Connect with us

Current Sources

Discussion in 'Misc Electronics' started by john, Jul 11, 2005.

Scroll to continue with content
  1. john

    john Guest

    hello,

    I have an interesting problem for you guys , I hope that you will
    advice me on it.
    I used the circuit diagram of the current sources given by texas
    instrument. The link is given below
    http://focus.ti.com/lit/ds/symlink/ina2133.pdf. If this link does not
    work then go to www.ti.com and type "INA2133" in the search box and go
    to its data sheet.
    I used figure# 16 on page#13.

    Now, the problem is that If you have two of these "voltage to current
    converters" having same gorund and power supplies but input signals are
    coming form different channels of the Digital to analog conveter. Now,
    If one current source is outputing some charge and other one is
    outputing zero charge then the current source who is outputing zero
    current will act as a return path for the current source who is
    outputing some current. Am I right or wrong?

    Regards
    john
     
  2. w2aew

    w2aew Guest

    Without consulting the datasheet you mentioned... two quick comments:

    1) I assume that you're putting both current-source outputs into the
    same load, right?
    2) Current sources by 'definition' are high impedance output
    devices/circuits. Thus, they should not present a significant load (or
    provide a current return path) to a current source in parallel with it.
     
  3. CWatters

    CWatters Guest

    You are wrong. Zero current means zero current (eg an open circuit). Its as
    if that output were disconnected from the load.

    Lets assume the current sources are called Ia and Ib and they are connected
    together to the load. Then the current through the load Il is:

    Il = Ia + Ib

    If Ib = 0 then

    Il = Ia - 0 = Ia

    It's as if Ib didn't exist and was disconnected.

    Try another example... If Ia were positive and Ib were negative then you
    could say that Ib is sinking some of the current provided by Ia.

    Colin
     
  4. john

    john Guest

    Hello,

    Thanks very much! Would you please advice me that If "Ia" is 1A then
    what should be the "Ib" to sink the total "Ia". How can I calculate it?
    I also would like to mention that the two current sources is providing
    charge to RAT's different tissues. so, the tissue is providing some
    kind of impedance between the outputs of the current sources.

    Thanks
    John
     
  5. CWatters

    CWatters Guest

    In any circuit if you look at a _node_ the sum of all the current going
    into/outof the node always equals zero.

    So your node _as_described_previously_ had three branches so...

    Ia + Ib +Il = 0

    Now substitute

    Ia = 1A
    Ib = ?
    Il = zero

    so substitute and re arrange and if Ia = 1A then Ib -1A.

    Simple really all the current going into the node comes out again and none
    goes through the load.

    But then you said...
    BOING... that changes everything. So the load is actually between Ia and Ib
    and the above analysis is junk.

    The circuit as you describe "does not compute" because you can't have two
    different current sources in series. (If you connect a garden hose in series
    with another one the water rate in each has to be the same.) In practice you
    need to model the load with THREE elements.....

    1) an unknown impedance connecting it to ground on the output Ia and
    2) an unknown impedance connecting it to ground on the output Ib and
    3) an unknown impedance between Ia and Ib.

    Any ideas how these compare?
     
  6. john

    john Guest

    Hello,

    The impedance between the current source#1 and ground will be around
    50Kohm and similar impedance between current source#2 and ground.
    Ground is common between the sources. It means two different loads
    having same grounds with two different current sources.

    Now, what would you suggest,inorder to get one current source absorbing
    and the other delivering charge.

    Thanks
    Regards
    John
     
  7. CWatters

    CWatters Guest

    Ok here's how I would do the analysis...

    1) Since the load to ground is the same at each end (a and b) I would keep
    the whole thing symetrical and make Ib = -Ia.

    2) Now if you were to measure the _voltage_ at either end of the load (eg
    "a" and "b") wrt ground the value would be the same but one would be +ve and
    the other -ve.

    3) Now imagine the load as a resistor with the +ve voltage at one end and
    a -ve at the other. The voltage at the _middle_ of the resistor/ load would
    be roughly 0V. Lets call it exactly 0V or Ground.

    4) This allows you (For the purpose of doing the analysis only) to draw two
    equivalent circuit diagrams each representing half the original. In this
    equivalent circuit - at point "a" you have 50K to ground in parallel with
    _half_ the value of the load.

    5) Now look at the current flowing into point "a". The current going in is
    Ia. The sum of the currents is zero but how much goes through the load and
    how much through the 50K? The answer is a simple ratio.. Iload =
    Rload/(Rload+50K.). Now you know the load current in terms of Ia. Since
    Ib=-Ia that's the problem solved.

    As an exercise when you get bored...

    Imagine a cube with a 1R resistor fitted along every _edge_ and soldered
    together at all 8 corners. What would be the resistance between measured
    between any two diagonally opposite corners of the cube? The method used to
    solve this problem is the same as used above.
     
  8. john

    john Guest

    Hello,
    _____________Va Vb _____________
    | | | |
    | | | |
    | | | |
    | 50kOhm 50KOhm v - 2A
    ^2A
    | | | |
    |____________|___________|_____________|
    |
    |
    _____
    ___ Gnd
    -
    I have following circuit in my mind and I did not understand the steps
    4 and 5 you adviced me in the last email. Two, current sources
    connected in parallel supplying current to two different loads having
    same ground. Va and Vb are the voltages across the loads ( 50KOhm).
    Now, would the -2A current source become the return path for the 2A
    current source. I hope ypu will forgive my drawing.
    Thanks
    Regards
    John
     
  9. john

    john Guest

    Hello,
    _____________Va Vb _____________
    | | | |
    | | | |
    | | | |
    | 50kOhm 50KOhm v - 2A
    ^2A
    | | | |
    |____________|___________|_____________|
    |
    |
    _____
    ___ Gnd
    -
    I have following circuit in my mind and I did not understand the steps
    4 and 5 you adviced me in the last email. Two, current sources
    connected in parallel supplying current to two different loads having
    same ground. Va and Vb are the voltages across the loads ( 50KOhm).
    Now, would the -2A current source become the return path for the 2A
    current source. I hope ypu will forgive my drawing.
    Thanks
    Regards
    John
     
  10. john

    john Guest

    Hello,
    _____________Va Vb _____________
    | | | |
    | | | |
    | | | |
    | 50kOhm 50KOhm v - 2A
    ^2A
    | | | |
    |____________|___________|_____________|
    |
    |
    _____
    ___ Gnd
    -
    I have following circuit in my mind and I did not understand the steps
    4 and 5 you adviced me in the last email. Two, current sources
    connected in parallel supplying current to two different loads having
    same ground. Va and Vb are the voltages across the loads ( 50KOhm).
    Now, would the -2A current source become the return path for the 2A
    current source. I hope ypu will forgive my drawing.
    Thanks
    Regards
    John
     
  11. According to your drawing above, the current sources _are not_ in
    parallel.

    What you show is two independent circuits, each one a current source
    and load, that just happen to have their ground nodes connected
    together. If your drawing is correct, there will be no current in the
    segment of the ground node between the two loads, and neither current
    source will have any effect on the other.


    --
    Peter Bennett, VE7CEI
    peterbb4 (at) interchange.ubc.ca
    new newsgroup users info : http://vancouver-webpages.com/nnq
    GPS and NMEA info: http://vancouver-webpages.com/peter
    Vancouver Power Squadron: http://vancouver.powersquadron.ca
     
  12. CWatters

    CWatters Guest

    He missed out the load between Va and Vb and the fact that he appears to be
    trying to get Va = -Vb (or at least that's my understanding)

    I'll try and explain it with diagrams....


    Note if Ia = -Ib then Vb = -Va so...


    _____________Va___Load___-Va_____________
    | | | |
    | | | |
    | | | |
    2A 50kOhm 50KOhm -2A
    | | | |
    |____________|____________|_____________|
    |
    |
    _____
    ___ Gnd
    -

    and this is the same as...


    _____________Va___0.5 _x__ 0.5___-Va_____________
    | | Load Load | |
    | | | |
    | | | |
    2A 50kOhm 50KOhm -2A
    | | | |
    |____________|_____________________|_____________|
    |
    |
    _____
    ___ Gnd
    -

    Note that the voltage at x = Ov

    So you can re draw the circuit like this

    _____________Va_________
    | | |
    | | |
    | | |
    2A 50kOhm 0.5 Load
    | | |
    |____________|__________|
    |
    |
    _____
    ___ Gnd
    -

    Now it's easy to calculate the current through the load.
     
  13. CWatters

    CWatters Guest

    In effect yes. If you need to know the current through the load (Rat!) then
    here is how you do the sums...

    I'll try and explain it with diagrams (change to fixed font if necessary)


    Note if Ia = -Ib then Vb = -Va so...


    _____________Va___Load___-Va_____________
    | | | |
    | | | |
    | | | |
    2A 50kOhm 50KOhm -2A
    | | | |
    |____________|____________|_____________|
    |
    |
    _____
    ___ Gnd
    -

    and this is the same as...


    _____________Va___0.5 _x__ 0.5___-Va_____________
    | | Load Load | |
    | | | |
    | | | |
    2A 50kOhm 50KOhm -2A
    | | | |
    |____________|_____________________|_____________|
    |
    |
    _____
    ___ Gnd
    -

    Note that the voltage at x = Ov

    So you can re draw the circuit like this

    _____________Va_________
    | | |
    | | |
    | | |
    2A 50kOhm 0.5 Load
    | | |
    |____________|__________|
    |
    |
    _____
    ___ Gnd
    -

    Now it's easy to calculate the current through the load.
     
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

-