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I have a signal (digital) which has output levels of 0V (logic 0) and 3.5V (logic 1).

I would like to know if it is possible to shift it between -2V (logic 0) and 1.5V (logic 1).

I found here on stackexchange and on google various example, but they all assume positive levels (0-5 or 0-3.3) and I don't really understand how to make them work in my case.

Thanks for all the help

EDIT after a couple of questions:

  • I have available easily on board: -2V, 0V, 1.5V, 5V
  • I am sending the signal to a CMOS logic gate

EDIT 2: Just to add a bit more to my question:

if you look at this thread Level shifting a +/- 2.5V signal to 0 - 5V

the first answer provides a very nice circuit to do level shifting.

My problem is that I do not understand how to modify the circuit to fit my needs (see above).

In particular, I am quite puzzled by the supply value of the OpAmp. It looks to me I should use +1.5 and -2, but then I get lost in understanding what is going on.

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  • \$\begingroup\$ Sure it's possible - what power supplies do you have available and what are you interfacing to? \$\endgroup\$ – Andy aka Mar 8 at 10:22
  • \$\begingroup\$ 3.5 is a bit strange. But anyway the simplest way is to use digital isolation and power supplies according to your needs. \$\endgroup\$ – Gregory Kornblum Mar 8 at 10:24
  • \$\begingroup\$ Edited with clarification, thanks! \$\endgroup\$ – Cesco Mar 8 at 10:48
  • \$\begingroup\$ What’s the speed & edge slew rate required for the signal? \$\endgroup\$ – Edgar Brown Mar 8 at 15:54
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schematic

simulate this circuit – Schematic created using CircuitLab

The LM358, being a single supply op amp, saturates in the negative direction within a few mVs of the negative supply rail.

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You did not mention any more specific requirements for the output signal, i.e. frequency and output impedance, but this is a very simple circuit that could get the job done for low to medium frequencies and with an output impedance determined by R1:

schematic

simulate this circuit – Schematic created using CircuitLab

V1-V3 represent your power supplies of +3.5V, +1.5V and -2V w.r.t. GND.

Using a P-channel MOSFET you can switch the +1.5V line to the output. The pull down resistor R1 provides -2V to the output when the MOSFET is off.

Because we need a P-MOSFET here, which turns on when the gate is driven low, an additional inverter is needed. This inverter however can be as simple as another MOSFET and a resistor.

Note that I just picked a random FET to illustrate the circuit. You would need one that already turns on sufficiently at -3.5V.

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