Connect with us

Detecting down to 25mA-150mA on 0-20A line?

Discussion in 'General Electronics' started by John, Nov 15, 2005.

Scroll to continue with content
  1. John

    John Guest

    I'd like to use a 20A CC/CV power supply to charge LiPo cells and want
    to terminate the charge when the charging current drops to 25mA-150mA

    Sense resistors are out, IMHO, because of the 20A max charging current
    (for multiple parallel cells). Any value that can give me a decently
    measureable voltage would burn up at 20A. And I can't find any
    Hall-Effect sensors (allegor, etc.) that can work down to those
    current levels.

    I don't need to measure the current, just flip a bit when the current
    drops below my 25mA-150mA setpoint. Or, I can monitor a voltage and
    just ignore anything over a certain level.

    Any recommendations for a part/method that could measure such low
    current levels while not burning up when the charge current is as high
    as 20A? Anything to +/-10% would be OK, +/-1% would be perfect.

    -- remove SPAMMENOT for e-mail responses --
  2. Fred Bloggs

    Fred Bloggs Guest

    You could use a sense resistor bypassed by a high current diode. This
    will clamp the voltage across the sense resistor to VF of the diode with
    a power dissipation of VF^2/Rsense. The way to proceed would be to set
    the Rsense voltage drop at "maximum Imin" well below the forward bias
    of the diode. So trying 0.4V at 150mA gives Rs~2.7 ohms. Then even a
    weenie 1/2W Rsense can handle sqrt(0.5W*2.7)=1.2V for VF, diode at 20A.
    Going with say a 2 Watt 2.7 ohm, the VF,max becomes 2.3V etc..There are
    ample precision sense resistors available in the range with very small
    tempcos like 50ppm, usually in a TO-220 type of package. Now all you
    need to do is find a 20A diode, and that's nothing special, you can
    manufacture one by paralleling lesser rated diodes with ballast
    resistors. Getting back to the original problem, using 2.7 ohms, gives a
    voltage detection threshold of 0.4V at 150mA to 0.05V at 20mA.
  3. Consider avoiding the problem!.
    Just have a normal current sense resistor, of perhaps 2R, and bypass this
    with a large Schottky diode. For currents in your required reading range,
    the voltage across the resistor will drop below the point where the diode
    starts to conduct, and you have a voltage output that can be read by a
    simple ADC. At higher currents, the diode conducts, limiting the maximum
    voltage across the sense resistor. Chose a diode, like the 52CPQ030, and
    the unit will withstand up to 50A, yet will give a nice linear voltage in
    the sort of region you require.

    Best Wishes
  4. default

    default Guest

    You could still use a sense resistor and an op amp or comparator
    across it. Op amps have ~200K open loop gain and will work in that
    application. Your sense resistor needs to drop about 3 micro volts at
    25 ma. That's .00012 ohms or a short length of heavy wire.

    Look inside your digital multimeter at the piece of wire they use for
    a shunt on the 10 amp range . . . you'd be doing the same thing by
    using a comparator of op amp to sense the small voltage drop.
  5. CWatters

    CWatters Guest

    or if you don't want to bur any power in the diode use a low Rdson
    resistance FET instead to bypass the resistor. Switch the FET off briefly to
    take the measurement every 10 seconds or so.
  6. Eric Sears

    Eric Sears Guest

    Whether you use the diode arrangement or a very low value resistor (=
    short thick wire), you will still need some circuit to do the
    While the diode solution suggested is a neat one if you need a higher
    sense voltage, the other solution suggested (eg 741 op amp as a
    comparator) means that a low voltage can be detected.
    One problem with the diode (even a Schottky at 0.4 v drop), is that is
    will still dissipate 8 watts at 20 amps, and will need some
    Granted you could use a simple transistor or darlington pair with the
    diode arrangement, but using a 741 with the thick wire would mean no
    problems with things getting hot!

    Just my tuppence worth.

    Eric Sears

  7. Some of the performance specifications for some of today's op amps are
    pretty impressive.

    It may be possible simply to use something like a one milliohm sense
    resistor (perhaps a "dog bone" style resistor built onto your PCB using the
    copper trace(s) as the resistive element, with larger pads on the ends for
    heatsinking and kelvin connections), and use some insanely low offset op-amp
    like the MAX4238 to amplify your 25uV (at 25mA) signal to something more

    Obviously layout/routing/attention to parasitics and noise reduction in the
    rest of the circuit will be crucial in order to make this method succeed.

    As for the other approaches... The shottky diode in parallel with sense
    resistor may not work so well. At 20A, the shottky diode will still need to
    dissipate large quantities of power, but at 25mA a diode that large will
    have very tiny (likely off the datasheet) forward voltage drop, quite
    possibly well under 100mV depending upon temperature. You would still need
    a decently precise op-amp for this case, but you'll have the additional
    disadvantages of an expensive shottky diode, heatsink, and substantial
    wasted power. Attention to routing/layout/noise reduction would still be
    important, although nowhere near so as the above mentioned method. The
    characteristics of a silicon diode would probably be more ideal, but with
    the serious disadvantage of substantially more power waste.
  8. default

    default Guest

    You're preaching to the choir Eric . . . I wouldn't use a diode in
    that application for the power dissipation reason.

    The only drawback is that trying to use a single 741 op amp might
    require a separate power supply. The poster didn't mention what
    voltage this thing was running at - but if it were me, I'd find a way
    to make it work, or put in another supply just for the op amp if
    necessary - 8 watts of heat is too wasteful IMO.

    Then he's talking about using a hall effect sensor . . . that might be
    more complicated, expensive and bulkier, than an op amp, and harder to
    adjust - assuming he wants to adjust the set point..

    If his source is a switcher and he's able to get at the high frequency
    output a current transformer might be the best choice.
  9. Jim Thompson

    Jim Thompson Guest

    Why not shunt your sense resistor with a power-MOS (with control
    circuitry) that kicks in over-range?

    ...Jim Thompson
  10. Jim Thompson

    Jim Thompson Guest

    On Tue, 15 Nov 2005 18:30:50 -0700, Jim Thompson

    What is the value of "CV"?

    ...Jim Thompson
  11. John

    John Guest

    Thanks for the responses!

    The CV value will be 4.20V or multiple thereof, depending on the
    number of LiPo cells being charged. This is considered the
    full-charge value for LiPo cells. But, I'd like to be able to do use
    other voltages from 3.70V to 4.5V per cell for testing purposes.

    I also have available regulated 9.3VDC-9.5VDC and 5.0VDC.

    One reason to avoid a sense resistor is that its voltage drop has to
    be added to the voltage of the battery at full charge to arrive at the
    power supply voltage to set. For each new end-of-charge current value
    (from 25mA to 150mA), I'll need to have a different power supply
    voltage for the LiPo cell/pack to come up to. I'd love to avoid this
    if possible. Especially since a LiPo cell is very easily damaged
    above full-charge voltage values of 4.235V to 4.250V/cell (depending
    on the manufacturer).

    This is why I was wondering if there was a hall-effect or other method
    to sense the current?

    I could start a timer when the voltage reached 4.19V or 4.20V
    (whatever) and just let the battery get charged until the timer stops
    but I want to graph the charging curve of these cells and don't want
    to guess at what the timer value should be. I could either waste a
    lot of time or end a charge early which would affect the capacity of
    the cell during the discharge testing.
    -- remove SPAMMENOT for e-mail responses --
  12. Fred Bloggs

    Fred Bloggs Guest

    Sounds like you're doing a manual adjust on a power supply with
    adjustable current limit. Since you are over your head here, just buy a
    different power supply with a current readout.
  13. default

    default Guest

    Would this be the same sense resistor that's supposed to tell when it
    is over range? And how does it recover? Presumably the thing starts
    off drawing tons of current so the MOSFET shunts the sense - what
    mechanism do you propose to enable the shunt to measure current after
    the current drops if the sense is shunted?

    My idea is to just amplify the piss out of the Vdrop across the sense
    resistor so a very low value can be used. I do it all the time with
    high current supplies and it works very well.
  14. default

    default Guest

    Not so. It is common practice when sensing voltage for regulation
    purposes to sense voltage after the current sense resistor. That's
    the way it is done in linear supplies.

    And if the current sense resistor is .0012 ohms that amounts to a 24
    millivolt drop, you'd have that much in the on-board or off board
    wiring and could be using that wiring as you sense resistor.

    It works very well . . . and is a lot easier than hall effect. For
    the hall effect system, you'd need a linear hall effect sensor or
    would have to settle for a single fixed current with no easy
    adjustment. To get enough flux down at 25 milliamps you'd need
    several turns of wire that would drop more than the amplified sense
    resistor scheme - and/or you'd have to worry about magnetic shielding
    down at low flux levels.

    Seems like a no-brainer to me. If you have the 9 volt supply it is
    easy to get an op amp functioning single supply for that app.
  15. default

    default Guest

    That would work. Have a timer kick out the MOSFET for a short period
    of time on a regular basis and read the sense resistor. We used
    schemes like that for recovery on foldback current limiting supplies

    It is more hassle and complicated than just making the sense resistor
    smaller than the lowest R MOSFET you're likely to find.

    I used the open loop gain of a single op amp for an illustration. Use
    a dual amp and you'd lower the drop further. Low impedance across the
    input so there's little noise to pick up - get too carried away with
    amplification and it might require a integrating cap to slow the
    response and keep it from oscillating in the RF range.

    Interesting project.
  16. Jim Thompson

    Jim Thompson Guest

    It should be obvious to the most casual observer ;-)
    How about amplifier offset voltage?

    ...Jim Thompson
  17. default

    default Guest

    Not a problem. I haven't encountered an op amp with specs so poor
    that it has any significant offset when the differential input is
    shunted. I've used a single 747 (dual 741) without offset
    compensation. One section regulates the pass transistor(s) the other
    works as the current limit. The 741(7) is a relatively primitive op
    amp and it works.

    It was a scheme we used for multiple linear outputs on made to order
    logic supplies in the 70's. If there is a problem with the scheme it
    was when we had high inductance wire wound resistor for sensing. On
    the low current supplies it was never an issue - on the high current
    10-20 amp 5V supplies we used a two turn phosphor bronze wire wrapped
    around a one watt resistor for a coil form. That two turns would have
    it oscillating in the MHZ region if the leads to the load were longer
    than a few feet - for that 100 pf cap from the inverting input to the
    output would neutralize it.

    That wasn't really a problem, but more a matter of the variable load
    we used for testing. Our spec was 0-20 amps and at an amp or two,
    with long leads, and the active transistor load, it could oscillate if
    the op amp was fast enough. Most of the op amps were too slow to
    oscillate, but we started using the neutralizing caps just to be safe.

    In the real load application they had no tendency to oscillate because
    of the bypass caps used in the load. Our dummy load was a low value,
    high wattage resistor with a self-biased NPN transistor with pot to
    control current. That load could be unstable.

    This thing is probably going to be working as a comparator with some
    small amount of positive feedback to lend hysteresis, or just latch
    the output off when the current falls. He doesn't say what he wants
    full charge condition to be.
  18. Jim Thompson

    Jim Thompson Guest

    The fundamentals are shown at...

    I'll leave (as an exercise for the student ;-) where to manipulate
    ground placement and add your various sensing and control circuits to

    ...Jim Thompson
  19. Jim Thompson

    Jim Thompson Guest

    On Wed, 16 Nov 2005 12:58:24 -0700, Jim Thompson

    Oooops! The MOSFET power should be 4 watts... I missed a resistor in
    the subcircuit model provided by IR.

    ...Jim Thompson
  20. Jim Thompson

    Jim Thompson Guest

    Simulation now corrected...

    ...Jim Thompson
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