I am trying to build a high frequency (~ 5MHz) level shifter circuit. For this purpose, I designed a comparator circuit based on TI's THS3491.
The circuit schematic is shown below:


simulate this circuit – Schematic created using CircuitLab

The D, D_bar voltage levels are 3.3V.

The circuit is working well until the ground of the square waveform generator and the power supply circuit is kept different. As soon as I connect both the grounds, high frequency signals gets pulled up to a digital 1 (24V), at lower frequencies, i.e. below 500KHz, I am still able to receive a waveform (though not perfect).

I am not able to figure out the reason for the same. Any help would be appreciated.

  • 1
    \$\begingroup\$ You must never consider designing any circuit with > 100MHz BW until you understand Transmission Line theory*. All input source , path and load impedances must be matched low impedance e.g. 50 Ohms . You must also consider traces and wires are 0.5~1nH/mm and C depends on ground plane signal track width/gap ratio ~1 or computed properly*. This is a complete newb question, but you will learn to balance L/C ratio with R \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 25 '18 at 13:20
  • \$\begingroup\$ Thanks for the direction, will surely take these factors into account the next time I design. \$\endgroup\$ – Ayush Agrawal Jul 3 '18 at 11:05

Any help would be appreciated.

The THS3491 is not a conventional voltage feedback op-amp but a current feedback op-amp and, because of this, it has a very low input impedance (circa 15 ohms) at the inverting input. It also requires "headroom" on the non-inverting input of greater than 4.3 volts from either supply rail. See page 6 of the data sheet.

The fact that you may have it working with signals that don't share a common ground with the op-amp's power rails is pure luck. You are likely to destroy the device the way you are driving the inverting input (fed from D_bar).

To use this device correctly you should study what current feedback amplifiers require from an input source.

I would be tempted to use a split rail of -5 volts (to give input headroom) and +26 volts then operate the op-amp with a non-inverting gain of about 7 to convert a 3.3 volt 5 MHz to about 24 volts p-p. You will also need a bias voltage to offset the output positively to make an output waveform going from approximately 0 volts to 24 volts.

  • \$\begingroup\$ This was exactly the problem. I didn't see the current feedback part. As I was using the pin of an FPGA for sourcing the current, it was not able to provide the current and hence the circuit was working without connecting the ground. After connecting the ground, the current had a path to flow. Used a voltage feedback amplifier for solving the problem. \$\endgroup\$ – Ayush Agrawal Jul 3 '18 at 11:04
  • \$\begingroup\$ Pleas feel free to mark this answer as accepted @AyushAgrawal \$\endgroup\$ – Andy aka Jul 3 '18 at 12:54

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