Feeding multiple 50 ohm terminated outputs from a single 50 ohm source without signal degradation

I'm having an issue in which I have a trigger from an instrument meant to be sent to another instrument with a 50 ohm termination. I want to use that signal to distribute it to several outputs that are also terminated with a 50 ohm resistance. Because these terminations are now effectively in parallel, the signal voltage level drops significantly (because of the 50 ohm output resistance of the source) and the triggers are missed.

Because of the BNC connectors, it is not very clean to interface to an op amp buffer, that I would have to design a whole PCB for.

Note: The voltage levels are not those of standard coax video formats (e.g., SDI), so commercial inline buffers/splitters are not really relevant. Instead, it's $5 \text V_\text{p-p}$.

• what are the voltage levels, then? Commented Nov 4, 2017 at 16:53
• Voltage levels are 5 VPP. Commented Nov 4, 2017 at 16:54
• so, pretty standard. Commented Nov 4, 2017 at 16:54
• Kevin raises a good point: what's the slope and delay accuracy you need? Commented Nov 4, 2017 at 16:56
• It's a TTL-compatible pulse with 2-5 V for high and 10 us separation between pulses. Commented Nov 4, 2017 at 18:59

So, energy conservation says that if you have N loads that you want to put the same energy into (which is synonymous with the same voltage, given constant termination), you need N times the energy.

You'll have to buy or build a distribution amplifier.

Note that termination is not necessarily the same as wave impedance – you can, using one of many architectures, build a splitter that conserves the transmission line impedance (i.e. 1× 50Ω in, N× 50Ω out), but that will, too, divide the power.

What is the speed of this trigger signal, what is the sensitivity of the instruments?

At low speed you can just use a tee connector to send the trigger to multiple instruments (and possibly disable the terminator at the instruments).

A passive splitter may be ok also if the trigger channel is adequately sensitive.

If not the signal will need a distribution amplifier as in Marcus' answer.