I have questions for transmission line

Discussion in 'General Electronics Chat' started by Dong-gyu Jang, Dec 10, 2015.

1. Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Hello.

I have questions for transmission line and please see the image below.

This is the circuit what I made in which Z_O and Z_L are output and load impedances, respectively. The commercial voltage probe is used to measure voltage over Z_L with oscilloscope. V_O, the voltage difference measured between output of the OP-AMP and ground by the probe is square signal of positive 13 V with ~ 10 μs pulse duration. Characteristic impedance of the transmission line Z_C is 50 Ω.

I have used this configuration with varying output and load impedances to test my knowledge of the transmission line theory. The followings are the results what I got. (V_L is measured voltage over the load)

1. Z_L: 50 Ω, V_L: 3.2 V of ~ 8.8 μs
2. Z_L: 384 Ω, V_L: 10.6 V of ~ 19.7 μs
3. Z_L: 510 Ω, V_L: 10.8 V of ~ 19.44 μs
4. Z_L: 2376 Ω, V_L: 12.6 V of ~ 17.89 μs

All signals are quite clean square and the results doesn't matter whether Z_O is 0 or 50 Ω.

It seems the original pulse duration is only achieved when Z_L and Z_C are matched and amplitude of V_L becomes closer to the original value as Z_L rises.

My original expectation before doing this test is that clean signal of half amplitude is seen when Z_O and Z_L are matched to Z_C and for other cases I should be able to some ringing due to signal reflections. However, in measurement, about quarter of the original amplitude is observed at 1st case and for other cases I don't see any observable ringing and the pulse duration is extended about twice. I'm even surprised to see signal transmission when Z[SUB]O[/SUB] = 0 (reflection coefficient at boundary from source to transmission line is 1 as output impedance of the feedback loop OP-AMP is zero as far as I know)

I really don't get how to explain these results and I would like to receive some comments.

Mod edit:
Please refer to another same topic was closed --

Last edited: Dec 10, 2015
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2. alfacliff Well-Known Member

Dec 13, 2013
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Zo should be paralell, ot series

3. RRITESH KAKKAR Senior Member

Jun 29, 2010
2,831
89
Hello Dong,
What are you building?

4. Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Hi.

It is actually to give trigger signal transmitted through the line to the some instrument. Is purpose of the circuit matter to my question?

5. Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Z_O should be parallel? Well, As far as I know, Thevenin's equivalent circuit of the non-inverting OP-AMP is of zero equivalent impedance. If it is true and Z_0 is in parallel to the ground then equivalent circuit covering Z_0 is also of zero equivalent impedance.

Am I wrong?

6. MikeML AAC Fanatic!

Oct 2, 2009
5,450
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You are right, Alfa is confused...

Last edited: Dec 10, 2015
7. RRITESH KAKKAR Senior Member

Jun 29, 2010
2,831
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ok, i have studied it but don't know sort of transmission line will be this.
is that electrical kv line?

8. MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
Practical questions:
How long is the coax cable between the source (opamp) and the instrument being triggered?
What is the natural input impedance of the instrument being triggered?
What is the pulse rep. frequency?

9. Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Hello.
1. I've tested with 1 m and ~10 m cables but seems doesn't make difference.
2. The datasheet tells it is at least greater than 490 Ohm. Specific value is not shown. I'm also now digging of its true value.
3. single shot

10. Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
What is kv line? The cable is typical BNC 50 ohm cable.

11. MikeML AAC Fanatic!

Oct 2, 2009
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My suggestion is to forget about the 50Ω termination at the instrument being triggered as long as the coax is less than a couple of meters long.

That unloads the opamp at the driving end so you do not need a 50Ω driver.

The reflections in a short coax due to an unterminated line die out so quickly (few ns) that they will not have an noticeable effect the triggering...

Here, shorter is better...

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12. RRITESH KAKKAR Senior Member

Jun 29, 2010
2,831
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i was thinking of overhead cables.
anyway what this mean 50ohm in wires?
if length will increase then the resistance will also increase.

13. MikeML AAC Fanatic!

Oct 2, 2009
5,450
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@RRITESH KAKKAR
TS is talking about 50Ω shielded coaxial cable, which has nothing to do with what you are commenting on...

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14. RRITESH KAKKAR Senior Member

Jun 29, 2010
2,831
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OK, BNC mean Base net connect it is used in CRO.
coaxial cable are used in TV

15. alfacliff Well-Known Member

Dec 13, 2013
2,449
428
bnc stands for "bayonet constant impedance" bayonet is for the twist lock style.

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16. RRITESH KAKKAR Senior Member

Jun 29, 2010
2,831
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Sorry , i was not aware.

17. Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4

You said unloading, however, the measurement shows that V_L depending on Z_L rather than matter of cable length (I've tested upto ~ 10 m cable). I still don't know how to explain this dependency.

And you mentioned reflected waves are died over ns scales. Could you please explain this more? Is this resistive loss? How it is done?

18. MrAl Well-Known Member

Jun 17, 2014
2,438
492
Hi,

Quick question for you...

Did you do the math or do a simulation of this system yet?
If so, what results did you see there and did they differ from what you measured?
What is "commercial voltage probe", is that the oscilloscope?

19. dl324 Distinguished Member

Mar 30, 2015
3,244
622
Reflections occur in cables with a characteristic impedance when they aren't properly terminated. This behavior can be used to find discontinuities (mainly in the form of opens or shorts) in cables by injecting a pulse into the cable and monitoring any reflected signal.

20. MikeML AAC Fanatic!

Oct 2, 2009
5,450
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What you observe has (almost) nothing to do with pulse propagation on an improperly-terminated coaxial transmission line; instead it is all about the voltage divider created between R1 and R2, and the shunt capacitance of the coax (93.5pF/m for RG-58/U cable).

If we can assume you have a perfect pulse source (with very fast rise and fall times, and a 50Ω source impedance), look what happens at the receiving end V(y) as you change the terminating resistor in steps: 50, 100, 200, 400 and 800Ω.

The simple voltage divider, and shunt capacitance accounts for the first order effects, while the transmission line concepts are almost insignificant second-order effects provided that the line is only a few meters long...

What opamp are you using? If you consider the actual slew rate of your opamp (compared to the near-ideal pulse generator I used in the sim), the transmission line effects will become even less significant.

Last edited: Dec 10, 2015