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MPC506: Circuit Design for MUX MPC506

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Replies: 10

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Part Number: MPC506

Hi TI

I am going to use MPC506 to select up to 15 channels, analog, and digital signals will be passing to the output pin.

Is it possible to power it with a single power supply +5V and leave the negative supply pin floating?

Can I also connect the Vref pin to the same +5V?

  • The datasheet says:

    If the +15V and/or –15V supply voltage is absent or shorted to ground, the MPC507A and MPC506A multiplexers will not be damaged; however, some signal feedthrough to the output will occur. Total package power dissipation must not be exceeded.

    The datasheet does not appear to state what the minimum supply voltage is.

    VREF should be connected only if it is larger than 5 V.

    Why not use the CD74HC4067?

  • Hi Basel,

    It is not recommended to leave the VSS pin floating. The part of the datasheet the Clemens is referring to is how the device operates when there are no power supplies on either the VDD or VSS pin. Grounding the VSS pin will ensure proper operation of a single sided supply - when left floating the device will not act as expected.

    Is it possible for you to GND the VSS pin in your application?

    Also is there a specific reason you chose the MPC506 for this application?

    Please let me know!

    Best,

    Parker Dodson

  • In reply to Clemens Ladisch:

    HI 

    Thanks for your answer, so no need to connect Vref to +5Volts.

    I can't use CD74HC4067 because there is no OVERVOLTAGE PROTECTION feature in it, and the Mux circuit designed to work as a signal sniffer.

  • In reply to Parker Dodson:

    Hi Parker

    So the negative supply pin is also a VSS pin? no there is no problem connecting it to GND, but it is a problem to connect it to -15Volts.

    Yes, there is a specific reason to use this Mux, is that MPC506 has an OVERVOLTAGE PROTECTION feature in it, and I don't want to let the signals interfacing while the MUX turned off, and this is a common case in my application.

  • In reply to Basel Tahboub:

    Hi Basel,

    To Clarify this part is intended to be a bi-polar supplied part - however it has been shown to work when connecting V- to ground. This comes at the cost of some more leakage current that could cause the chip to consume power. 

    However what are the expected overvoltage conditions you expect to see at the input and what is nominal signal levels - I want to make sure this part will work for the application. At (V-) = 0V and (V+) = 5V you will have a protection range of [0 - 20V, 5V + 20V] -> [-20V, 25V]. 

    One more question on the application, you say you want the mux to act a signal sniffer, but in this application the mux would protect the watchdog MCU from an overvoltage event - it will only pass voltages in the range (V-) to (V+). Maybe I am misunderstanding the application, if you could elaborate a little more that would be helpful. Mainly I want to know what information about the signals that are being passed. Is the goal that if they surpass a predetermined threshold the watchdog timer will react? If the supply is the same for both the watchdog and the Mux an overvoltage event may not be passed through the device - however I could just be misunderstanding the application!

    Please let me know!

    Best,

    Parker Dodson

  • In reply to Parker Dodson:

    Hi Parker

    those signals are output and input signals to my board and will have a maximum voltage of 4 volts (digital (0V-+3.3V), analog (0V-+4V)).

    this application mainly will work under two conditions. in the first one, the mux will be powered on and will pass all the signals sequentially to the Watchdog MCU that will work on logging it as data on a MicroSD card, the second is the critical condition while the mux is turned off  (V-) = 0V and (V+) = 0V, but there are signals on its pins which is still transmitting between the other two MCUs.

    however, I have designed an evaluation board that uses this mux ADG406BNZ, but in the off condition, signals were interfacing between each other.

    I hope you can now understand it if this didn't help I will make another explanation by a detailed drawing.

    Thanks

  • In reply to Basel Tahboub:

    Hi Basel,

    That makes perfect sense, thanks for the explanation! So what you need is powered off protection which this part does provide. 

    This part should work in the application you described. However there is one caveat, this part will have a very high Ron at a 5V supply, in the kill-ohm range so depending on the input impedance there could be some insertion loss that could degrade the signal going over the part. That would be my only concern using this part in the application.

    There are other options but that are cheaper, use less PCB area, and fit the use case of the application better - the trade off is that it is a multi IC system - so it would increase the needed supply bypass capacitors by 2. One option could be to use 3 SN74CBT3253C devices. It supports the powered-off protection functionality needed. Each IC has 2 4:1 muxes inside that are controlled via the same two control pins. Two of these devices can be used to monitor the 15 signals needed and one device on the backend to select from the incoming signals. It would have the same number of GPIO to control the system as the MPC device. If this is a possible option for you please let me know if you have any questions.

    Best,

    Parker Dodson

  • In reply to Parker Dodson:

    Hi Parker 

    Sorry for the late reply but it was the weekend here and couldn't reach the office.

    I will try first to stick with the MPC part since it will reduce the complexity and device number.

    However, is there any solution to decrease Ron or to avoid any voltage drop on the signal?

    Thanks

    Basel 

  • In reply to Basel Tahboub:

    Hi Basel,

    No need to apologize, I understand !

    That makes sense and I see where you are coming from. 

    The only way to decrease Ron is to increase the supply voltages. The larger the range of VDD - VSS the lower Ron will be. 

    Depending on how precise the analog signals need to be you have a couple options:

    1. Leave the solution as is - even though it will have high on resistance, it generally will still be much smaller than the input impedance of an ADC/MCU which is typically around 1M Ohm or higher. While you will still lose some signal across the mux, if the load impedance is very large and draws a small current the error might be negligible. Essentially the voltage seen at the output of the mux , assuming the signal frequency is low enough where the parasitic caps are open circuited,  is V_in * R_load/(R_on + R_load) --> so the larger R_load is w.r.t. to R_on the less insertion loss will occur due to Ron.

    2. If you don't want to sacrifice analog accuracy - a high impedance buffer would be recommended. This will draw very little current through the mux allowing for a low insertion loss. Essentially you are looking for a buffer that has low offset and bias currents and I'd focus on CMOS architecture devices as they tend to have very low input bias currents as well as typically can swing rail to rail if you want to go down this pathway.

    If you have any more questions please let me know!

    Best,

    Parker Dodson

  • In reply to Parker Dodson:

    Hi Parker

    Thanks for your understanding, I will supply the mux with 12Volts hope it will work fine and get a good signal accuracy.

    Thank you
    Basel Tahboub