Connecting Multiplexer/Demultiplexer input to Vcc of the Same Chip,will this make any performance issues

Thread Starter

hoyyoth

Joined Mar 21, 2020
192
Dear Team,


In one of my circuits, I am connecting the I/P pin of Mux/Demux to Vcc (5V) as shown below.I have 2 questions.

Question-1

May I know when switching, will the current draw cause performance deviations in this connection.

Question-2

The Can I Connect EN Signal(Coming from Arduino) to D also instead of connecting to VCC

1636650759184.png
 

MisterBill2

Joined Jan 23, 2018
10,555
If you are actually operating magnetic relays by using a multiplexer as the driver, that seems like a rather poor choice.
So just what are you attempting to do? It is really not very clear. There are decoder/driver IC devices made for that purpose. A CD4028 will decode 4-bit binary and give good logic outputs.
 

AnalogKid

Joined Aug 1, 2013
9,631
Question-2
The Can I Connect EN Signal(Coming from Arduino) to D also instead of connecting to VCC
I recommend against this. No matter how fast is the EN signal risetime, it still is finite. There is a point somewhere in the middle of the edge when it starts to affect the mux internal circuits. If the D input also is changing state at the same time, there might be unintended, very brief output signals. I would leave D connected to Vdd.

Also, you can delete R94 and connect D directly.

ak
 

Thread Starter

hoyyoth

Joined Mar 21, 2020
192
Dear Team,

Thank you very much for your comments.I decided to change the connection as shown below.

1637143471066.png
Regards
HARI
 

MisterBill2

Joined Jan 23, 2018
10,555
I recommend against this. No matter how fast is the EN signal risetime, it still is finite. There is a point somewhere in the middle of the edge when it starts to affect the mux internal circuits. If the D input also is changing state at the same time, there might be unintended, very brief output signals. I would leave D connected to Vdd.

Also, you can delete R94 and connect D directly.

ak
Looking at the circuit it does not seem the "D" signal, Vcc thru a resistor, will be changing very much. So that is probably not an issue in this application, since the ANALOG multiplexer is being used as a binary to decimal decoder/driver. That choice is a puzzle.
Just what is the "Mod REL1 EN supposed to do? If we have more information there will be a lot less guessing and more useful advice.. If the multiplexer output is supposed to drive a relay I predict the circuit shown will not perform as hoped.
 

AnalogKid

Joined Aug 1, 2013
9,631
Looking at the circuit it does not seem the "D" signal, Vcc thru a resistor, will be changing very much. So that is probably not an issue in this application,
Actually, I was responding to his question about disconnecting the D input from Vdd and connecting it to the EN signal out of the microcontroller. This smells like it could create a problem where there isn't one.

However, his latest schematic shows this change.

ak
 

MisterBill2

Joined Jan 23, 2018
10,555
Once again, it is not the best design to use an analog multiplexer for digital decoding like this. There are quite a few binary to decimal decoders around that are better suited, smaller, use less power, and cheaper. A CD4028 is the first one that comes to mind. And for the driver IC, use a darlington transistor such as the MPSA13 to avoid needing the resistors.
 

Thread Starter

hoyyoth

Joined Mar 21, 2020
192
Hi Bill,

Thank you very much in next revision I will consider CD4028 but CD4028 don't have any enable pin.

Could you please tell what you mean by" use a darlington transistor such as the MPSA13 to avoid needing the resistors. "
You mean I can remove base resistors.

Regards
HARI
 

AnalogKid

Joined Aug 1, 2013
9,631
Don't think so. Many power darlington transistors have base pulldown resistors built-in. Something like a TIP110 would eliminate R36. Or, for driving multiple relays, a ULN2003 darlington transistor array.

The MPSA13 does not have these internal resistors.

ak
 

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MisterBill2

Joined Jan 23, 2018
10,555
No, the CD4028 does not have an enable pin, but it does decode all 16 of the 3-bit binary codes, including zero, so if no channel is selected none will be active. That may work, depending on the controlling code.

The Motorola MC14512 also is a "data decoder" and it includes both an inhibit input and a disable input.
The MPSA13 is a darlington transistor with a gain of several thousand, so it will switch into full saturation with just a bit of base drive current. And because it has essentially two base/emitter junctions in series the base voltage at saturation is higher. It might be that it could still use the resistors, but I do not recall using them when driving it with the outputs of a CD4017 counter IC.
Because the output is both active high and active low you would not need the pull-down resistor, though.
There are other similar decoders by other makers and so a bit of research is possibly in order. The twwo I mentioned are mature products and should also be available from other makers as well.
 
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AnalogKid

Joined Aug 1, 2013
9,631
Because the output is both active high and active low you would not need the pull-down resistor, though.
To be clear, this is true for the CD4028 and MC14512, but not for the TMUX1208. Its output is not a CMOS totem pole output stage. Effectively, it is a series resistor with a value that is either very high or very low. When an output is "off", as in not selected, it is an open circuit, essentially floating. This is not good for the relay driver, especially if it is a darlington type; its high gain makes it more sensitive to environmental noise.

The MPSA13 is a darlington transistor with a gain of several thousand, so it will switch into full saturation with just a bit of base drive current. And because it has essentially two base/emitter junctions in series the base voltage at saturation is higher. It might be that it could still use the resistors, but I do not recall using them when driving it with the outputs of a CD4017 counter IC.
It sounds like you were relying on the relatively high output impedance of a CMOS B-series gate to limit the transistor base current to a safe value. For any device that is trying to make 5 V, 9 V, 12 V, or whatever Vdd was, to be clamped to 1/2 V, is a significant over-stress of the output stage.

That brings up another point. Because the 1208 is an *analog* mux, a selected output is connected to the D input with a 5 ohm resistor. This means that the relay driver device is connected directly to the microcontroller output pin, and any current required by the driver must be sourced by that pin. With a darlington driver transistor and no base series resistor, the uC output pin will see a two-diode-plus-5-ohm-resistor path to GND. At any output voltage above an idealized 1.2 V, the uC pin will see a 5 ohm resistor. So, if it is trying to make a typical high output signal of 3 V, the required output current would be almost 1/4 amp. This is an unnecessary over-stress of the uC output stage.

Based on this, I recommend returning the D connection to Vcc, to isolate the microcontroller output pin from the relay driver current requirements. Also, if you connect D to Vcc through a resistor, as in post #1 but with a lower value, that one resistor is the base current limiting resistor for whichever output is selected, replacing all instances of R35. You still need an R36 pulldown resistor for each driver, or a driver with one built-in.

ak
 
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