Latch with a three-state input

[jnoche]

I need a latch (whose output Q is initially 0) that has a three-state input D and if that is at a high impedance, then the output is unchanged. For example, if the input D goes 0, Z, 1, Z, 0, 1, Z, 1, 0, Z, then the output should go 0, 0, 1, 1, 0, 1, 1, 1, 0, 0. My attempt is shown below.

But this doesn't work as I want it. What happens is if the input is Z, then the output is unpredictable. I tried using the built-in RS latch (see below)

but it also doesn't work as I want it. (A high-impedance input causes an unpredictable output.)

Is there a way to create the component that I want using the Digital software?

[mengstr]

I think this contraption does what you want... maybe.... ;-)

[hneemann]

The "Bus-hold" or "pin keeper" is not possible in this simulator because there are no real resistors. But you can build it like this:

This would possibly be an even more elegant solution to the original question and would also actually be possible to build in reality.

[mengstr]

Ah yes, that's definitely the best solution here!

[jnoche]

How do I modify this simpler circuit so that its initial value (in Digital) is 0?

[hneemann]

Somewhere on the bus a reset logic must be added:

[jnoche]

Thank you for this.

[hneemann]

When a real circuit is built in CMOS technology, an open input is unpredictable and can damage the circuit. And also in TTL an open input is not advisable, although it is recognized as high. The simulator simulates this "unpredictability" by generating random values.
But you can detect high-z in the simulator with a special circuit, and build an according logic.

[jnoche]

Thank you very much for this. Is there any particular reason the transistor at the top is floating gate and the one at the bottom is not?

[hneemann]

Is there any particular reason the transistor at the top is floating gate and the one at the bottom is not?

Thanks for the hint! No, there is no reason! I made a mistake, which I have now corrected!

[jnoche]

I'm writing this information here so that future readers can benefit from it. I've decided to use the Neemann circuit with the RS latch (which I'll call Neemann1) instead of the simpler Neemann circuit (Neemann2). It's because the simpler circuit results in an error when it is placed in the complex design I have.

Neemann1 with D=Z looks like this:

Note that the input box is gray with a white background; it remains at high impedance.

Neemann2 with D=Z looks like this:

Note that the input box is gray with a light green background; it used to be high impedance but is now pulled up. The external circuit connected to the input is affected, causing an error.