0
\$\begingroup\$

I'm trying to design a clock circuit to convert a 50 MHz 5-volt clock signal down to 3.3 volts without inverting it. The concept is relatively simple, but as I never designed anything like this from scratch before, I'm beginning to look into concepts that I never considered in designing circuits. For the scope of this question, lets limit it to the output signal.

Assuming that I manage to generate a 3.3 voltage clock source at 50 MHz, the signal has to travel across a SMB cable, for now at most 6 feet (1.83 meters), to a NIM module, whose connector has a 50 ohm impedance at its inputs. This got me thinking about transmission lines in general. When using other devices that require a coaxial connection, such as an oscilloscope, I've used various lengths of cables to connect from a source to them, and the signal stays relatively intact with minimal voltage drop. At 50 MHz in the very high frequency range, I would assume that there will be some voltage drop from the generator output to the NIM module input, but aside from matching the impedance at the output of the generator, is there anything I can do to mitigate the probable voltage loss across the line?

Sorry if the question is unclear. I was just thinking about this since cables come in various lengths, and signal generators tend to produce waveforms within significant tolerance since there's no guarantee way to match the characteristic impedance for every length. I'm just trying to design it in such a way that I don't have to rely on a certain cable of a certain length, so if the situation arises, my team can use a different length cable.

For a rough reference, here's a quick and dirty schematic of how I roughly imagine this problem to be. The input port of the module may have a 50 ohm impedance, but I'm not sure if you could treat it as a 50 ohm load in a schematic.

schematic

simulate this circuit – Schematic created using CircuitLab

If you really want specifics, I'm working under the assumption that the cable is coaxial, with copper for the conductors and teflon for the insulator. The cable itself is 2.489 mm in diameter while the center conductor is 0.508 mm in diameter.

\$\endgroup\$
  • \$\begingroup\$ if the coaxial cable has a characteristic impedance of 50 ohms, it will be 50 ohms no matter the length. The voltage drop will depend on the length of coaxial cable and the quality, although it will probably be quite low at 50 MHz \$\endgroup\$ – Joren Vaes Sep 28 '17 at 19:55
  • \$\begingroup\$ @JorenVaes, even at DC there's a voltage divider formed by the source impedance and the load impedance, which reduces the signal voltage by half (assuming you do have source termination). \$\endgroup\$ – The Photon Sep 28 '17 at 20:01
  • \$\begingroup\$ @ThePhoton ofcourse, but the cable does not change this and assumign a lossless cable (which I think is valid for a few meters length at 50 MHz) the lenght of the cable will not impact this behavior. \$\endgroup\$ – Joren Vaes Sep 28 '17 at 20:06
  • \$\begingroup\$ OP, are you stuck using full-swing CMOS logic signals? Often when working with 50 ohm lines it makes more sense to use ECL or some other logic family that is designed to work with such a heavy load. \$\endgroup\$ – The Photon Sep 28 '17 at 20:10
  • \$\begingroup\$ @ThePhoton Thanks for your reply. I'm not entirely sure if you would consider me 'stuck'. The NIM module input is limited to 3.3 V, so not necessarily, as long as 0 V is still low, and 3.3 V max (as seen by the module) is logic 1. Like Joren Vaes mentioned, with the voltage drop, I could increase the voltage to 6 V to get the 3 V output at the other end, but this is still on paper for now. What worries me is the load impedance (module's internal circuit). You can't assume that its 50 ohms; I would imagine its much greater than that, so the voltage wouldn't necessarily be 3.3 V there? \$\endgroup\$ – user101402 Sep 28 '17 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.