Connect with us

AD620 INPUT PROTECTION

Discussion in 'General Electronics Discussion' started by lordmag, Mar 3, 2012.

Scroll to continue with content
  1. lordmag

    lordmag

    11
    0
    Nov 27, 2011
    in the ad620 datasheet the max differential input is 25 V so i want t o make protection for my input amplifier what do you think about this circuit is that the right one ?

    my application is that i want to measure the VOLTAGE of a contact RESISTANCE like what we see in the attach 2
     

    Attached Files:

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,500
    2,840
    Jan 21, 2010
    That will limit either input to prevent them from exceeding either supply rail by more than 0.7V.

    If you need better than this, you can replace the diodes with zener diodes, and as long as their voltage is significantly greater than 1/2 of your supply voltage (greater then say, 12V in this case), you can limit the excursion to some number of volts away from either supply rail.

    Another approach is to place a pair of diodes across the op amps inputs. This will prevent them from differing by more than 0.6V, which is plenty in your application. You still need a way of ensuring that both inputs are not taken further than the supply rails though. You could add these diodes to your existing circuit.

    Beware leakage currents through the diodes though. In our case the input resistance should be low enough that this won't have any substantial impact.
     
  3. lordmag

    lordmag

    11
    0
    Nov 27, 2011
    Can you explain for me more
    this type of amplifier protection :) !!
    and why we use here resistance 10 k
     
  4. lordmag

    lordmag

    11
    0
    Nov 27, 2011
    hi ,
    any one can explain for me this type of ad620 amplifier protection !!!
    protection input.png
     
  5. lordmag

    lordmag

    11
    0
    Nov 27, 2011
    i' m using 1N4148 diode
     
  6. lordmag

    lordmag

    11
    0
    Nov 27, 2011
    this is my application
     

    Attached Files:

  7. duke37

    duke37

    5,364
    772
    Jan 9, 2011
    As *steve* says, a diode will conduct if the input exeeds the voltage of the power supply by 0.7V. This limits the input voltage to the amplifier.

    That circuit looks very complicated, the voltage measured will be 0V when the contact is closed and 13.5V when it is open. Depending on the detail of what is going on, you could do the same with a resistor and a LED.
     
  8. lordmag

    lordmag

    11
    0
    Nov 27, 2011
    duke , you want to say some thing like this
     

    Attached Files:

  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,500
    2,840
    Jan 21, 2010
    Why did you remove three of the four protection diodes in that last circuit?

    Also why a gain of 495? Each volt measured on the voltmeter will be 0.00202020202 volts across the switch, hardly a neat number.

    I would go a gain of 100, or 1000, or some similar number giving a more easily interpreted number.

    With the current setup, the voltmeter will read about 12V with the switch open simply because that's likely your supply voltage. It would try to swing the output to about 6000 volts.
     
  10. lordmag

    lordmag

    11
    0
    Nov 27, 2011
    oki

    look at this circuit :
    if R contact is ON
    R_contact = 0.2 ohm
    current I = 10 mA
    ==> U_ contact = 2 mV.

    for the AD620 RG = 100 ohm ===> G = 495

    so that Vout = G * U_contact = 0.99 V
    if R_ contact is OFF

    ?? what is happening :))
     

    Attached Files:

  11. lordmag

    lordmag

    11
    0
    Nov 27, 2011
    so what do you think steve
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,500
    2,840
    Jan 21, 2010
    I explained what I thought was wrong in my previous post. Nothing has changed
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-