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Seeking assistance with (seemingly) simple circuit analysis

Discussion in 'Misc Electronics' started by Eng. Tech., Jul 17, 2007.

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  1. Eng. Tech.

    Eng. Tech. Guest

    Greetings all,

    Perhaps I'm just getting old, and losing my powers of concentration (if I ever had them), but the analysis of the circuit shown below eludes me. This is a work-related test fixture where I must set a test current and test voltage external to the device under test, so most of the circuit elements are fixed values. My attempt at an ascii schematic (mono-spaced font needed for viewing) is pasted below, along with it's notes. TIA,

    An engineering tech who should be better at this by now

    +14.5V<---/\/\/--+----+ +5V
    90 | | ^
    | \ |
    | /Ra _ schottky
    | \ ^
    | | | /
    \ +--/\/\/---+---o o--/\/\/---+---0 Vtest out
    Rc / | 180 S1 47 |
    \ \ \
    | /Rb /500
    | \ \
    | | |
    | | +---0 Vtest/2
    -+- -+- |
    - - \

    The resistor values shown are constraints I can't change.
    Vtest must be 4.0V with S1 closed, so current through the 1K divider will be 4mA.
    Total current through 90 ohm resistor Rs must be 10 mA with S1 open.
    Ra + Rb = 10K, but this isn't a requirement.
    The schottky shouldn't conduct when S1 is open.
    I only have control over Ra, Rb, and Rc. All else is pre-existing equipment.
    If Ra and Rb can set the 10 mA current, then Rc is not required.
    S1 could be closed for both test cases, but when open the schottky shouldnt conduct
    (not sure the +5V supply can "sink" current from 7 idle channels).
    This shows the first of 8 test cases. Next will be 3.5V and 15 mA, 3.0V and 20 mA, etc.
  2. The requirement that is impossible to satisfy with
    just resistors is that the no-load voltage at the
    junction of Ra and Rb does not exceed 5V.

    All specs can be met by going sideways though.....

    90 |
    860 /Ra schottky clamping
    \ minimised.
    | Rc | /
    +--/\/\--/\/\/---+---o o--/\/\/---+---0
    | 23 180 S1 47 |
    | \
    \_|_ /500
    5V band gap ref /_\ "Rb" \
    | |
    | +---0
    -+- |
    - \
    Use a precision 5V shunt regulator instead of Rb. /500
    Ra sets the req'd constant 10mA, S1 open or closed.
    The 23 ohm Rc trims the output current to 4mA.
    Schottky clamping is minimised.

    Schottky clamping will happen whenever the new "Rb"
    voltage is above the internal 5V. However, the
    voltage differences are small and the resultant current
    is limited by the 23+180 ohms connecting resistance.
    eg, A 5% tolerance difference results in only 1.23mA of
    clamping current.

    Simple changes to Ra and Rc?
  3. No, you also have to add in the current drawn by the
    Schottky clamp when S1 is open.

    A slightly better resistive circuit is to make
    Rb= infinity and just work with Ra and Rc.
    The calculated values for Ra and Rc below assumed
    an exact 5V and Vdiode= 0.2V.

    | |
    Rc \ \Ra -+-5V
    2119.2/ /2164 _|_
    \ \ /_\ SD, 0.2V Fwd drop.
    | | |
    | | 180 | S1 47
    | +---/\/\-+-+/+--/\/\--+-->Vout
    | |
    | [500+500]
    | |

    With S1 closed there is 4mA of output current and
    about 10.4mA load current through the 90 ohm.

    With S1 open there is 10mA of load current through
    the 90 ohm, and about 3.58mA Schottky current.
    Whatever the simplification the sums end in a quadratic
    equation with one sensible root. There are *probably*
    values for Ra/b/c that gets the 10mA with S1 open or
    closed, and the 4mA with S1 closed, but also probably
    at a higher Schottky current.
    The low output resistance of a 5V reference is the thing
    that removes the interaction between output current and
    load current. It also produces minimum Schottky current.
  4. Eng. Tech.

    Eng. Tech. Guest

    Colin & Tony,

    You guys are amazing! I stopped by sci.electronics.misc after skipping a day or two, to see that you'd predicted precisely what I'm seeing on the bench. While first trying a 5.1V zener in place of Rb, and seeing that it begins to conduct near 4.0V (which the plot in the datasheet appears to indicate is normal for the available diode), I set this aside for the moment in favor of the infinite-Rb approach, in the interest of having some results to show for time spent (I'd started to feel a bit self-conscious having spent so much time thus far). I approximated the values of Ra & Rc (no 5V reference) for a first-pass estimate, and then empirically iterated a couple times to zero-in on more precise values. With some working values verified, I attempted a closer estimate by (as Tony already pointed out) taking the schottky current into account. When this appeared to yield very good first-pass results, I transferred my steps into a spreadsheet and duplicated the approach for remaining test voltage/current combinations. While this provides "target" hardware for the test software developer, I can't help regarding it as cheating.

    At the risk of parading my ignorance, what led you to conclude the solution would be a quadratic? (I don't disagree, I'm just clueless.)

    I'll paste in the spreadsheet sections once I've verified it for a couple more test cases.

    Thanks again for the constructive feedback,
  5. Here goes, although the low output resistance band
    gap reference is still the easier/better way to go.

    Lucky I didn't throw the scribbles away. Please note
    that I am also known as Tony (Typo) Williams.


    1. S1 closed, current in the 1227 must be 4mA.

    90 Rx
    14.5V---/\/\---+----+ Vx--/\/\----+
    | | |
    Rc \ \Ra \Ra
    / / /
    \ \ \
    | | -----> |
    | | |
    | \ \
    | /1227 /1227
    | \ \
    | | |
    -------------+----+ -------------+

    The 90+Rc attentuator can be redrawn to an equivalent
    Vx and Rx.

    Vx = 14.5*Rc/(90+Rc), and Rx = 90*Rc/(90+Rc).

    The three resistors in series must draw 4mA from Vx.

    So Vx/(Rx + Ra + 1227) = 0.004 amps.

    Do some algebra on that and you should get.....

    Rc = (90.Ra + 110430)/(2308 - Ra) ....... eqn.1

    2. S1 open, current through the 90 ohm must be 10mA.

    14.5V---/\/\---+----+ <--- 13.6V when I90= 10mA.
    | |
    Rc \ \Ra -+-5V
    / / _|_
    \ \ /_\ SD, 0.2V Fwd drop.
    | | |
    | | 180 |
    | +---/\/\-+ <-- 5.2V estimated.

    13.6/Rc + (13.6 - 5.2)/(Ra + 180) = 0.01 amps.

    Expand and again reduce to Rc in terms of Ra.....

    Rc = 1360*(Ra + 180)/(Ra - 660) ....... eqn.2

    Combine equations......

    (90.Ra+110430)/(2308 - Ra) = 1360*(Ra+180)/(Ra-660)

    Flog through yet more algebra to get.......

    Ra^2 - 1960.7*Ra - 439919 = 0.

    There's the quadratic. (b^2 - 4ac) is positive.
    Sensible root is Ra = 2164. Rc then also falls out.
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