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Thread: Protecting an analog output

  1. #1
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    Protecting an analog output

    I am working on a custom curcuit that has 2 different voltage outputs i need to protect from accidental voltage input up to 18v.
    1. My project has a 5v ouput that runs to a sensor and i would like to protect this from higher voltages flowing back to the voltage regulator and microcontroller. I realise a diode would be very easy but would really like the 5v output( yes 4.7v after diode voltage drop would operationally be fine but im mostly interested in how this can be done as i just cant figure it out).
    2. I am using the teensy analog pin to controll an opamp to then output a 0-5v signal. I would like to provide the same protection from current flowing backwards with this too.

    I can find lots of designs using mosfets to prevent reverse current but im under the impression that in my use these will not work as although the drain of the fet might have 12v and be over voltage the source would still be at 5v and be over voltage.
    The best solution i could find was this http://www.ti.com/lit/ds/symlink/lm74610-q1.pdf which i wasnt sure of its operation but then i read that it has no ground pin so without a ground pin it really has no option but to prevent reverse current does it ?

  2. #2
    Senior Member PaulStoffregen's Avatar
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    Quote Originally Posted by 100 View Post
    I realise a diode would be very easy but would really like the 5v output( yes 4.7v after diode voltage drop would operationally be fine but im mostly interested in how this can be done as i just cant figure it out).
    You can use an opamp to turn a regular diode into a "super diode" with zero forward voltage drop. Or actually, the result is the diode's forward drop divided by the opamp's gain, but since opamps have huge gain, that results in pretty much a zero forward voltage drop.

    http://ecetutorials.com/analog-elect...ave-rectifier/

    Of course, this circuit requires feeding the output back to the opamp's negative input, so you'd need an opamp which can withstand the 18V possible output.

    The other big disadvantage of this circuit (or a regular diode) is your output is not longer actively sinking current, only sourcing. The load resistor provides all current sinking, which might not give you as good an output signal as active push-pull drive from an opamp.

  3. #3
    Senior Member PaulStoffregen's Avatar
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    A more common way to make rugged analog outputs involves a low value resistor in series with the opamp output.

    This sort of design is very common in modular synthesizers, where analog outputs need to withstand accidental connection to +/- 12V. I'm sure you can find many published designed if you search for "modular synth schematic".

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    That device is designed for this use (active diode) but it's doing some voltage multiplication magic to power itself. See the drawing on page 9 for how it's actually got the FET off for some of the time with a full diode drop present giving a pulsed output. Still far less heat dissipation in the diode than the classic solution but may not be what you want if this is supposed to be a stable output.

    Suspect searches for 'active diode' may net you some useful hits
    I got this one
    http://au.mouser.com/new/Diodes-Inc/...-controller%2F
    Which I think does what you want (see note about use in redundant power supplies) but you can probably find similar devices from whichever supplier is your choice.

    Less certain how these will handle an analog output being reverse driven.

  5. #5
    Senior Member+ Theremingenieur's Avatar
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    This is how Bob Moog did it :
    Click image for larger version. 

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    Quote Originally Posted by PaulStoffregen View Post
    A more common way to make rugged analog outputs involves a low value resistor in series with the opamp output.

    This sort of design is very common in modular synthesizers, where analog outputs need to withstand accidental connection to +/- 12V. I'm sure you can find many published designed if you search for "modular synth schematic".
    Thanks Paul but for both methods if i only have a 5v supply to the op amp wont the maximum output still be 5v- diode forwards voltage or for the method with the resistor the drop would be would be v = ir. I guess with either even if it didnt go all the way to 5v the op amp would compensate for any temperature effects on the diode or resistor.
    Im researching the other options now.

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    After looking at alot of different devices related to the gremlinwranglers post which look suitable im not wondering if all i really wanted was a pct fuse and a zener diode or will the responce of the pct be to slow to protect everything ?

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    Quote Originally Posted by Theremingenieur View Post
    This is how Bob Moog did it :
    Click image for larger version. 

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    This is the classic way to protect outputs - clamp the output to a range that the drivers can tolerate and limit the current to within the output drive capability of the drivers. CAVEAT: people forget that V+ is a common power supply rail and a sustained over-voltage condition on that protected output can result in V+ increasing. Essentially, you can pull the power supply out of regulation and increase V+ to all your parts. This might result in a protected driver but a roasted micro! Put an overvoltage clamp on V+ that limits V+ to the max that all the parts can tolerate OR create a separate Vclamp that is a shunt regulator or hefty zener diode and lots of capacitance.

  9. #9
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    Just updating for any one else looking for an answer, in the end after heaps of reading and learning for the voltage output this time a simpler option was better so i just ran the 5v output through a diode but also used an analog input with divider and over voltage protection to read what that output voltage was so any changes could be compensated for in the signal returned from the sensor which is ratiometric from 4.5 - 5.5v and i will just do that analog outputs through the diode with feedback from the other side as paul sugested. I think i got it stuck in my head that i needed 5V when it really wasnt the case but i learnt alot

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