I have questions for transmission line

MikeML

Joined Oct 2, 2009
5,444
Here is another sim where the source is replaced with an opamp with a finite slew rate, and also a finite current-source, current-sink capability. Note the huge increase in rise and fall times at the load end. This is primarily due to the limited current available from the opamp to charge and discharge the coax capacitance.... Also, most opamps cannot supply enough current to drive a resistive load that is as low as 100Ω (when R1=50Ω and R2=50Ω). That is where my comment about "unloading" the opamp came from.

155o.gif
 

Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
@Dong-gyu Jang
So do you see that all of your observed waveforms are the consequence of the slew-rate and output current limitations of the opamp, and not attributable to the coax other than its capacitance???
Thanks to show me simulation result.

As your comments here and other comments from my other question, in terms of output amplitude, loading effect looks clearly come from current limitation of the Op Amp. My Op Amp is very fast one which bandwidth of about 80 MHz so I don't think slew rate is problem.

Still, one problem remains; why pulse duration becomes twice with higher load resistance? For 50 ohm load, input and output pulse durations are the similar. But higher load impedance gives extended pulse duration. I still don't know how to dig this issue..

Anyway, as Op Amp has limited current capability, I'm now thinking about replacing Op Amp to MOSFET driver, which is easier to use and gives ~ 1 A current at peak. I'm just no sure that this is good way to drive load through which 10 to 200 mA current is expected to flows with voltage over load of 18 V.
 

MikeML

Joined Oct 2, 2009
5,444
You would do better to transmit a 1V pulse from your experiment to the instrument on the coax cable. Then put a voltage amplifier between the cable end and the trigger input to the instrument to amplify the 1V pulse to 10V. This is because of how long it takes to charge the coax cable capacitance from the limited current available from the amplifier at the experiment end through a deltaV of 10V. It is much faster to charge the cable through a deltaV of only 1V, and then do the amplification to 10V at the other end...

Look at this simplistic case: Note how long it takes the 25mA current source to charge 1nF of cable from 0V to 1V compared to 0V to 10V.

158.gif
 

Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
You would do better to transmit a 1V pulse from your experiment to the instrument on the coax cable. Then put a voltage amplifier between the cable end and the trigger input to the instrument to amplify the 1V pulse to 10V. This is because of how long it takes to charge the coax cable capacitance from the limited current available from the amplifier at the experiment end through a deltaV of 10V. It is much faster to charge the cable through a deltaV of only 1V, and then do the amplification to 10V at the other end...

Look at this simplistic case: Note how long it takes the 25mA current source to charge 1nF of cable from 0V to 1V compared to 0V to 10V.

View attachment 96434
Hm... Can I ask one question? Could you give the reason why you treats transmission line nothing but capacitor? For lossless line, certainly there is inductance also. Besides, I always think that even although transmission line is short in length, treating it as the line rather than lumped element is good way of circuit design. In fact, I hardly distinguish exactly which case the line can be considered as lumped element, especially when speed of signal propagation in the cable is significantly lower than free space due to insulation material.
 

MikeML

Joined Oct 2, 2009
5,444
Hm... Can I ask one question? Could you give the reason why you treats transmission line nothing but capacitor? ...
I thought it was obvious.

Velocity of propagation in a RG-58 coax line is about 0.66 of the velocity of light, or about 0.66*3e8m/s. The time it takes for a pulse to propagate 10m in RG-58 is (10m/(0.66*3e8))s = 50ns, which is a tiny fraction of how long it takes to charge the ~1000pF cable capacitance if the current available is 25mA, and you are charging that capacitance through a delta of 10V.
 

RRITESH KAKKAR

Joined Jun 29, 2010
2,829
It is actually voltage amplifier (Op Amp) and output transmitted to the instrument to be triggered. Our commercial trigger system can give TTL signal of 4 V amplitude however, the instrument requires at least 10 V for optimal performance. That's why I'm building the circuit.
So, What type of signal you well sent is that optical laser?
 
Top