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Deciding factor with potential dividers

Discussion in 'Electronic Basics' started by Ian Tedridge, Jan 31, 2006.

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  1. Ian Tedridge

    Ian Tedridge Guest

    Idiots guide to potential dividers. Please help

    If I want to produce a potential divider with a 5Vdc supply. Wanting 2Vdc
    out. I guess i would use 2 equal rated resistors in series.

    How do I decide what size resistors 10K or 20K etc ?

    I dont seem to understand the importance of resistor size.

    Please Help
  2. Fred Stevens

    Fred Stevens Guest

    How much current does your load need to draw at 2V?

  3. Ralph Mowery

    Ralph Mowery Guest

    You just can not say you want 2 volts out. You must specify the load
    resistance. If the 2 volt load is very high in resistance you usually will
    use a high value of resistance. If the load is low in resistance , you will
    have to use low values of resistance. It will also depend on some extent of
    the source to supply the needed current. If for example you want to get
    2.5 volts from a simple two resistor devider into a high resistance load and
    you use a pair of 5 ohm reisitors in series across the load , the source
    must be able to supply 1/2 of an amp. This is usually more wasted power
    than you want. If you keep the resistors so they are more than 100 times
    the load resistance the error is usually small enough not to be noticed.
    As with most electronic circuits , there is a trade off in the components
    used with the power and price of them.
  4. Ian Tedridge

    Ian Tedridge Guest

    Looking to pull 250mA Max
  5. Ralph Mowery

    Ralph Mowery Guest

    To reduce a 5 volt source to a 250 ma load you should not be looking at
    resistors but another way. Maybe 6 silicon 1 amp diodes in series with the
    load. The reisitors will not hold the voltage constant under a changing
  6. You won't really be able to use a resistor divider to provide 250mA,
    that is *way* too much current. You are better off using say a LM317
    voltage regulator to drop the 5V input to give you 2V output. Then it
    will be 2V output regardless of the load you put on. A resistor divider
    voltage output will change with the current your load draws.

    Dave :)
  7. If by "equal rated" you mean equal valued, as in ohms, no. That would
    provide 2.5V, not 2V.
    I'm assuming you imagine this:

    : +5V
    : ,----------,
    : | |
    : | |
    : | \
    : | / R1
    : | \ ???
    : | /
    : | |
    : --- | +2V
    : - V1 +--------,
    : --- 5V | |
    : - | |
    : | \ \
    : | / R2 / Rload
    : | \ ??? \ 2V @ 250mA
    : | / /
    : | | |
    : | | |
    : '----------+--------'

    In this case, your Rload resistance is as little as 2V/.25 or 8 ohms.
    But let's assume that it can draw a lot less current (for example, 1mA
    would suggest 2000 ohms, 1uA would suggest 2M ohm, etc.) So one of
    the first key questions to ask yourself is:

    What is the range of voltage I want to accept at the +2V node?

    I think you can see that if the answer is 2.0000 to 2.0000, it's going
    to be practically impossible to achieve. No tolerance of a voltage
    variation is impossible, if the load will itself vary it's loading.
    You have to have some finite, non-zero range you accept.

    So let's say it is from 2.2V to 1.9V, with 2.2V being the voltage when
    Rload isn't even connected (no loading) and 1.9V being the case when
    Rload requires 250mA. First off, with no load, you just have R1 and
    R2 and they must develop 2.2V: V1*R2/(R1+R2) = 2.2V, V1 = 5V.

    With Rload drawing current of 250mA, it's easier to imagine that the
    circuit has been transformed into this Thevenin equivalent:

    : +2.2V
    : ,----------,
    : | |
    : | |
    : | \
    : | / Rth
    : | \ ???
    : | /
    : | |
    : --- |
    : - Vth +-- +1.9V
    : --- |
    : - |
    : | \
    : | / Rload
    : | \ 1.9V @ 250mA
    : | /
    : | |
    : | |
    : '----------'

    I'm not sure if you have already read about Thevenin, but the basic
    idea is that the more complex combination of V1, R1 and R2 can be
    replaced by a new, simpler Vth and Rth, which is what your Rload
    "sees" looking back into the V1, R1, and R2 circuit. If you haven't,
    this all may seem a bit strange and I'd recommend doing some reading
    and thinking on the idea on your own. It's not hard to follow, but I
    don't want to side-track the discussion right now with it. So just
    accept the above equivalent.

    Now Rth is just R1 in parallel to R2, so we have Rth = R1*R2/(R1+R2).
    The difference in the 1.9V node and the 2.2V value for Vth is 0.3V.
    And we already know what current Rload is drawing in this case, 250mA.
    So that means Rth must be .3V/.25A or 1.2 ohms total. So we then know
    that: R1*R2/(R1+R2) = 1.2 ohms.

    This gives us:

    (1) 5*R2/(R1+R2) = 2.2 --or-- R2/(R1+R2) = 2.2/5
    --and also that--
    (2) R1*R2/(R1+R2) = 1.2

    So, substituting (1) into (2), we get:

    (2) R1*2.2/5 = 1.2

    Which works out as R1 = 1.2 * 5 / 2.2, or 30/11 or about 2.7 ohms.

    R2 is then simply R2/(30/11+R2) = 2.2/5, or R1*(2.2/5)/(1-2.2/5),
    which works out to 15/7 or about 2.1 ohms.

    This would mean that your divider would be normally consuming about 1A
    of current. In other words, about 5 watts of power would be
    dissipated in R1 and R2, without ever hooking up your load. With the
    load in place, more than 5.5 watts total.

    This is usually impractical, which is why other suggestions were

  8. redbelly

    redbelly Guest

    To put it another way, and more briefly:

    Suppose you want to regulate the voltage to within, say, 10% of the 2V.
    That means choosing divider resistors that will draw TEN TIMES the max
    load current: 10 x 0.25A or 2.5A.

    This means using a power supply that is considerably costlier than what
    you'd need by using a different method.

    Best to use a voltage regulator, so that the power source need only
    supply slighly more current than the load's 250 mA. AND the regulation
    will be way better than 10%!.


  9. Suppose you transform this, using an NPN transistor:

    : +5V
    : ,----------+-------,
    : | | |
    : | | | almost 250mA
    : | \ | |
    : | / R1 | |
    : | \ ??? | v
    : | / |
    : | | |
    : --- | |/ Q1
    : - V1 +-----| NPN
    : --- 5V | |>-,
    : - | | <---- supposed to be about 2V
    : | \ \
    : | / R2 / Rload
    : | \ ??? \ 2V @ 250mA
    : | / /
    : | | |
    : | | |
    : '----------+--------'

    Now, in this case you can use far less current in your R1,R2 divider
    because the transistor's base will not divert much current away from

    Q1 will bring pretty much all of the 250mA used in Rload via the
    collector that connects directly back to the +5 side of the 5V battery
    or power supply. As a 0-order estimate of Q1's base voltage, you can
    just assume that the base-to-emitter voltage (Vbe) is somewhere
    between 0.6V (lower Vbe voltages occur with smaller base currents,
    larger Vbe with larger base currents, and it is unusual for 0.6V to be
    the case unless we are talking about fairly low currents overall) to
    about 0.9V or slightly more (for times when the transistor is involved
    with fairly high currents.)

    Normally, this variation of Vbe is about 60mV for each factor of 10
    change in current, so you can see that it varies only slightly. So
    let's say it runs at about 750mV with a base current of 100uA, just to
    pick some figure to start. If the base current went to 1mA, you'd
    expect the Vbe voltage to then be about 750mV + 60mV or about 810mV.
    That should give you a rough picture, here.

    Getting back to the diagram... Rload needs about 250mA per your
    specification. This means a collector current of about the same
    value. But what base current? (We need to have a rough idea of the
    base current to estimate Vbe -- not that it is terribly important to
    do, as it isn't because of the log() behavior, but let's do it

    Well, if the transistor is to be operated in saturation that would be
    one thing. But it isn't in this case. The collector voltage is known
    to be 5V and the emitter is at 2V, roughly. What causes a transistor
    to go into saturation, so to speak, is when the voltage potential at
    the collector is such that it begins to even slightly forward bias the
    base to collector diode in it (it's supposed to be normally reverse
    biased.) In this case, we already know that the base won't be more
    than about 3V (we are still trying to figure exactly what, but we can
    delay this for now) and with the collector at 5V the internal base to
    collector diode will definitely be reverse biased. So no saturation
    to worry about.

    Okay, so that means that we can anticipate a transistor beta of say
    100 or more. Very old transistors, like point-contact types, might
    have betas in the 25-30 range. But none of those I've used recently
    are that "bad" unless they are in reasonable saturation (when the
    forward biased base-to-collector diode demolishes any decent beta
    computation.) You could probably even be safe thinking 200. But
    let's call it 100 to be extra safe in the other figures we derive.

    So the base current will be about 1/100th of the 250mA. This means...
    2.5mA. Going back to the 100uA starting point I mentioned, this is a
    factor of 25 times more or about a fourth of 10x10. The first 10x
    gets us another 60mV. Let's not get into log() functions and just
    accept that the next 2.5 times will about to 1/4th of the 60mV more,
    or 15mV. So that means 75mV or so above our guess of 750mV for the
    NPN, or 825mV. Call it 850mV.

    Okay, so you want the base to be about 850mV above 2.0V or 2.85V. But
    notice now that your load on the R1,R2 divider has been reduced to the
    base current which we estimate at 2.5mA -- maybe even less than that.
    This means that all of the same calculations performed before apply
    here again -- except that you now what the divider to provide 2.85V
    instead of the original 2V or so and that you now only need to supply
    2.5mA or so instead of 250mA.

    With no load at all, or better for thinking about, close to no load
    but with a very slight load, the Vbe will be around 0.6V or so. This
    means that if we want there to be 2.2V unloaded and 1.9V loaded down,
    we want the base to be 2.2+0.6 or 2.8V unloaded and then about 1.9+.85
    or 2.75V loaded. A change of 50mV total. Remember this figure, as
    I'll use it shortly in (2) below.

    Referring back to the old equations and the Thevenin equivalent,
    updated now with new values, we have:

    (1) 5*R2/(R1+R2) = 2.8 --or-- R2/(R1+R2) = 2.8/5
    --and also that--
    (2) Rth=R1*R2/(R1+R2) = dV/dI = (50mV/2.5mA) = 20

    The 50mV comes from the voltage change noted above. The 2.5mA comes
    from the change in base current between an unloaded case and a loaded

    This solves out to R1 = 20*5/2.8 = 100/2.8 = 35.7 ohms and to R2 =
    R1*(2.8/5)/(1-2.8/5) = 45 4/9 = 45.5 ohms.

    Now we are talking about only 60mA or so through the divider with
    250mA through Rload and the wattage in the divider is about 300mW
    instead of 5W.

    There will be another (5V - 1.9V)*250mA or 775mW in the transistor, as
    well. But that waste is unavoidable in linear (non-switching) designs
    because you simply _have_ to throw away about 3V at 250mA no matter
    what you do. So that 3/4 watt is going to be wasted somewhere.

    So this design is about 300mW more than the theoretical best case for
    a linear arrangement (which must be 750mW, as I argue) rather than
    being more than 4W beyond, like the simple divider was. And all you
    had to do was add a transistor to get there and redo a few
    calculations for the resistors.

    You could plan on further reductions on the base current by adding
    another transistor. But all you could hope to do is eat into that
    300mW and reduce it somewhat. But that 750mW waste will always remain
    no matter how many transistors you stack up, unless you consider a
    switcher arrangement.

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