# How to properly terminate this circuit (50ohm coaxial to oscilloscope)

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 50ohm resistor.

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

Is it necessary to put 50ohm resistance like this?
a)

Or the 50ohm resistor should be like this?
b)

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

Or just put 50ohm termination on the "other side"?
d)

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:

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

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

b)

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

c)

Output sees 5V pulse. Current peaks at 100mA. No bounces.

d)

Output sees 10V pulse. Current peaks at 200mA. No bounces.

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

• a is fine, in b move the resistor to the end of the transmission line – Sorenp Dec 20 '19 at 9:50
• @Sorenp I have just added even option d). So you say even a) and d) are properly terminated? – Chupacabras Dec 20 '19 at 9:54
• 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 – Sorenp Dec 20 '19 at 9:59

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.

• 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? – Chupacabras Dec 31 '19 at 11:47
• @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. – Andy aka Dec 31 '19 at 11:54

(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.