n00b needs some help with Ohms law @work

Discussion in 'General Electronics Chat' started by stormBytes, Sep 4, 2010.

  1. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
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    I've included (hopefully it will work this time!) a diagram of two simple circuits. The objective here is to test/verify Ohms law.

    Circuit 1 consists of a 5V source, a 680 ohms resistor and a 5mm green LED. Current was measured with a Fluke 87V to be 4.44 mA. Using Ohms law, E=IR, I divided 5V by 0.0044 Amps and got a total resistance of 1136 Ohms.

    Next, in circuit-2 I introduced a potentiometer with unknown resistance into the circuit. I measured the total current and got 2.76 mA. Again, applying Ohms law I divided 5V by 0.00276 A and got 1811 Ohms for total resistance.

    Subtracting Resistance-initial (1136 Ohms) from Resistance-final (1811 Ohms) I got 675 Ohms which, by my understanding, should be the resistance setting of the potentiometer.

    However, measuring resistance across the pot's terminals I got a reading of 433 Ohms!?

    What am I doing wrong?

    ** EDIT:

    I just tried the same experiment using a regular (second) resistor in place of the pot. I guess my math must be wrong because the way I did it, I got the wrong answer again for the 2nd resistor value.
     
    Last edited: Sep 4, 2010
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Perhaps you should re-do the test without the LED which may be exhibiting a non-linear resistance with varying current.

    Is this a test you devised yourself or something done as part of a classroom experiment?
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Your 433 Ω makes some sense.

    In the first case you have 4.44mA, meaning the LED is taking about 2V and the 680Ω the remaining 3V from the 5V supply.

    In the second case assume the LED voltage is roughly the same - about 2V or something slightly less. Let's say it's 1.9V. The two resistances (680Ω + pot) then share 3.1V.

    In that case the total series resistance of 680Ω + pot would be 3.1/2.76mA=1123Ω

    Subtracting 680 from this gives Rpot=1123-680=443Ω which is somewhere near what you have.
     
    Last edited: Sep 4, 2010
    stormBytes likes this.
  4. eblc1388

    Senior Member

    Nov 28, 2008
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    To OP:

    The voltage drop of the LED depends on the current value but does not change in a linear fashion as a normal resistor does.
     
  5. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
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    I just spent the better part of 2 hours building variations of this circuit on a breadboard, and taking measurements. This particular example is not as important. I'd much rather simply understand the basic idea.

    I built a circuit using (1) LED, (1) resistor, and a voltage source of 5-7V. Each time I ran the available numbers to predict the variable, using Ohm's law, and each time everything worked exactly as it should have.

    My follow up questions:

    1) I know that resistors are used to reduce current as well as voltage, but I'm a bit fuzzy on how this is *useful* in practical terms. Could you provide an example?

    2) What do you do if you've got 5V and you need all 5V but only want to limit current? Is there some variation of Ohm's law that can be used to figure out what resistor value to use? Does this question even make sense!?

    3) How do you predict how much voltage a given resistor will consume in a circuit you haven't built yet? (I've thus far been deriving lots of my numbers through sheer measurement)

    Thanks for the help!
     
  6. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
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    I'm fuzzy on this 'voltage drop' concept. Why is it that no voltage remains in a circuit after the components? I'm using a simple circuit as an example, consisting of an LED light, a resistor and a 7V source. When I measure the voltage across the 680 Ohms resistor, I get 5V, then across the LED, I get 2V. If I try to measure the voltage from the positive leg of the LED (the last component) and the positive terminal on my breadboard, i get zero volts. I don't understand how this works, could you shed some light?
     
  7. eblc1388

    Senior Member

    Nov 28, 2008
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    You are measuring between the ends of the connecting wire. Being a conductor the resistance is very small and the voltage drop according to ohm's law is also small so a common DVM will give zero volts.

    If you use a better DVM with more digits, you WILL see a non-zero reading.

    [​IMG]
     
  8. Markd77

    Senior Member

    Sep 7, 2009
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    In most cases you can't set current and voltage at the same time. If you have a 1 ohm resistor there is no way to have 5V at 1A across it's terminals. You can build a circuit that supplies 5V and limits the maximum current to 1A , but in most cases when 1A is reached the voltage has to drop to compensate.
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The very circuit you have been playing with is a good simple example.

    If you connect an LED directly to a "stiff" DC supply of 10V you will burn out the LED. That's why you need the series resistor to limit the current in the LED to a safe value. If you look up a manufacturer's data sheet for a particular LED range you will most likely be given a typical case of forward current vs forward voltage drop - at the very least. Say you are given the data that a particular LED has a forward voltage drop of 2.3V at 10mA. You are intending to drive the LED at around 10mA from a 10V DC source. You will need to work out what resistor will limit the current to 10mA.

    The resistor needs to drop 7.7V at 10mA. So R=7.7/0.01=770Ω. Now, 770Ω isn't a typical preferred value. If you are comfortable with going a little above 10mA you could use a 680Ω which might give you around 11mA. Alternatively if you want the current to be no greater than 10mA you could use a preferred value of 820Ω which might give you a current of say 9.5mA.

    I don't really understand what you are getting at. I can think of an example where you might want to set up an exact current to check the calibration of a current indicating instrument.

    One of the most common tasks students learn in their semiconductor electronics studies is how to correctly bias the operating point of various transistor amplifier configurations. If you follow this Forum for a while, you'll see this particular topic can be a source of much discussion - and often perplexity for the beginner. Presumably this will be something you will master with time if you are pursuing studies in this area.
     
  10. marshallf3

    Well-Known Member

    Jul 26, 2010
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    2) What do you do if you've got 5V and you need all 5V but only want to limit current? Is there some variation of Ohm's law that can be used to figure out what resistor value to use? Does this question even make sense!?

    That's what a fuse or circuit breaker is for. You can make current limiter circuits but you will lose some of your 5V due to the internal resistance of the circuit. Any time you pass current through a resistor there is a corresponding voltage drop.

    Think about it in ohms law. 5V in and 5V out = 0V drop. R=E/I and if E=0 then R is going to equal 0 as well.

    Current limited supply circuits do exist but they start with a few volts more than your desired output voltage.
     
  11. curry87

    Member

    May 30, 2010
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    Sort of a related question if i had a power supply of 12v and had 4 led hooked up in series each having vf of 3 would i need a resistor to limit the current or do the leds just taken what they need ?
     
  12. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
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    I'm having a hard time putting it all in perspective. If you have a water pipe with 60psi and you add a valve to it, there *will* be some pressure after the valve. The valve may cut 20-50-90% of the initial pressure, but it surely won't be at 0psi unless the valve is 'off'.

    Perhaps I should change books. The way this book explains stuff is just leaving me confused.
     
  13. Ghar

    Active Member

    Mar 8, 2010
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    If you're not regulating current you should have a resistor.

    Every 50mV to 100mV is another factor of 10 in current, it depends on the diode. So if you don't know the exact voltage of the diode for the current you want it will be way off from what you expect.
     
  14. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
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    That's actually an interesting question and I'd be curious to hear what the more experienced folks have to say about it -

    As I understand it, each LED drops say... 2V. So after your 4 LEDs there should be 4V left in the system, or so I'd gather?

    But as I've been told, LEDs don't have 'linear resistance' curve, so why not put it in terms of incandescent bulbs?
     
  15. Markd77

    Senior Member

    Sep 7, 2009
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    Incandesent bulbs don't have a linear curve either. Resistance increasess with current.
     
  16. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
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    Now I know why I opted OUT of engineering......
     
  17. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
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    I wish I'd understand the mechanics of current & resistance. The example posted above with the 4 LEDs would make a lot more sense. So far breadboarding example circuits has been an exercise comparable to tossing darts in pitch darkness.
     
  18. Ghar

    Active Member

    Mar 8, 2010
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    You look at voltage-current relationships of components.

    Check out the diode V-I curve on the bottom of this page:
    http://www.allaboutcircuits.com/vol_3/chpt_3/1.html

    It's a very steep slope, so small changes in voltage lead to large changes in current.

    By using a resistor you exploit this fact. You can approximate the diode as a constant voltage giving you a constant voltage across the resistor. Constant voltage across a resistor is a constant current.
    You can apply a voltage to a diode but every diode is a bit different and you need to be exactly right to get what you want because small errors lead to large changes in current.
     
  19. curry87

    Member

    May 30, 2010
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    So in this case of having a 12 volt dc supply and 4 led with vf of 3 in series you want no more than 20ma to flow so what do you do:

    Take one led out and just use a resistor to drop the 3v and limit it to x amount of current ?

    Increase the power supply use a resistor to drop the excess and limit it to x amount of current ?
     
  20. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    As in solving certain non-linear math problems you may have to apply non-linear solution methods. One would start with the LED diode equation - presumably something akin to the Shockley diode equation. The solution obtained would have to satisfy the condition that the sum of the individual diode forward voltage drops equaled 12V at a current as yet to be determined. A simplifying assumption would be that the diodes are ideal and that each has exactly 3V drop. A significant issue of the solution would be the temperature instability of the current in a practical circuit - the forward voltage is temperature dependent.

    Introducing a series resistance would reduce the instability but not necessarily significantly, since the bulk of the supply voltage would still be across the diodes.

    One might envisage a design in which each diode has ~2.5V and a total added resistance of 200Ω - with ~10mA current. Again one would have to solve the diode equation for 2.5 V and 10mA (or thereabouts) - it would require an iterative approach to arrive at a satisfactory or predictable solution.

    EDIT : Another method I was taught in "the old days" was the graphical / load line technique - which the OP might like to research.
     
    Last edited: Sep 4, 2010
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