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Capacity test - Li-ion Sanyo vs UltraFire (graph)

  • PolarCircle Sunday, January 24, 2016 11:44 AM Reply

    @sheepish

    ebpDoug probably would do it differently, but two ways you could do it:

    1) As you do know the voltage measured with the ADC, and because a power resistor is fairly linear, you can adjust the voltage at he other end of the resistor to drain or inject current. So if you have a 3.3 ohm resistor connected to the positive terminal and you keep it 3.3 V below the positive terminal it will drain 1A.

    To enable the DAC to sink the current through the resistor, it probably needs help. If you connect the output of the DAC to the + (non-inverted) input of an op-amp, the output of the op-amp to the base of a N-P-N transistor, the resistor and collector of the transistor to the - (inverted) input of the op-amp, and the emitter of the transistor to ground you should be set.

    Like this

    0 ground, battery -, NPN emitter
    1 battery +, ADC sense, resistor +
    2 resistor -, NPN collector, op-amp -
    3 DAC out, op-amp +
    4 op-amp out, NPN base

    And with many ADC and DAC channels you can easily operate many in parallel


    2) A second way of making the drain would be to connect the output of the DAC to the gate of a FET transistor, and a multimeter directly from the + of the battery to the FET and be done with it! But I would highly recommend to connect a resistor in series with the multimeter to limit the current, and to offload the transistor.


    Both ways would enable "the contraption" to be used as a programmable load. Constant current, constant power, constant resistance. And if you want, variable current, power and resistance.


    @ebpDoug

    Do you concur?

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    post edited by PolarCircle on 1/24/2016 at 11:47 AM
  • ebpDoug Monday, January 25, 2016 4:41 AM Reply

    The simplest voltage controllable current sink requires a resistor and a bipolar transistor. We're sinking from positive, so we need an NPN.

    The resistor goes between circuit common (I'll call it ground) and the emitter of the transistor. The collector is the sinking node (so battery [+] to collector, battery [-] to ground). The control voltage is applied between the transistor base and ground. This is exactly the same as a "common collector amplifier" or "voltage follower" - we just get to select the load as we wish and are concerned with current, rather than voltage, as such.

    Assume a transistor that has a base-emitter voltage of 0.7 volts and very high current gain.

    The voltage across the resistor in the emitter circuit will be (Vbase - 0.7), so the current through it will be (Vbase - 0.7)/R. The behavior. of the transistor is such that the 0.7 V between base and emitter is maintained. Most of the current is from the collector circuit. With a fixed voltage, relative to ground, at the base, if the current through the resistor started to drop, Vbe would start to rise, turning the transistor on harder, increasing the collector current, raising the voltage across the resistor and returning Vbe to 0.7V (there is some impliction of time dependence here - one event, then another - and there is a little truth to that, but you can pretty much think of it as all happening instantaneously).

    So, let's put 1 ohm between emitter & ground.

    - at any base voltage (base to ground, not Vbe) less than 0.7V, the current will be zero.

    - at any higher base voltage, the voltage across the resistor will be (base voltage - 0.7) and will will get 1 ampere per volt e.g. 1 V input, 0.3 A; 4 V input 3.3 A

    There are several issues. The base-emitter voltage varies from transistor to transistor, it varies with temperature, and is influenced by collector voltage to some extent. The current through the emitter-ground resistor is the sum of the base current and the collector current. If you want more than a few hundred milliamps, the available bipolar transistors will have current gain that is quite low, so the error will be fairly large. With a microcontroller with an ADC and a DAC you could measure the current and tweak the base voltage to compensate (frequently enough to compensate for change with temperature and collector voltage), but you still have the base current error. If there is no collector power supply, the current into the base will be very large, so extra bits may be required to prevent that.


    Refer to National Semi Ap Note 31, page 13, lower left circuit

    http://www.ti.com/ww/en/bobpease/assets/AN-31.pdf

    (the ap note rev is dated 2002, but the original is many years older)

    This circuit uses a composite transistor made from a BJT and a JFET. The base drive for the BJT is from its collector circuit, and the gate drive of the FET is zero, in practical terms, so the error due to low current gain in the transistor is gone. The op amp makes the whole thing locally closed-loop. The voltage across the emitter resistor will be exactly equal to the voltage applied to the non-inverting input of the op amp (ignoring the usual sources of error such as offset voltage, amp open loop gain, etc. - all can insignificant with modern amps).

    The circuit was designed long before practical power FETs existed. These days, you would simply use a suitable N-channel power MOSFET in place of Q1 and Q2 (source via resistor to ground, drain is the sinking node). The emitter resistor would be chosen for convenience relative to input voltage range, dissipation you'd allow, etc. You might want to use a voltage divided at the input to the op amp (e.g. for 0 to 5 V from your DAC, you might want to attenuate by five for 0 to 1 volt full-scale across the emitter resistor; with care, you could go as low as perhaps 50 mV across the emitter resistor at full scale, but you'd need to pay careful attention to amp offset voltage, etc.). A trim pot could be part of the attenuator, to allow accurate calibration.


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  • ebpDoug Monday, January 25, 2016 4:56 AM Reply

    Forgot to mention this before:

    Lithium cells tend to "self passivate". I don't know the details of the mechanism and I don't know to what extent it applies to rechargable types. With primary lithium cells, it is a significant contributor to very long shelf life. However, a cell taken off the shelf may exhibit very high internal resistance, almost appearing dead. Briefly applying a high load or even a short circuit will reverse the passivation effect and bring the cell back to life. This might account for the apparent negative resistance.


    - - - -

    Aluminum and stainless steel get their corrosion resistance largely because of passivation. A thin layer of oxide forms very quickly and then blocks further oxidation. Certain metals (if I recall, titanium used in nuclear reactors) can be anodically protected, rather than cathodically protected.

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  • sheepish Top 10 Forum Poster Monday, January 25, 2016 5:44 AM Reply

    Briefly applying a high load or even a short circuit will reverse the
    passivation effect and bring the cell back to life. This might account
    for the apparent negative resistance.


    That might account for a sudden drop in internal resistance but it's not going to create higher voltage. Even if it was combined with test circuitry that could act like a boost circuit it's not going to be doing it rapidly and repeatedly.

    Remembering 30 years.
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  • ebpDoug Monday, January 25, 2016 8:14 AM Reply

    Sheepish, I'm sorry, I was looking at your curve with the little upward blip at the beginning, rather than thinking about your description of the "test results for 4.2 V cells that have the discharge graph starting from more than 4.2 V". I haven't a clue how that could happen unless the cells were overcharged or the test equipment was bad.

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