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I have questions for transmission line (impedance matching)

  1. Dec 10, 2015 #1

    goodphy

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    Hello.

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

    Non-inverting OP-AMP to load via transmission line.jpg

    This is the circuit what I made in which ZO and ZL are output and load impedances, respectively. The commercial voltage probe with oscilloscope is used to measure VL, voltage over ZL . VO, 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 ZC 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.

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


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

    It seems the original pulse duration is only achieved when ZL and ZC are matched and amplitude of VL becomes closer to the original value as ZL rises.

    My original expectation was that clean signal of half amplitude should be seen when ZO and ZL are matched to ZC 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 pulse duration is extended about twice. I'm even surprised to see signal transmission when ZO = 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 my results and I would like to receive some comments.

    Thanks for reading and please give me some help.
     
  2. jcsd
  3. Dec 10, 2015 #2
    The output resistance of the op-amp will not be zero, but a hundred ohms or so.
    The absence of pulse distortion is probably because the line is short. Remember 1 microsecond requires 300m of line.
    The line acts as a capacitor shunting the circuit for short distances when ZL is high, and this will give pulse distortion.
     
  4. Dec 10, 2015 #3

    CWatters

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    Goodphy.. Did you mean uS or nS?

    Perhaps try a faster logic gate instead of an op-amp?

    Re:

    You mention monitoring the voltage in two places..

    Assuming the driver and line are fast and long enough to behave like a transmission line I would expect to see a near full amplitude step in both of those two locations. The output the gate will swing more or less up to the rail if it is capable of driving 2*ZC. Then due to the potential divider effect of Z0 and ZC a half amplitude step will be launched down the line. At the far end the signal will be reflected and double in amplitude (to full amplitude). It then travels back down the line to the source. So at both of the two places you mention it should be a full step. The half step should appear on the right hand side of R0. The duration of the half step will be roughly the time it takes for a signal to travel down the line and back (eg could be very short, barely a kink in the edge not an obvious step).

    A problem with series termination is that any logic gates half way along the transmission line see the half amplitude signal for a short period. Only the destination sees nice clean edges.



     
  5. Dec 11, 2015 #4

    meBigGuy

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    How long is the transmission line?
    What is the rise time of the opamp output?
    I (and the previous posters) expect your signals are too slow to see reflections.

    Also, as mentioned, the output impedance of the opamp needs to be included.

    tech99 - it is bad form to quote the OP's complete post, especially twice. Why do you do that? Can you edit your post?
     
  6. Dec 13, 2015 #5

    goodphy

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    Hello.

    it is us, microsecond.

    I'm still reading your comments to understand but I would like ask one thing now.

    You mention logic gate. Is MOSFET driver one of its kind? I guess MOSFET driver can be good alternative to Op Amp for my purpose.
     
  7. Dec 13, 2015 #6

    goodphy

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    Hello.

    1. Transmission line in the test was ~ 10 m.
    2. Rise time of direct output from Op Amp would be ~ 166 ns.
    3. output impedance of Op Amp should be near zero as it is voltage feedback loop.

    Anyway, I guess loading effect for amplitude may come from Op Amp current limitation. 50 ohm is too much low to attempt to draw current higher than what Op Amp can give.

    But pulse duration extension for higher load is not...understandable so far.
     
  8. Dec 13, 2015 #7

    meBigGuy

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    The output impedance of an opamp is not zero ohms. It is actually much higher (open loop output impedance), and is effectively reduced by the feedback, which is slow.

    The drive capability of the driver is important. Look at the 74S140. (also, google "50 ohm line drivers")

    Here are some basics about transmission line drivers
    http://www.ti.com/lit/an/snla043/snla043.pdf

    In order to clearly see reflections, the risetime needs to be less than the roundtrip prop delay. 10m ~ 60ft round trip ~ 80ns.
     
  9. Dec 13, 2015 #8

    goodphy

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    Oh, nice material to fully understand what happens on the transmission line! Thanks and I'll get it into my mind:)

    For output impedance of Op Amp, could you tell me how slow feedback is? if it is slower than rising time of original signal (our signal has ~ 166 ns rising time to 13 V), it could be problematic.

    I've done google "50 ohm line drivers" and it looks many line drivers are for 5V TTL signal transferring via line but I need to give at least 10 V at the instrument and that's why Op Amp came up in my drawing. Due to Op Amp's current limitation, I'm now thinking to use MOSFET gate driver (it maybe too much current capability..) Can I simply replace Op Amp to gate driver? I'm afraid it is only used for capacitive load, not general load.
     
  10. Dec 13, 2015 #9

    meBigGuy

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    The closed loop response of an amplifier determines its output characteristics. (its closed loop frequency response)

    There is a simple way to roughly equate the bandwidth of an amplifier to the risetime it can support (ignoring slew rate for the moment).

    Think on a 10Mhz sinewave. It's fastest transition is about the zerocrossing. It turns out that the effective risetime is about 1/3 the period.
    10Mhz = 100ns, so a 10Mhz amplifier will support about 100/3 = 33ns risetime.
    Google "risetime vs frequency response" to find, for example
    http://www.edn.com/electronics-blog...-The-bandwidth-of-a-signal-from-its-rise-time

    But, with 166ns rise time and 10meters of cable, you will see very little distortion. Maybe a little bump in the middle of the pulse.
    But you still need an amplifier that can drive 50 ohms, and has 3/166ns = 18Mhz bandwidth

    Try google search for "50 ohm 10V driver".
    You can use a 50 ohm video amplifier.
    http://www.ti.com/lit/ds/symlink/lm6181.pdf

    Btw, if you series terminate, you don't need a terminator at the other end. What sort off load are you driving? Is this strictly a digital signal?

    If you terminate at the far end, you don't need a series termination (you drive with as close to 0 ohms as you can). That is, if power transfer is not the issue and you just want a clean pulse at the other end. All depends on what you are really trying to do.
     
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