Connect with us

DC block

Discussion in 'Electronic Design' started by john, Nov 13, 2007.

Scroll to continue with content
  1. john

    john Guest


    I am using the following circuit ( Figure: 16 , page 13 )

    I connected the INA 133'S pin#3 to the output of the DAC (voltage
    out). DAC outputs DC voltage of minus 2.5 volts at the power up.
    Inorder to avoid DC -2.5volts , I put a capacitor in between the DAC's
    output and the non inverting inout of the INA, but I am still getting
    -2.5 volts at the non inverting input of the INA. Can somebody advice
    whats really happening? Capacitor suppose to sblock DC.

  2. Tim Wescott

    Tim Wescott Guest

    Oy. Where to start...

    Capacitors don't _conduct_ DC, and they only _block_ DC in the sense that
    they insure that no DC current flows. If the amplifier pin wants to drift
    to -2.5V, then the blocking capacitor will just insure that that happens
    all the time.

    I suggest you find some place to post your whole circuit, or at least the
    part that includes the DAC and amplifier, and post it over on

    Tim Wescott
    Control systems and communications consulting

    Need to learn how to apply control theory in your embedded system?
    "Applied Control Theory for Embedded Systems" by Tim Wescott
  3. It is wired as voltage follower, input impedance at pin 3 extremely high.
    The cap will not charge if no current flows.
  4. Bob

    Bob Guest

    We need to see the whole circuit to see what is going on.
    What value is the capcitor connector to pin 3?
    What have you done with pin 2 on the INA2133?
    What value of R (as shown in figure 16) are you using?
    What is the load?
    What are the power supply voltages?
    Do you have the DAC ground connected to a 0v rail in between
    the pin7 +v and pin 4 -V rail?

    I don't think you are giving us the full story.

    Assuming the problem really is that the DAC output has -2.5V
    DC on it then you need to consider the switch on behaviour.
    The capacitor connected to pin 3 then it
    will take some time to charge up so there is 2.5V across
    the capcitor. The capacitor charging current is limited
    by the 25K resistor inside the INA2133
    For example a 220uF capacitor will take 5.5 seconds to
    charge up to 2.5V from a 0V initial charge.

  5. john

    john Guest


    1. What value is the capcitor connector to pin 3?

    2. What have you done with pin 2 on the INA2133?

    3. What value of R (as shown in figure 16) are you using?

    4. What is the load?

    5. What are the power supply voltages?
    +/- 15volts

    6. Do you have the DAC ground connected to a 0v rail in between
    the pin7 +v and pin 4 -V rail?

    7. It is wired as voltage follower, input impedance at pin 3 extremely
    The cap will not charge if no current flows.

    I also connected a 1kohm resistor on the pin 3 after the capacitor and
    grounded it but still the INA is getting -2.5 volts.


  6. Shorted cap, or you measured at the wrong side of it?
  7. john

    john Guest

    Shorted cap, or you measured at the wrong side of it?

    Cap. is not shorted and measured it on the right side.

  8. OK, you gave that diagram, and I looked it up first time.
    You say you connected it to ground.
    When and while you did, you MUST have read 0v?
    Else remove the cap, and measure voltages again?
  9. john

    john Guest

    I removed the resistor and just used the capacitor, I am measuring
    -2.5volts at the input of the capacitor and -10 volts at the output of
    the capacitor or at the input of the INA. SO, rigth now the setup is
    DAC ------ capacitor----------INA (PIN3)

  10. With load connected?
    I am still lookin gat that diagram with the OPA131,
    What is on pin 2?
  11. john

    john Guest

    Pin2 of INA is grounded. The problem has changed. I want to post the
    circuit diagram. how can I do it?

  12. There are several sites where you can upload a picture, I have my own server,
    so I cannot tell you which one :)
  13. john

    john Guest


    Please go to the following link to see the circuit diagram

    The problem is that at the power up, - 0.2 volts appears across the
    load ( 10kohm) which is not desriable for my application. The load is
    isolated by a 10uF capacitor. The voltage at the input of the
    capacitor is -2.4 volts ( at the power up) . I do not know that why
    this capacitor is not blocking the DC voltage completely. I tried
    polypropalene 0.5 uF capacitor and the voltage across the load drops
    to 0.0001mvolts ( desirable) . But the input voltage at the capacitor
    increased from - 2.4 volts to -14 volts.

    The capacitor 470uF is not blocking the -2.4 volts either, its output
    is -0.5 volts going to INA's pin 3.

    Please advice!
  14. John Larkin

    John Larkin Guest

    Supload, Imageshack, many others.

  15. OK I got it.
    The circuit is a voltage to current converter.
    The output current will at first flow in the 10uF capacitor,
    but after that capacitor has charged, only in the 1MOhm load.
    Even a VERY low current in a 1MOhm load, will create a huge output voltage!!!!!
    The output voltage comes back to the input via the resistors in the INA133.

    Nothing to do with those capacitors (but watch the polarity).

    What you perhaps COUKLD do (or I would do), is short the 10uF, remove the 1MOhm,
    short the 470 uF, remove the 385 Ohm, and apply a volatge of say 2.5V DC
    from some reference at pin 2 of the IN133.
    That way the DC path is correct, and the output will be (UADC - 2.5) / R.
    Select the 2.5V on pin 2 so you get the right swing.
    For example for a 0-5V ADC, output current would swing from negative to positive
    with 2.5V on pin 2.
    No capacitors needed.
    The 2.5 must come from a low impedance point.

    Well, it is late here, but something like that :)
  16. john

    john Guest


    What should I do about the leakage current? The put the 10uF capacitor
    to stop leakage current.

  17. I am not sure what you mean.

    If pin 2 and pin 3 of the INA133 are at same potential,
    then the output current (without capacitor) in a *low impedance*
    load will be (V3-V2) / R, so zero!

    I marked *low impedance* as lets take a numeric example:

    Say pin3 is +1.5V (some AD ouput).
    and pin 2 is +2.5V (the reference).

    Now if the load is a short-circuit (you amp meter),
    then the output current will be (1.5 - 2.5) / 5000 = -1 / 5000 = -200uA.
    If you made the load impedance 100 kOhm then that would cause a voltage
    drop of -.0002 x 100000 = -20V, and your circuit CANNOT supply that.
    So you load impedance has a maximum value where it will still work.

    I do not know why you want a current source, or what your load is,
    but that is one factor.

    So if your DA conbverter has for exampe an output from 0 to +5V,
    YOU have to decide where the zero point is of Iload,
    and you can do that by setting the apropriate voltage on pin2.
    With ADC midrange 2.5V and 2.5V on pin 2, there is zero output current.
    If you put 0V on pin 2, then 0V ADC out gives zero output, and 5V gives
    1mA output.
    If you have say 10V supply, then the max Rload is 5kOhm.

    If pin 2 is at 5V, the 0V on the ADC wil lresult in -1mA output,
    while 5V on the ADC will result in 0mA output.
    So the voltage on pin 2 sets your range, the 5kOhm your gain.

    Hope I made no mistakes, but do you get the idea?
  18. Jamie

    Jamie Guest

    Well, I'll guess. did you put the cap in backwards assuming it
    maybe polarized?
    did you confirm the INA outputting 0 volts for example with no
    You may want to ding around a bit here.
  19. Jamie

    Jamie Guest

    You need to use DC coupling with offset biasing to obtain your
    0.0 output..
    Using caps will just add to the problem.

    Also, the settling time on the DAC from the process control end
    will place a role in this.
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