Is it possible to make the output of this circuit 0V, when the input is floating as shown in the picture . I tried adding a pulldown resistor to the input, but it only worked at a very low value(100 Ohms.)


Can anyone provide a better schematic, one that does not sink current to the input and reflects floating input as 0V output? I have extra components available to me

  • \$\begingroup\$ Which side is the input? Why not use a better design? \$\endgroup\$
    – pjc50
    Commented Dec 12, 2013 at 16:34
  • \$\begingroup\$ Only if you reinvent TTL. \$\endgroup\$ Commented Dec 12, 2013 at 16:35
  • \$\begingroup\$ Left is the input, its floating. So there is absolutely no way to do this, with any combination of components? \$\endgroup\$
    – Ollusaurus
    Commented Dec 12, 2013 at 16:40
  • \$\begingroup\$ No, not with that circuit. You need a different circuit. \$\endgroup\$
    – Dave Tweed
    Commented Dec 12, 2013 at 16:48
  • \$\begingroup\$ You need a circuit like this. \$\endgroup\$ Commented Dec 12, 2013 at 17:00

2 Answers 2


It appears that your intention is to have the output voltage roughly mirror the input voltage when the input is driven to a voltage below 2.3 volts, and otherwise have it passively pulled up to five volts. As implemented, the circuit will also have the effect that any current sourced into the output must be sunk by the input; it is unclear whether that behavior is desired, or simply tolerated.

If you wish to have the output be low when the input is floating, then you must either have whatever would pull the output high refrain from doing so except when the input is high, or you must have something which can sink current to ground when the input is floating. It will be necessary to either use a transistor in an inverting-amplifier configuration, or use two or more transistors operating opposed; two transistors are going to be necessary in any case. If you need a two-transistor circuit whose output polarity matches the input polarity, the simplest approach may be to use a pair of common-emitter inverters, though an alternative might be to adapt your circuit to replace the output pull-up with a PNP current source (a pull-up resistor to 5 volts will source five times as much current into something trying to pull that pin down to ground, than it would invest trying to keep the input above 4 volts; a current source could supply the same current in both cases).

  • \$\begingroup\$ I came up with both of those 2-transistor circuits, too. The advantage of the NPN-PNP configuration is that it consumes no power when the input is low/open (except for whatever current the load is driving through the output pull-down resistor). The NPN-NPN configuration always consumes current from the power supply. \$\endgroup\$
    – Dave Tweed
    Commented Dec 12, 2013 at 20:36
  • \$\begingroup\$ I got around the original behavior by a manual switch, until the inputs could be pulled low (in software) I just kept the output unpowered. Because this is my first project involving electronics, this seemed to be an acceptable solution.You're right, the behavior of the input sinking the current was not desired and only tolerated, up until now - sinking too much current started crashing the controller and I need a new solution. Could You provide a schematic for the solution you came up with? \$\endgroup\$
    – Ollusaurus
    Commented Dec 18, 2013 at 20:45

The easiest way to accomplish this is to use a voltage translator such as the SN74LVC1T45, with a pull-down resistor on the input.

enter image description here

The transistor level shifter (and voltage dividers in the opposite direction) can be cost-effective in some situations, but their performance is not great, and they have issues such as the one you've discovered. Tie the DIR line high (3V), make \$V_{CCA} = 3.0V\$ and \$V_{CCB} = 5.0V\$ and put a 10K or so pull-down resistor to ground on the A input.


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