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How to drop a voltage using a series resistor?

Discussion in 'Electronic Basics' started by [email protected], Aug 9, 2005.

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  1. Guest


    I'm looking at a schematic here that drops a 5V TTL output through a
    series resistor into a 3.3V LVTTL input.

    I cannot figure out how the value of the resistor is chosen, and would
    like someone in the know to explain it (step by step, in small words,
    like you would to a software engineer).

  2. Byron A Jeff

    Byron A Jeff Guest

    A single series resistor isn't the way to go. Too many variables.
    Also I don't think that a 5V TTL output can be guaranteed to be 5V.
    In my trivial PIC programmer ( I
    use HCT parts because the CMOS output is pretty much guaranteed to
    be near the 5V rail.
    To be sure it would be better to use two resistors in a voltage divider
    arrangement. Then you won't have to depend on the 3.3V LVTTL input
    to present a set impedance to match up with the series resistor.

    Check out the theory here:

    Since 3.3 is 2/3 of 5V you want to generate a 2 to 3 ratio for your
    resistors. Input current for LVTTL seems to be in the 10's of microamps
    so using 10k and 20k resistors should provide more than enough current.
    Create the voltage divider by tying one end of a 10k resistor to the
    5V TTL output. Tie one end of the 20K resistor to ground. Finally tie
    the two loose ends of the two resistors together. At that point the voltage
    will be 3.3V when the TTL output is high. Tie the junction of the two
    resistors to the LVTTL input.

    When the TTL output is high, the resistors will consume 5V/30k = 160 uA of

    Hope this helps,

  3. Guest

    Thanks for the reply Byron. I understand how to do it with two
    resistors, but that wasn't the question. I am not designing a circuit,
    rather I am trying to understand how an existing circuit was derived.

    Someone mentioned to me needing to look at the IBIS models of the part
    being driven. This flew so far over my head it didn't move a hair.
    They also said series isn't the way to go, but didn't have time to
    elaborate on how R was determined.

    So I'm still at a loss, and seeking knowledge.

  4. JeffM

    JeffM Guest

    A single series resistor isn't the way to go. Too many variables.
    They looked at the threshold spec and the typical input current spec
    and hoped that their guess would hold for all iterations.

    The repeatability of the 1-resistor technique
    from one manufacturer to the next will not be predictable.
    Even chips from different batches from the same manufacturer
    may not work.

    A series-resistor/shunt-resistor pair should be chosen
    to swamp out the variables.


    I'm not sure if I understand the question exactly but it seems to me if
    your 5V source is being dropped to 3.3V then 1.7V is being dropped across
    the resistor in question. The value of R would be determined via Ohms law.

    The device which needs the 3.3V source will draw a certain amount of
    current and for this to work that current must be constant.

    I am not familiar with LVTTL but just for the sake of discussion say it
    draws 10mA at 3.3V. You can then think of it as a 330 Ohm resistor. 3.3V /
    ..010A = 330 Ohm.

    To solve for your R, You know you need to drop 1.7V and you know the
    current is .010A so, 1.7 / .010 = 170 Ohms

    As I said, your LVTTL device must draw a constant current for this to
    work. If it does not your 3.3V will fluctuate as the current draw

    Ohms law, E=I*R, I=E/R & R=E/I

    To know what you device really draws for current 1.7V/R1 = I

    I hope this answers your question
  6. Guest

    Sounds like you understood perfectly :)
    It was the 10mA at 3.3V bit that confused me. JeffM suggests this is
    unreliable, and my friend's comment about looking up IBIS models made
    me think so too.

    Any other notes, besides "lousy engineering"?



    So, now you understand how the size of the resistor was determind? I agree
    about the lousy engineering and un relable part, Just want to make certain
    your question has been answered.
  8. John Fields

    John Fields Guest

  9. Guest

    Yes, the only outstanding question is where the input LVTTL current is

    Specifically for this question I'm looking at the Xilinx Spartan-3 FPGA
    with LVTTL inputs. I don't see the input current documented in the
    data sheet, but I may simply not be seeing it.

    The schematic I'm looking at is Digilent's Parallel I/O board for their
    Spartan-3 starter kit. PC Parallel is TTL, and Digilent use 100 ohm
    resistors to drop a TTL output down to something suitable for the 3.3V
    Spartan-3. (

    Can you see the data that led them to 100 ohms?



    I see what you are saying


    Case 1 addresses a 5-V TTL device driving a 3.3-V TTL device. As shown in
    Figure 1, the switching levels for 5-V TTL and

    3.3-V LVC are the same. Since 5-V tolerant devices can withstand a dc input
    of 6.5 V, interfacing these two devices does not

    require additional components or further design efforts.


    But earlier in the text,


    The Case for Low Voltage

    LVL, or low-voltage logic, in the context of this application report, refers
    to devices designed specifically to operate from a

    3.3-V power supply. Initially, an alternative method of achieving
    low-voltage operation was to use a device designed for 5-V

    operation, but power it with a 3.3-V supply. Although this resulted in 3.3-V
    characteristics, this method resulted in significantly

    slower propagation time. Subsequently, parts were designed to operate using
    a 3.3-V power supply.

    A primary benefit of using a 3.3-V power supply as opposed to the
    traditional 5-V power supply is the reduced power

    consumption. Because power consumption is a function of the load
    capacitance, the frequency of operation, and the supply

    voltage, a reduction in any one of these is beneficial. Supply voltage has a
    square relationship in the reduction of power

    consumed, whereas load capacitance and frequency of operation have a linear
    effect. As a result, a small decrease in the supply

    voltage yields significant reduction in the power consumption. Equation 1
    provides the dynamic component of the power

  11. John Fields

    John Fields Guest

    Apples and oranges.

    One is saying that there is no problem with getting LVTTL inputs
    powered from a 3.3V supply to switch when driven by TTL outputs
    powered from a 5V supply, while the other is saying that that
    combination will cause LVTTL to consume more power than if its
    inputs were driven by LVTTL powered from a 3.3V supply.


    So the OP's mystery resistor, (he was wondering how the designer chose it's
    value) Serves what purpose? Current limiting? Seems to me there must be a
    reason for it. Maybe not...
  13. John Fields

    John Fields Guest

    Here's what the OP's circuit looks like connected to a typical TTL

    | |
    [130R] |
    | |
    C |
    B NPN |
    E [1K2]R1
    | |
    [DIODE] |
    | | R2
    B NPN

    TTL, only being able to source 400µA or so and guarantee 2.4V out,
    (with a 5V supply) is pretty weak, and that's what the 1.2k resistor
    in parallel with it is for, to provide extra current into the load.
    How much? assume the load needs to see 2V and we have:

    Vcc - Vil 5V - 2.4V
    I = ------------ = -------------- = 0.002 ~ 2.0mA
    R1 + R2 1200R + 100R

    Also, those resistors will keep any downstream circuitry from
    floating if there's nothing plugged into the connector and will
    provide everything downstream with known logical staes if nothing is
    plugged in

    The resistance of the 100 ohm resistor is only [really] important
    when the load is being pulled down, and since TTL can sink 1mA and
    stay below 0.1V, that current through 100 ohms will cause an
    additional 0.1V drop, resulting in a total drop of 200mV with 1mA
    out of the load. With a maximum current of 75µA out of the LVTTL
    load and a 0.1V drop across the bottem totem-pole transistor of the
    TTL sink that comes to a drop of:

    Et = (Il Rs) + Vol = (7.5E-5A * 100R) + 0.1V = 0.1075V ~ 0.1V

    Which is well below LVTTL Vil(min) of 0.8V.

    Bottom line? the 100 ohm resistor isn't really needed but is in
    there in case someone does something stupid like connect 5V with no
    current limiting to the connector.

    To find out, check the maximum input current allowed into LVTTL and
    check if that's exceeded with raw 5VCD on its inputs when it's
    running with a 3V supply.
  14. Rich Grise

    Rich Grise Guest

    Probably not. According to the data sheet, the input leakage current of
    an LV chip is 1.0 uA - that's microamps. So to drop 1.7V at 1 uA, you'd
    need 1.7M. And depending on what kind of Voh your TTL is putting out,
    (It's typically 2.7-3.3), the value was probably picked to give a worst-
    case amount of current that can be shoved into the pin.

    Yup. It's a bad design.

    Hope This Helps!
  15. Rich Grise

    Rich Grise Guest

    In the LVTTL data sheet. Ii, input leakage current. For the 74LV00, it's
    1.0 uA (microamp).
    Page 4, table 6: Il, "Leakage current at user I/O, Dual-Purpose, and
    Dedicated pins."
    For some reason, I'm not able to open PIO1-sch.pdf - but I can tell you
    what led them to use 100 ohms - they pulled it out of the air. There
    is no calculation involved at all - it's more of a rule-of-thumb, seat-
    of-the-pants kind of thing. ("Oh, got a parallel port output? Stick a
    100R resistor in series, 'cuz that's what everybody else does." :) )

  16. Rich Grise

    Rich Grise Guest

    Just a generic Q on this, since I'm too lazy to look it up - are there
    LVTTL chips that _can't_ handle a 'regular' TTL level at its input?

  17. John Fields

    John Fields Guest

    Dunno. I only checked into it for the OP's sake and got my info
    from TI's 1994 Low-Voltage Logic databook, so since it's not
    something I need to pursue for my own ends, I'm too lazy to look
    into it any deeper. :)
  18. Guest

    Thanks for all the responses - more help than I expected.

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