Connect with us

differential amp

Discussion in 'Electronic Basics' started by kell, Dec 22, 2006.

Scroll to continue with content
  1. kell

    kell Guest

    If you had, say, a current sensing resistor in the top rail, could you
    use a circuit like this:
    (view in monospace font)


    I-->-+------Rs------+---->
    | |
    | |
    100k 100k
    | |
    | |
    +--, ,--+
    | | | |
    | | | |
    | 100k 100k|
    | | | |
    | | | |
    | | 10k | |
    | '-/\/\/\-' |
    | ^ |
    | | | ,-------,
    | | | | U1A | ,-10k--,
    | gnd | '-|-\ | | |
    | | | >--+--1k--+-|-\ |
    '----------|+\ '---|+/ | >-+--out
    U1B | >--+-----1k----+----|+/
    ,-|-/ | | U1C
    | | 10k
    '-------' |
    gnd
    U1: TLV274

    The resistors connected to the 10k pot bring the signal down in the
    range where the TLV274 can measure it, and the pot nulls out error.
    Workable?
     
  2. It is workable, but requires that the 4 divider resistors
    drift the same way. If you use a pair of transistors in
    common base configuration (emitters to the current sense
    resistor), you can cut down on this requirement and get some
    voltage gain out of the voltage translation.
     
  3. Emanuele

    Emanuele Guest

    kell ha scritto:
    Search:
    High Side Current Sense in Google
    You'll found integrated circuit with all and more....

    Emanuele


    --
    *** YOUR ELECTRONICS OPEN SOURCE ***

    http://dev.emcelettronica.com

    ;---------------------------------------------------------
    Progetti Completi (Full Projects):

    [IperKEY-ResKEY] Chiave elettronica a microcontrollore
    (Microcontroller Electronics Key and key reader)
    //TAGS:pic, Eeprom, ADconverter, I2C

    [IperCODE] Ricevitore RADIOCOMANDI e decodifiche
    (Remote Control Receiver and decoder)
    //TAGS:pic, Radio control, routines, rolling code, UART, CCP1, LCD
     
  4. kell

    kell Guest

    It is workable, but requires that the 4 divider resistors
    I'm trying to picture the circuit from your description. Does it look
    like

    I-->-+------Rs------+---->
    | |
    \e e/
    pnp |----------| pnp
    /c c\
    | |
    | |
    100k 100k
    | |
    | |
    +--, ,--+
    | | | |
    | | | |
    | 100k 100k|
    | | | |
    | | | |
    | | 10k | |
    | '-/\/\/\-' |
    | ^ |
    | | | ,-------,
    | | | | U1A | ,-10k--,
    | gnd | '-|-\ | | |
    | | | >--+--1k--+-|-\ |
    '----------|+\ '---|+/ | >-+--out
    U1B | >--+-----1k----+----|+/
    ,-|-/ | | U1C
    | | 10k
    '-------' |
    gnd
     
  5. You are getting close. You need to put the offset
    correction at the emitters. You also don't need the upper
    half of the voltage dividers, since the two collectors act
    as variable current sources. Oh, and you need a mechanism
    that applies a voltage to the 2 bases that produces an
    average voltage at the collectors that you need for the
    opamp inputs. You can linearize the emitter voltage to
    collector voltage gain by adding a pair of series emitter
    resistors.
     
  6. kell

    kell Guest

    Next try


    I-->-+------Rs------+---->
    | |
    1k 1k
    | 10k |
    +--/\/\/\/\/\--+
    | ^ |
    | | |
    | | |
    \e | e/
    pnp |----+-----| pnp
    /c | c\
    | | |
    | | |
    | | |
    | 10k |
    +--, | ,---+
    | | | | |
    | | | | |
    | 100k | 100k |
    | | | | |
    | | | | |
    | | | | |
    | '---+---' |
    | | |
    | | | ,-------,
    | | | | U1A | ,-10k--,
    | gnd | '-|-\ | | |
    | | | >--+--1k--+-|-\ |
    '----------|+\ '---|+/ | >-+--out
    U1B | >--+-----1k----+----|+/
    ,-|-/ | | U1C
    | | 10k
    '-------' |
    gnd
     
  7. kell

    kell Guest

    U1: TLV274
    Or instead of having the pot wiper connected to the transistor base
    node, I could connect the pot wiper to ground and use a fixed
    base-emitter resistor on each transistor.
     
  8. Something like this might be made to work. Do you see how
    it gives you a lot more voltage gain than the original
    divider method? think about what you would have if you
    replaced the two collector resistors with a current mirror
    (to subtract one collector current from the other). Then
    you wouldn't need a subtractor output, but only a current to
    voltage converter.

    There is also a much simpler way to produce the ground
    referenced single ended output with a single opamp that has
    a common mode range that includes its positive rail.

    http://www.maxim-ic.com/appnotes.cfm/appnote_number/746/

    Figure 8 shows the general concept.

    Sorry if you just wanted a schematic of a finished high side
    current sense amplifier. I am more interested in exploring
    possibilities and circuit concepts with you.
     
  9. kell

    kell Guest

    I'm totally on the same page with respect to intellectual curiosity.

    I already knew about jfet op amps like the TL082 or LF353 where the
    common mode goes all the way to the top rail. In fact I've seen the
    very same circuit you pointed out in Figure 8 of the maxim app note in
    regard to high side current sensing, but kind of forgot. And well, to
    tell you the truth, this thread isn't really about high side current
    sensing. I do want to amplify a differential voltage at or very near
    the top rail, but there will be more impedance behind it. As a matter
    of fact, this is about the self-regulating heater we discussed in
    another thread on s.e.d. I'm still working on it! And here's the
    latest concept:




    14v
    |
    ,----+---+----,
    | | | |
    .01 | | |
    | | | _
    ,------+---------1k--------+ 15k / ^
    | | | | \ |
    | = 1000uF | | / |
    | | ,-----+---10k--|----+-->\200k|
    | gnd | | | / |
    +---------, | =100uF load \ |
    | | | | 0R5 | |
    \c /+|-' | gnd 25uH | |
    |--< | | PTC heating | |
    /e \-|---' coil | |
    | U1 | | |
    | '--------+----+
    | |
    | ,----------------+-------100k--------+
    | | | |
    | | = 100n |
    | | | |
    | | gnd |
    | '-|-\ |\U3 |-'
    | U2 | >---| >o---------------------||
    +-----|+/ |/ |-,
    | |
    '------------------Rg--------------------+
    |
    gnd


    U1: TL082
    U2: LM393
    U3: inverting driver

    Still using the concept of the resistance bridge to detect the
    temperature of the heating element, but in a different way. Before, I
    tried to do it going cycle by cycle, using the comparator to turn the
    heater off when it reached the setpoint and the big problem was, when
    you turn the heater off, everything stops. And even if there's a way
    to get around that issue, the heating coil has inductance, and when the
    comparator tries to turn the heating element off, it produces a huge
    spike of negative feedback.

    In the latest version of the circuit the comparator keeps running as a
    voltage controlled oscillator that regulates the power through the
    heating element. The RC network connected to the heater resitance
    bridge filters out ac, smoothes out the voltage across the resistance
    bridge and the diff amp sees this mean dc voltage that follows the
    actual temperature of the heating element.
    The resistance of the heating element changes about 2 or 3 tenths of a
    percent for each degree change of centigrade. To keep the heater
    within a ten degree window, we want the diff amp to go to the rails
    when the voltage across the sense resistor changes about 2 or 3
    percent. Now we have 25 amps going through 0.01 ohms for .25 volts,
    and 2 percent of that is 5 mV. Amplifying that to 15 volts requires a
    gain of 3000. So I guess Rg would need to be... 3 megs?
     
  10. kell

    kell Guest

    speaking roughly of course. And the circuit should look more like
    this.
    Maybe it doesn't have too many mistakes this time.
    I'll have to stick a pot in somewhere to null op amp error.
    Gain = Rg/Rd
    14v
    U1A |
    /+|-----+--------, ,----+---+----,
    ,--+-< | | | | | | |
    | | \-|--, | 100k .01 | | |
    | | | = 10uF | | | | _
    | '-------' | '---+ 15k / ^
    Rd gnd | | \ |
    | | | / |
    | ,-----+--100k--|----+-->\200k|
    | | | | / |
    +---------, | =10uF load \ |
    | | | | 0R5 | |
    '-| /+|-' | gnd 25uH | |
    ||-< | | PTC heating | |
    ,-| \-|---' coil | |
    | U1B | | |
    | '--------+----+
    | |
    | ,----------------+-------100k--------+
    | | | |
    | | = 100n |
    | | | |
    | | gnd |
    | '-|-\ |\U3 |-'
    | U2 | >---| >o---------------------||
    +-----|+/ |/ |-,
    | |
    '------------------Rg--------------------+
    |
    gnd


    U1: TL082
    U2: LM393
    U3: inverting driver


    Here's wishing you and yours a very Merry Christmas, John, and many
    more to come.

    Kell
     
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

-