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newb: I'm trying, but need this cleared up...

Discussion in 'Electronic Basics' started by SklettTheNewb, Mar 31, 2005.

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  1. I have been messing with CircuiMaker and doing "spice simulations"
    I have also been reading from these different sites:
    http://www.kpsec.freeuk.com/ohmslaw.htm
    http://www.play-hookey.com/digital/ripple_counter.html
    http://www.allaboutcircuits.com/vol_1/chpt_2/index.html

    I'm slowly getting my head around the current and voltage ideas. I
    have encountered a strange situation w/ a spice simulation like this:


    LED 680
    ..--|<-----|___|-----.
    | |
    | |9V
    | ---
    | -
    | |
    |-------------------'
    (created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

    if I place the probe before the 680 ohm resistor, I get 9v, if I place
    it between the resistor and the LED I get 1.79V and 10.60mA

    I'm trying to understand how I got that number, I know that;
    V
    I = -
    R

    if I use that formula with my values:
    9
    - = 0.013
    680

    - that makes no sense.
    I'm missing something VERY basic and simple, but I can't figure it out.


    Any help or guidance would be REALLY appreciated at this point.
    Thanks,
    Steve
     
  2. Andrew Holme

    Andrew Holme Guest

    The voltage drop across the resistor is 9 - 1.79 = 7.21 V

    7.21 / 680 = 10.60mA
     

  3. LEDs aren't resistors and don't obey Ohm's law. They usually have an
    approximately fixed voltage drop across them almost independent of the
    current flowing through them. Your simulated LED has a forward voltage of
    1.79 Volts. This is a fairly typical voltage drop for most LEDs except for
    blue and white ones (which usually drop closer to around 3.5V)

    You have a voltage loop. The sum of the voltages around the loop (if you
    respect polarities) is always zero. Starting from the bottom of your
    schematic and going counterclockwise you gain +9V up through the voltage
    source, then you drop 7.21V across the 680 ohm resistor, and then drop
    another 1.79V across the LED. So +9V - 7.21V - 1.79V = 0V. If in your
    simulation you were to measure the voltage across the 680 ohm resistor
    (instead of just measuring voltages with respect to ground) you would indeed
    find it to be 7.21V. The resistor obeys ohms law and you will indeed find
    that yes: 7.21V/680 = 0.01060A.
     
  4. Bob Eldred

    Bob Eldred Guest

    Your equation is right but your reasoning is wrong. You measured the voltage
    across the LED as 1.79 Volts. What is the voltage across the resistor? Well,
    it has to be 9 - 1.79 = 7.21 Volts, right? Therefore the current is I = V/R
    = 7.21/680 = 10.6 mA. As you can see, ohms law applies to the voltage across
    the resistance in question and not some other voltage like the battery,
    etc. Now, given that, what is the resistance of the LED? You should get
    168.8 Ohms. What is the series resistance of the resistor and the LED, You
    should get 848.8 Ohms. Now what is the current in the circuit, It should be
    (total voltage/total resistance) 9V/848.9 = 10.6 mA which checks.
    Bob
     
  5. Great answeres all, I was missing the "voltage across the component"
    part, I was trying to base everything off of the supply voltage.

    This makes sense now, thank you for the great responses.
    Take it easy,
    Steve
     
  6. Active8

    Active8 Guest

    Argghh! I just looked at that third link. That current convention
    thing (from menu on the left) is backwards to anything I've ever
    seen in circuit analysis texts and will f you up if you pusue
    electronics seriously - like Kirschoff's laws and node and mesh
    (loop) analysis.

    It's this way:

    + V -
    ___
    -|___|-

    I -------->

    Note that this is the same from the perspective of voltage
    generators as opposed to drops. An example might clear this up:

    +------------------+
    | |
    | |
    | |
    + | ^ I |
    | | | .-. +
    --- | | | |
    V - | | | | V
    | | | '-' -
    - | | V |
    | I |
    | |
    | |
    +------------------+
    |
    ^ call this the reference node or O V

    Kirschoff's Voltage Law (KVL) says that the sum of the voltage drops
    around a closed loop equals zero.


    Sum(V_i) = 0. or

    n
    ___
    \
    /__ V_i = 0
    i=1

    Starting at the negative terminal of the batt, and noting that we're
    going with the flow, i.e., current (charge flow), from negative to
    positive, we have a negative drop or positive voltage rise. Say the
    drop is -9V, i.e. 0 - 9 = -9 You take the starting value and
    subtract the ending value.

    At the resistor, we're going from positive to negative, which is a
    positive drop from 9V to 0v, i.e., 9 - 0 = 9

    so -9 + 9 = 0

    Another way of stating KVL is that the sum of the generator voltages
    is equal to the sum of the voltage drops. So in the same manner,
    going from the reference node to the positive terminal of the
    battery, we have a positive generator and

    9 = 9 which is the same as 9 - 9 = 0 or -9 + 9 = 0

    I'd recommend a better site if I knew of one for sure. Maybe that's
    the only thing wrong, though.
     
  7. Active8

    Active8 Guest

    Yes. You're missing the fact that the voltage across the resistor is

    9 - Vf or 9 - Vd, the drop across the diode, V_d, or its forward
    voltage, V_f. Plug that in. Hint: V_f is 1.79 V since you're
    measuring it relative to 0 V AKA "common" sometimes AKA ground AKA
    GND.

    I might as well mention that in this group you'll be seeing
    notations like Vf and V_f which just means that "f" is a subscript.
    We try to do what it takes to get the point across. Look at some
    data sheets and parts catalogs.
     
  8. Colin Dawson

    Colin Dawson Guest

    Maybe I can help a little. I'm using Livewire, to simulate the circuit.
    Let's add a few points to the circuit to help make things a little easier to
    explain...


    Probe B Probe A
    | |
    | |
    LED | 680 |
    .--|<-----|___|-----.
    | |
    | |9V
    | ---
    | -
    | |
    |-------------------'
    |
    |
    Probe C


    The current for this circuit is 10.44mA.
    The Voltate at the power source is 9v exactly (as I'm using a circuit
    simulation this won't change unlike real life, but let not complicate things
    with that)

    The measured voltages are as follows

    Probe A is 9V
    Probe B is 1.9V
    Probe C is 0V

    The difference in voltage between Probe A and Probe B is 7.1V Remember
    there's still 1.9V left to drive the LED

    So from that

    9V/680R = 0.01044A

    This is in Amps. My Simulation shows the current in Milliamps so I need to
    multiply my figure by 1000

    0.01044A*1000 = 10.44mA


    So what's the resistance for the LED?

    V/I=R

    The measured voltage for the LED is 1.9V (Probe B) so

    1.9V/0.01044A = 181.992R (182R will suffice)

    Add that to the 680R and lets do the entire circuit.

    9V/(680R+182R) = 9V/862R
    = 0.01044A
    = 10.44mA


    I hope this helps a little.

    Regards

    Colin Dawson
    www.cjdawson.com
     
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