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5v 20mA max output driving a 5v 500mA max circuit.

Discussion in 'Electronic Basics' started by Daniel Pitts, Apr 30, 2013.

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  1. Daniel Pitts

    Daniel Pitts Guest

    I have a signal at 0 or 5v. The source of that signal can only handle
    20mA of current.

    I have another circuit which needs to have the same voltage level with
    reference to ground, but it could have up to 500mA going through it.
    The actual current is controlled by a current sink, and it may vary over
    time.

    In other words, when my signal is low, I want no current. When my signal
    is high, I want to source as much current as possible for the sink to
    take in.

    I have some BJTs laying around, both PNP and NPN, both can support the
    max Ic I need. From what I figure, I'll need to hook my signal up to
    the base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v for the
    resistor. 4.3v/20ma=215ohms).

    Will a circuit like that work? Is there an easier way? I actually have 8
    of these signals (although, only one is active at a time), so I was
    hoping to avoid adding the 8 resistors and 8 transistors to my project,
    though that seems to be the cheapest way.

    Any advice is appreciated.
    Thanks,
    Daniel.
     
  2. An emitter follower? That should be fine. The transistor might use
    less than the whole 20mA. (depends on the current gain.) Do you care
    about speed? (it might take some time to turn off.) I'm a bit
    confused about the load. What are you driving?

    George H.
    A
     
  3. It's not clear to me what the directions of the currents are both for output
    and input. Modern digital outputs can often handle 20mA both sink and source
    but an input that requires op to 500mA sink or source? Is it a digital
    input? What are the voltage limits of that input? Crucial things to know for
    providing a usefull answer.

    petrus bitbyter
     
  4. Daniel Pitts

    Daniel Pitts Guest

    That's more than fine, 20mA is the maximum the signal source can handle,
    but it could be pico-amps for all I care.
    I care a little about the speed. are we talking more than 10's of
    milliseconds?
    The load is several parallel LEDs, all controlled by another chip (a
    shift register with latch constant-current sink). Each "led" will get
    20mA, and I can have up to 24 on at the same time, (that is, between 0
    and 24), so a maximum of 480mA (rounded up to 500ma)

    I've posted about this project in the past, its basically my own LED
    display, based on an RGB LED matrix.
    <http://www.seeedstudio.com/depot/datasheet/2088RGBMatrix.pdf>

    I need to switch on exactly one of the rows, and set up the column shift
    registers. I switch rows at about 8KHz, so I suspect most transistors
    can handle that no problem.

    There might be another alternative. On digikey, I found "PMIC - MOSFET,
    Bridge Drivers, Internal/External Switch", which look like they do what
    I want as well, with fewer discrete parts. It costs a bit more, and the
    interface is "serial" rather than a simple 3-to-8 decoder I currently
    have. Overall, I'm not sure which is the better approach. I think
    overall I'd use a little less power when using the mosfet chip, which
    could be useful if I make this a battery powered device.

    What's the difference between internal and external for these "bridge
    driver" chips? Also, what's the difference between High-Side and
    Low-Side in these chips?

    Thanks,
    Daniel.
     
  5. Hi Daniel, First I may say something wrong.. and then hopefully those
    with more knowledge will correct me.
    No. when you saturate a bjt it takes a microsecond or so to come out
    of saturation.
    (hmm I'm not real sure about that number... maybe 100's of ns?)
    Yup you could use a FET too. You've got I*R losses in the FET where R
    is the on resistance. In the BJT you'll have (maybe) 300-500 mV of
    Vce drop... that might be an issue if you need all of the 5 volts for
    the LEDs.
    Well low side is when the switch (transistor) is on the ground side of
    the load, (so the load floats above ground) and high side is when the
    switch sits between the power supply and load with the load
    grounded.

    George H.
     
  6. Tim, stupid question, why is that better?

    George H.
     
  7. Usually there are more ways to skin a cat and I drew some of them below.
    (View using fixed font like Courier)

    -----+----+------------------+-----+--------------------------+----
    | | | | |
    | | .-. | |
    | | 56R| | | |
    | | | | | |
    | | '-' | |
    | | | |/ |
    | | +---| BC635 |
    | | | |> |
    | | | | |
    .-. | .-. | |
    270R| | | 220R| | | |
    | | | | | | |
    '-' | '-' | |
    |\ | | |\ | | |\ |
    | \ | |/ | \ | | | \ ___ |<
    | >--+--| BC635 | >---+ | | >---|___|---| BC369
    | / |> | / | | / 270R |\
    |/ | |/ | |/ |
    +------ +------ +------
    | | |
    | | |
    V V V
    - LED - LED - LED
    | | |
    | | |
    | | |
    O___ O___ O___
    O O O
    controlled| | |
    current | | |
    limiter | | |
    | | |
    -----------+------------------------+--------------------------+----
    (created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

    The one to the left uses a 270R resistor to limit the current through your
    driving circuit below 20mA when its output is low. As we do not know the
    voltage on the anodes of the LEDs we cannot calculate what current is
    driving the base of the BC635 when the output is high. That current may not
    be enough to drive the BC635 into saturation as it has a limited gain. If so
    you can look for a transistor that provides more gain (hard to find) or a
    Darlington for example a TIP122. Pretty much a canon to swat a mosquito and
    its saturation voltage may be to high.

    A better solution seems to split up the current limiting resistor as in the
    example in the middle. If you still have not enough base current you can
    lower the 56R as long as you raise the 220R accordingly.

    The example to the left uses a PNP-transistor. Note that this circuit lights
    the LEDs whem your driving output is low. You only have to be sure that this
    output when in high state is raised to at least 4.4V. If not, you may use a
    pullup resistor to that output but that will increase the current to be
    sinked through the output so you may have to raise the 270R.

    petrus bitbyter
     
  8. Daniel Pitts

    Daniel Pitts Guest

    I don't care as much about the voltages at the various parts of the
    transistor. I'm using a 74HC238 to select the row, though if necessary
    I could swap that out for 74HC138 (active low instead of active high).
    As I've said elsethread the 20ma is a maximum draw from the signal, not
    a desired amount. Less would be better actually.
    Datasheet for the 74HC238:
    Here's the current-sink (to actually drive the LEDs):
    <http://www.ti.com/lit/ds/symlink/tlc5916.pdf>

    BTW, I believe its a "conventional current sink", meaning an electron
    source. Why does convention have to be backwards from physics!

    So, I would either have a PNP with the emitter on the +5v line, or an
    NPN with the collector on the +5. The opposite end would be on the LED
    line, and that line may be somewhere between 5v and 0v depending on the
    current sink on the other side of the LEDs. The end of the transistor
    not tied to +5v would likely be between 2v and 5v. 5v means "all the
    LEDs are off anyway", so I would expect no current. 2v means I'm
    dropping 3v across the transistor at potentially 500mA. Hmm, 1.5 watts.
    I wonder if my components can handle that. Unless (and likely) my
    analysis is wrong.
    I'm glad he asked then ;-)
     
  9. Daniel Pitts

    Daniel Pitts Guest

    That was a lament, not a question ;-)
    Right, I realized this after I posted. Indeed, if I'm in saturation, the
    voltage drop should be acceptable.
    I have plenty of SS8550DBU (PNP) and PN2222ATA (NPN). I was also
    looking at some FETs on digikey last night.

    I'm not opposed to ordering other parts. I know what I'm making isn't
    novel, but if I can "do it cheaper", I might consider trying to sell my
    end-product. In any case, I'm enjoying the learning experience.
     
  10. You'll have to drive a PN2222A next to its limits and that's never a good
    idea. If you want to give it a try, look at the datasheets of the
    manufacturers. They have different specs for the same type of transistors.

    That SS8550 suits better. A 15-18mA base current should be enough to turn
    the transistor on. If the transistor is not driven into saturation, it may
    become (too) hot. You have to account for that.

    petrus bitbyter
     
  11. Daniel Pitts

    Daniel Pitts Guest

    Here's the datasheet from my supplier:
    <http://www.fairchildsemi.com/ds/PN/PN2222A.pdf>. 500ma is %5 above my
    "worst case" (a pure white display) Looking at the datasheet though, the
    current gain is 40 there, meaning I'd need at least 12.5ma to the base
    to drive the thing to saturation. I could do that, but as you said that is
    I was hoping for that, though I'm not seeing it. I'm seeing plenty of
    N-channel mosfets for < $0.30 (although none that cheap can handle the
    amperage I'm expecting), but the cheapest P-Channel is $0.42. That
    would add $3.20 to my project.

    I was also considering a Darlington, but that seems to be as expensive
    as a similarly suitable P-channel MOSFET. It would be cheaper to hook up
    two transistors myself.

    <http://www.onsemi.com/pub/Collateral/2N6040-D.PDF> could be suitable. A
    bit overkill (8A, 75W), but only $0.30 each. Comes in both PNP and NPN
    flavors.


    Actually, there is an octal array that might also be worthwhile...

    <http://www.st.com/web/en/resource/technical/document/datasheet/CD00000179.pdf>

    Still only 500ma max, but again, that is my worst-case scenario. This
    array is only $0.27. It also says outputs could be parallel to increase
    max current, so I could in theory add two to my project to support up to
    one amp, and it would only add a half dollar to the cost.
    The Toshiba PNP version:

    <http://www.semicon.toshiba.co.jp/docs/datasheet/en/LinearIC/TD62783AFG_en_datasheet_091116.pdf>

    Seems like one of these might be the way to go. Have an array of 8
    simplifies a lot for me too, though it does add one big chip to the
    project ;-)

    Thanks again for all the advice.
     
  12. Daniel Pitts

    Daniel Pitts Guest

    Yeah, I remember considering that at some point, but then it slipped my
    mind again. Too much going on in my life outside of this hobby ;-)

    So basically, if I have 480mA through one device with a duty-cycle of
    12.5%, then as long as it can handle the "peak" without damage, the
    average is about 60ma. That greatly improves my wattage as well. I
    could also consider lowering my current by a little over a 3rd, and just
    use the original IC I have now. This might be feasible since current
    relates to LED brightness, and they may be bright enough at 6mA.
    Especially for an indoor portable display, which was my original intent.

    Something to ponder.
    On the NPN Darlington array I linked to, the current gain is 1000. Well
    more than enough for my needs. I think this really is the way for me to go.

    The Toshiba one I linked to (the TD62783APG) doesn't list current gain,
    but the I-in(on) at with VIN = 3.85 V is a max of 260uA. Well below my 20mA.
    Yeah, pretty much through-hole only. This whole thing is on a
    solderless breadboard. I've considered designing a PCB, but that's a
    whole other experience. I have soldered SMT stuff before (had a kit I
    bought at a workshop). It was interesting, but I don't think I'm ready
    to design an actual circuit board from scratch.

    Anyway, Thanks again for the feedback and advice. I really appreciate it.
     
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