I'm currently working on a project that requires interfacing to a clock and data line at a configurable voltage between 0V-5V with nonstandard logic levels, this voltage is generated elsewhere. To be able to receive data the lines needs to be able to go into a high impedance state.

I'm currently playing around with a 4066, I have two lines of two pairs hooked together, one channel of the 4066 is hooked to the voltage and the other to ground and I simply drive them inverted. To get a high impedance state I drive both switches low. This being said it doesn't work real well since I'm having trouble hitting the ground rail.

All the buffers I found with tri-state outputs are for digital logic levels, could anyone recommend a analog one or another approach?

(For anyone interested I'm investigating the lowest voltages that I can use to interface to various IC's)

I eventually found this bug, I replaced the the 4066 with one from another manufacturer and it worked...

  • \$\begingroup\$ Perhaps you could accomplish something by paralleling the analog switch with an open drain buffer that would do a better job of driving to ground? \$\endgroup\$ – Chris Stratton Jan 5 '12 at 18:59
  • \$\begingroup\$ Can you use that variable voltage to supply the buffer? If so, couldn't you actually use a digital logic level buffer supplied with your target voltage? \$\endgroup\$ – AndreKR Jan 5 '12 at 19:01
  • \$\begingroup\$ The problem with using a digital buffer in this way is that I wouldn't be able to use the entire 0V-5V range, specifically the lower portion which is what I'm interested in, I have yet to see a digital IC that can operate on such low voltages. \$\endgroup\$ – s3c Jan 5 '12 at 19:05
  • \$\begingroup\$ From TI's logic selector guide (ti.com/litv/pdf/sdyu001z) it looks like they have an AUC family that runs from 0.8 V up to 2.5 V, and several families that run from 2.0 up to 5.0 V; but I don't see anything that covers from sub-1-V up to 5 V. How low a logic voltage do you need? \$\endgroup\$ – The Photon Jan 5 '12 at 19:31
  • \$\begingroup\$ Also, it's not clear why you need to disconnect both the signal line and the ground (return) line. If you had a tristate digital buffer, it would only decouple the signal line. Do you really need to disconnect ground as well? \$\endgroup\$ – The Photon Jan 5 '12 at 19:34

If I understand correctly, you are looking for an analog multiplexer with the option of setting the channels to high impedance.
If so, the 4052 or 4053 should do the job, they have an enable input which sets the ports to high impedance.

As mentioned in the comments, it might be a good idea to post a rough schematic of what you are doing, as it's a little unclear at present.

| improve this answer | |
  • \$\begingroup\$ If I were in the same situation the most useful part to me would be knowing that the device I need is called an analog multiplexer. I have used them and knew that, but trying to find one looking for an analog tristate buffer I see as a daunting task. \$\endgroup\$ – Kortuk Jan 6 '12 at 3:50
  • \$\begingroup\$ The on resistance of these chips seems to be lower than the CD4066B but several times higher than the CD74HCT4066. Still, there are probably even better devices out there. \$\endgroup\$ – Chris Stratton Jan 6 '12 at 4:33

Different `4066 devices have very different on-state resistances under various conditions. For example:

TI CD4066B is around 400 ohms at 0v input with 5v supply, but 100 ohms for the same input voltage if you increase the power supply to 15v. Cool it from room temperature to -55C and it's down to 70 ohms.

TI CD74HCT4066 is only 25 ohms for 0v input with 4.5v supply, at room temperature.

There are of course other manufacturers and other switch chips.

If you had a slightly negative power supply (-1V or so), you could use the FET switch with an op-amp follower circuit driving it and taking feedback from it's output, both powered from that negative rail and thus attempt to achieve unity gain from input to output of the assembly.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.