1
\$\begingroup\$

I have this switch datasheet.

I also have a connector with default pins arriving to switch D point. S points are connected to another circuit point.

Note: D side connector is an output connector and matches with another input connector that will connect D points with IN logic gate inputs.

When the set of switches is ON, S will be connected to D point. But I am interested in looking at D during the OFF state.

I'm assuming that D is high impedance state during off. I want to turn this high impedance state into a "high level state" signal, 12V, in order to be a logical gate input for the next stage.

I have two questions:

  1. Could be said that D is really totally disconnected (high impedance) or I am missing something about the devices that can influence that they are not fully electrically isolated to say that they are in high Z?

  2. Is it a correct way for turning high impedance into 12V?

  3. How can I choose an appropriate value for this Resistance?

enter image description here

\$\endgroup\$
1
\$\begingroup\$
  1. Assuming the switch is a standard metal contact switch then yes D would be completely disconnected from S and can be considered high impedance as its impedance would be the air space inbetween the two contacts

  2. Youre not 'turning high impedance into 12V'. This is because regardless of what you do to point D, so long as the switch is not ON, its still high impedance. All your doing is connecting a 12V source to a high impedance node such that current will not flow from the 12V source back to S. You would be correct in saying that you have a method for controlling a logic signal by changing the impedance seen at point D

  3. That entirely depends on the logic chip. If the inputs of the chip are high impedance you would only need a small amount of current to drive it. A typical value would be something like 10k which would provide 1.2mA. You would need to read the datasheet of the logic chip to get a better idea of what requirements it has

\$\endgroup\$
  • \$\begingroup\$ Hi @TheAndyEngineer.. Thanks for your efford. If I understood you mean that my words are wrong, but I really will have 12V at IN node, when switch is off. Is it? And yes, I'm connecting 12V to a disconnected point such the current can' t flow because this circuit point is open. I will try to follow your instructions about R calculation. If I was lost with it, I will come back with the datasheet information. \$\endgroup\$ – Eugenia Suarez Sep 9 '19 at 7:46
  • 1
    \$\begingroup\$ When the switch is off you would have 12V at in node. Current always flows to the path of least resistance. If point D is high impedance and the logic chip offers lower impedance then the current from the 12V source will flow to the logic chip. Please dont just put in a 10k resistor and try it you need to read the datasheet otherwise youre guessing and may fry the chip \$\endgroup\$ – awsem_eng Sep 9 '19 at 7:49
  • \$\begingroup\$ thanks @TheAndyEngineer I will take care about this. I won't do anything without reading datasheet. DC input current any one input = 10mA, says the datasheet: ti.com/lit/ds/symlink/cd4071b.pdf \$\endgroup\$ – Eugenia Suarez Sep 9 '19 at 8:53

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.