characteristic impedance of 75Ohm. The pulse from the generator has an amplitude of 1V and it's long enough to interfer with the reflecting pulse (if any). What is the proper resistance of the termination at B?


Homework Equations


[itex]V_r = V_0 \frac{Z-Z_0}{Z+Z_0}[/itex]

The Attempt at a Solution


At A there will be a reflection because of the higher impedance. The reflection amplitude will be
[itex]V_r = 1\text{V} \frac{75-50\Omega}{75+50\Omega}=\frac{1}{5}\text{V}[/itex]

Now, would this reflection increase (decrease seems more reasonable...) the amplitude of the pulse arriving at B, with 1/5V?
Would then a proper calculation of the termination be using V0= 1.2V, Z0=75Ohm and Z=-0.2?

I agree with lewando. The question asks for the match at B, and that is just adding the 75 ohms in parallel with the 'scope.

hadoque said:
Ok, but putting a 25Ohm resistor in series between cable and generator would give a match at A aswell, right?

You can add this "back termination" resistor at A, but it does not do any matching in the forward direction. It is only useful if there is a [STRIKE]mismatch at A[/STRIKE] mismatch at B (like just the 'scope), because you will not get a re-reflection at A when the reflected wave comes back and sees 50+25=75 Ohms looking back into the signal generator.
 
Last edited:
Ok, I got a bit confused there, but now it's clear. Thanks
 

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