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I'm a bit confused how this type of circuit should be terminated. I have no experience with transmission lines, so I do not know where to put the 50 Ω resistor.

It's basically output from an op-amp, there will be a BNC connector, and I want to connect it to an oscilloscope.
Bandwidth would be 100 MHz.

Is it necessary to put a 50 Ω resistor like this?
a) enter image description here

Or should the 50 Ω resistor be like this?
b) enter image description here

Or should there even some termination at the "other end", between the end of the coaxial cable and the input to the oscilloscope?
c) enter image description here

Or just put 50 Ω termination on the "other side"?
d) enter image description here

My feeling is that option c) is correct, but I'm really not sure.

UPDATE:
I have simulated those circuits in LTspice (it is able to simulate transmission lines).
It looks like this:

enter image description here

I have captured voltages at the ends of that transmission line and current going down to transmission line (actually current through R7). The voltage source sends a 10 V pulse.

a)
enter image description here

The pulse has 10V peak at the end (output from transmission line), but it bounces back to the input of transmission line. There are 2 current peaks: +100 mA and -100 mA.

b)
enter image description here

It looks like the output of the transmission line sees many bounces (with 20 V peaks), and the current peak is 200 mA.

c)
enter image description here

The output sees a 5 V pulse. Current peaks at 100 mA. No bounces.

d)
enter image description here
The output sees 10 V pulse. Current peaks at 200 mA. No bounces.

So it looks like a) is the best option (the output sees 10 V, and there is only a 100 mA current peak), but I don't know whether that pulse bouncing back to the input of the transmission line could cause some problems or not.

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  • \$\begingroup\$ a is fine, in b move the resistor to the end of the transmission line \$\endgroup\$
    – Sorenp
    Dec 20, 2019 at 9:50
  • \$\begingroup\$ @Sorenp I have just added even option d). So you say even a) and d) are properly terminated? \$\endgroup\$ Dec 20, 2019 at 9:54
  • \$\begingroup\$ Come to think of it, b is fine as well.. Your goal is to have 0 voltage drop across the transmission line. The difference is just if its a source or end termination \$\endgroup\$
    – Sorenp
    Dec 20, 2019 at 9:59

3 Answers 3

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Option (a) is the normal approach because: -

  • The op-amp output sees the least loading effect due to the 50 ohms (R7) being in series with the output.
  • The voltage seen at the scope-end is ideally the same voltage as delivered by the op-amp (after a delay of course).

Option (b) just shunts the very low output impedance of the op-amp with 50 ohms i.e. it is ineffectual.

In option (c) R8 can be used (compared to option (a)) but it usually doesn't bring a whole lot to the party other than halving the signal amplitude.

Option (d) loads the op-amp with 50 ohms and this might be a little too "heavy handed" for most op-amps.

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  • \$\begingroup\$ I have updated my question (added simulations). It looks like a) is the best approach. But there is pulse bounced back to the input of transmission line. Could it cause any problems? \$\endgroup\$ Dec 31, 2019 at 11:47
  • \$\begingroup\$ @Chupacabras the pulse that bounces back is fully expected and is totally dissipated in the series source resistance i.e. no more bouncing back to the load to disturb things. \$\endgroup\$
    – Andy aka
    Dec 31, 2019 at 11:54
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(Assuming the scope has a high input impedance, say 1Mohm)

As Andy aka has written, option b) is useless since the source resistor R7 is in parallel with the output impedance of the opamp, witch is very low. Then it’s like if you have no resistor at all.

The base of transmission line (TL) theory: 1. If the load impedance is not matched with the TL impedance, the signal arriving on the load is partially reflected. 2 (corollary of 1) If the source impedance is not matched with the TL impedance, the signal arriving on the source is partially reflected.

From this, option b) is the weakest solution: reflections occurs both at the load and at the source (signal goes form the source to the load, get partially reflected, then goes back to the source, get partially reflected, then goes back to the load, and so on…). You’ll be able to see this on the oscilloscope if you apply a square wave signal in the input of the opamp (with frequency of say 20MHz): at each signal edge you’ll see oscillations that are in fact the superposition of all the reflections.

In option a) you still have reflection on the load but hopefully, the reflected signal is absorbed on the source since its impedance is matched with the TL one: you’ll not measure the reflection (nevertheless, be careful with option a) since on some circuits the returned signal can have energy that is sufficient to destroy the source).

In option d) you still have reflection on the source but hopefully, there is no reflection on the load since its impedance is matched with the TL one (then no signal is returned to the source).

Option c) is definitively the best option for the signal integrity but at a price of measuring half the signal amplitude which is not an issue for me, this occurs when you use a 50ohm input oscilloscope.

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This is correct termination. R1 conform output impedance to 50 ohm line, C1 cancel DC and R2 conform high impedance input to 50 ohm line. Please check

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  • 2
    \$\begingroup\$ Welcome to Electrical Engineering. Please explain why you think this is the correct connection, especially since you've added a series capacitor which will block DC signals from the oscilloscope. \$\endgroup\$
    – Null
    Feb 15, 2023 at 22:32
  • \$\begingroup\$ DC being canceled while THS4631 has bipolar supply ) But not required in some cases. \$\endgroup\$ Feb 15, 2023 at 22:59
  • 1
    \$\begingroup\$ The DC blocking capacitor is a bad idea, you're changing the function of the circuit. C1 will also change the output impedance, so it won't be matched anymore (though that may not be a problem; you can get away with the impedance at the source not being matched if the sink impedance is well matched). \$\endgroup\$
    – Hearth
    Feb 15, 2023 at 23:09
  • \$\begingroup\$ For 100MHz output 0.01uF capacitor has 0.15 ohm impedance but are you right, capacitor change the function of the circuit. \$\endgroup\$ Feb 15, 2023 at 23:13
  • \$\begingroup\$ Putting feedback after the resistor also both obviates the resistor (it is no longer terminating the line, at low frequencies where feedback is active) and makes the amp prone to oscillation (such as when an unterminated cable is connected). \$\endgroup\$ Feb 16, 2023 at 1:25

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