For my scientific work I wish to register signals coming from a photon detector which generates 50 Ohm TTL signals:

Tau Spad Specs

The expected count rate will never exceed 4 MHz.

The devices that need to record these signals are an NI-6343 of which the input impedance is orders of magnitude higher than 50 Ohm and a specialized time tagging device with 50 Ohm impedance on its input.

The time tagging device (TCSPC) and the detector are designed to be used together so there is no issue there. However, for reasons not relevant here, I also need the NI device to register the TTLs.

Everything is connected using Mil spec RG58 coaxial cable and I use a good quality 50 Ohm splitter to run the TTLs towards the NI device and my TCSPC.

Naively connecting things won't work: obviously the input impedance of the NI device is such that no current will flow to it and it 'sees' nothing as long as the 50 Ohm TCSPC is also connected on the other terminal of the splitter.

To address this I think I will use an OPA 2350 of which we happen to have a few in the lab. This opamp is 38 MHz so it might work just fine.

I think I need a unity gain buffer:


simulate this circuit – Schematic created using CircuitLab

However, since my knowledge of electronics doesn't go much further than Ohm's law I am not sure I am not missing something obvious:

  1. Will the fact that the output impedance of the opamp is never really 0 mess with the voltage that is 'seen' by the NI-6343 as it is effectively in series with the NI-6343? Should I be wary of any other voltage division effects that might occur, effectively lowering the voltage of the output below TTL spec?
  2. I do not have a -5V supply voltage in the NI-6343, only +5V. Therefore, I would ideally put the supply of the opamp between 0 and +5 but people tell me a 'symmetrical' supply might be better.

Any feedback would be much appreciated!

  • 1
    \$\begingroup\$ Op amps can work with single supplies just fine. But I don't know if you're addressing this properly. The voltage divider of the two 50R would make a TTL '1' (5V) appear as 2.5 V to you data acquisition device. But in the documentation it says, it would accept a '1' as low as 2.2V. Actually , it's weird to match impedances in this case where the input impedance of the counter should be very high. \$\endgroup\$
    – John Wick
    Commented Mar 18, 2016 at 8:52
  • 1
    \$\begingroup\$ For a unity gain buffer you want the feedback going to the "-" input, (not the +). \$\endgroup\$
    – Nedd
    Commented Mar 18, 2016 at 8:58
  • \$\begingroup\$ Check that your NI card can actually count or respond 100% to pulses as thin as 10 ns. \$\endgroup\$
    – Andy aka
    Commented Mar 18, 2016 at 9:04
  • \$\begingroup\$ @JohnWick thanks for the speedy feedback. At the risk as coming across a bit dense, could you elaborate on the 'voltage divider' bit? I'm not sure which two 50R you refer to: 50Ohm on the input and ... Also, why is it 'weird': if I connect both counting devices without an Opamp, nothing is seen on the NI-6343 side. Decoupling the TCSPC immediately resolves this. I can only assume this is because the signal likes to flow to the lower impedance counter. \$\endgroup\$
    – Kris
    Commented Mar 18, 2016 at 9:04
  • 1
    \$\begingroup\$ @JohnWick does the voltage division stem from the fact that I am taking the opamp input between the 50Ohm detector and the 50Ohm TCSPC, which are thus two 50Ohm devices in series, hence they are effectively a divider of the voltage? \$\endgroup\$
    – Kris
    Commented Mar 18, 2016 at 9:14

1 Answer 1


There isn't a need to do anything clever with opamps. Put a Tee piece on your high impedance instruments, daisy chain 50 ohm cables from the source, through the Tees, and finish the daisy chain at the 50 ohm instrument like this (it's a diagram for a 75 ohm system, but the principle's exactly the same)

Note these Tee pieces I refer to are not 'splitters' that contain resistors, but simple Tee connections, zero ohms between all ports and no loading to ground.

This will work whether your source is a low impedance output that will drive valid TTL levels into a 50 ohm load (as I suspect), or a 50 ohm output impedance that will drive TTL levels into a 50 ohm load and >5v levels into an open circuit. I have looked at the data sheet, and it doesn't clarify which, so I suspect the former. If you want, you can tell the difference by running it into a scope with a switchable 50 ohm / high impedance inout and seeing whether the level changes a little, or by a factor of 2.

The way this works is the source launches a TTL level signal edge down the 50ohm line. This is essentially not loaded by each high impedance instrument it passes. Note, fit the Tee pieces to each high Z instrument directly, so the capacitive stub of 50 ohm line is very short and doesn't load the signal significantly. Don't use a Tee to connect 3 bits of cable, and run a cable to the high impedance input, that will get you into trouble! Finally, the edge arrives at the last instrument, and is absorbed in the 50 ohm termination. Each instrument sees one TTL level edge, and one only, and all voltages meet TTL level specs.

There is obviously a progressive time delay down the cable of about 5nS/metre, which has to be allowed for when you analyse your results if that level is significant.


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