Amplify .7Vpp square wave to 5Vpp square wave buffered to go over a 15-50 foot cable

Question

I've got a sensor that outputs a .7Vpp square wave at 36kHz-60kHz. I need to take that square wave and amplify it to 5Vpp, as close to 0V to 5V rails as possible. Then I need to send that signal over a long cable, 6-50 feet depending on where I'm using the sensor and have that square wave still be nice and square when it reaches the other end.

I have a single +5VDC supply to work with.

Currently I can get a good square wave by using an LM311P comparator, but when I try to drive it over a long cable I end up with a wave that looks like a shark fin.

Should I be taking a different approach all together instead of the comparator, or is there something I can put between the comparator and the long cable to help the signal?

This is a little bit of a simplification of what I'm doing, but it's basically what I'm doing now. NOTE: the BJT shown is actually part of the LM311P comparator. Also, the 1V source is actually 5V.

^{simulate this circuit – Schematic created using CircuitLab}

Here's the shark fin output at 36kHz, about 5Vpp, that I was describing:

Answer 1

The circuit you're using is able to drive the cable very quickly to the low level, but only pulls up slowly to the high level because R1 limits the current available to charge up the capacitance of the cable. The maximum current you're going to be sourcing into your cable is about 3.3 mA.

One solution is, instead of buffering with the Q1/R1 combination, use a buffer/driver IC such as SN74LVC1G126. This can supply up to 32 mA current either sourcing or sinking. You'll have to work out from your cable length and capacitance per foot whether that will give you adequate edge speeds for your application.

Answer 2

The comparator you are using is an open drain output, which means it only sinks current and uses the resistor to pull the line high.
This means you have an uneven output impedance, with the output high impedance effectively being being the same as the pullup resistor (1.5k in your circuit)

So to put it simply, you need a lower impedance drive, and it helps if the OH/OL drive is similar. The reason for needing a lower impedance is the longer the line, the more capacitance (between the line conductors) the output needs to charge.

You can either swap the comparator to one with a decent push-pull output (i.e. a PNP and NPN or P-ch and N-ch MOSFET working together) or you can buffer the output with a logic gate with decent drive, or even parallel a few together.

Here's a rough simulation with the gate representing your comparator, with a 36kHz square wave input, and a worst case line capacitance estimate of 2nF (Cline).
In the top trace the gate is set to an output high of 1.5kΩ, and in the bottom trace it is set to 20Ω. You can see the expected "shark fin" slope as it takes longer to go high, and then a considerable difference when the impedance is dropped.

Answer 3

Maybe you should rethink this - why not just send the 0.7Vp-p signal down the cable then use a comparator at the receiving end to restore it. I'd suggest a MAX999 comparator at the far end because it's very fast and swings close to 0V and +5V.

If you can't have any circuit added at the far-end then still, I'd use the MAX999 for driving down the cable, or, if you are insistant on 5Vp-p use an AC series CMOS driver like TI's 24mA drive output 74AC86 - I picked on this because I designe dit into something the other day and I believe you can get it in an equivalent single gate package.