Series and Parallel/Series Circuits

Discussion in 'General Electronics Chat' started by teesee, Apr 5, 2015.

  1. teesee

    Thread Starter New Member

    Jan 29, 2015
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    I have been reading as much as I can about the basics e.g. Ohms law, I,V,R and P so far and most of the tuition seems to be based on resistor circuits in which in order to calculate current when V and Rtotal are known is merely Ohms law of I=V/R. So I wondered what happens when I put a few basic components into the mix and it seems to change some things and adds to my confusion.
    My series circuit in a clockwise direction is a 9V battery, 1N4001 Diode, 120 ohm, red 3mm LED, 120 ohm, red 3mm LED.
    The only way that I can see how to calculate the current is to subtract the total of the forward voltage of the diode and the two LEDS from my 9V and divide that by the total resistance.
    So if i'm antwhere close to being right this means that although there is a Vdrop across resistors they don't count in this calculation since there are "real" components in the circuit. Is that correct.
    Also is there a different /more professional method of calculating current in this instance?
    I will pose my parallel circuit question on a different thread.
    Many thanks.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Very good. I = (Vsup - All diode voltage drop)/Total_resistance
    You meant to say that we did not include a diode resistance ??
    No
     
    Last edited: Apr 5, 2015
  3. teesee

    Thread Starter New Member

    Jan 29, 2015
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    Thanks Jony130 for your reply. Actually I didn't mean to mention the "diode" resistance. Since your response I have googled the topic just to find it a very confusing issue. Could you refer me to somewhere that I can get information on this issue. It didn't seem to affect the current calculations of my circuit.

    Thanks again
     
  4. WBahn

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    Mar 31, 2012
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    Up to this point you are doing great. In the future, try throwing together a simple schematic even just using Paint. It really aids in communicating ideas.

    What do you mean by "they don't count". Your method is entirely based upon them counting.

    The sum of the voltage drops across the resistors, the diode, and the LEDs must be equal to the battery voltage. After you subtract the diode/LED voltage drops from the battery, what you have left IS the total voltage drop across the resistors. That is why dividing that voltage by the total of those resistors give you the current through those resistors.

    For almost all purposes, the way you did it is just fine and is how it would be done in practice. In rare cases you might be concerned with the fact that the voltage across the diode/LEDs is not a constant but is, in fact, dependent on the current. This is a highly nonlinear relationship and there are a number of ways to tackle that problem.
     
  5. Jony130

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    But to be honest I do not see what the issue is ? Can you explain it bit more? Whats bother you? The diode voltage?
    For your circuit we can apply II Kirchhoff's law (Kirchhoff's voltage law, KVL)
    10.PNG


    So we have:
    9V = 0.6V + VR1 + 2V + VR2 + 2V (The sum of voltage drops across all components must be equal to the battery voltage).
    VR1+VR2 = 9V - 0.6V + 2V + 2V = 4.4V
    And now we can use Ohm's law to find current
    I1 = 4.4V/(120Ω + 120Ω) = 18.34mA

    As you can see we treat diodes as a short circuit with "voltage drop". Why ? Because of the diode property when forward biased. Take a look at diode I to V characteristics

    [​IMG]

    As you can see the voltage drop across a diode (in the forward biased state) is almost constant regardless of the current running through it. (diode act almost just like a short circuit).
    For example if diode current will increase by a factor of 10 (from 1mA to 10mA) the diode forward voltage drop increased only by 60mV. This means that diode when forward biased has very low resistance. And this is why diode always need a current limiter device (a resistor) in series with a diode.
     
  6. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Don't

    Do it here until you are clear on the difference between series and parallel.
    This is basic and covers much more than just resistors.

    Current passes through devices :

    Voltage appears across devices:

    This is easy for two terminal devices like a resistor.

    Curent passes into one terminal and out of the other.
    In a series circuit the current is the same in all devices in series. It goes out of one device and into the next in the chain.
    Series circuits are characterised by a chain of devices.

    Voltage appears across or between the two terminals.
    So each voltage is in series and the chain of voltages add up to the total applied voltage.
    A series circuits splits the voltage.

    Each resistor in the chain obeys Ohm's law so that the current through each device times its resistance equals the voltage across it.

    A parallel circuit is the other way round.

    The voltage is the same across each device in the parallel combination.
    Thus it is equal to the applied voltage.
    Some of the total current passes through each device .

    Again if each device is a resistor they obey Ohm's Law so the amount of current times the resistance equals the common voltage across them.

    These facts can be used to deduce the series and parallel resistor combination formulae.

    Two resistors connected together terminal to terminal acrossa single battery are in parallel.

    Two resistors connected end to end across a single battery are in series.

    One resistor connected to a single battery can be considered either in series or in parallel with the battery.

    I suggest you draw up for yourself these simple cases. It will help cement the ideas for you.
     
    Last edited: Apr 6, 2015
  7. teesee

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    Jan 29, 2015
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  8. teesee

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    Jan 29, 2015
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    Hello Studiot .... Thanks for your response. I will be away till late tomorrow but will digest your post and respond with questions then.

    Thanks
     
  9. studiot

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    Studying to get a good grasp of the basics like you are doing is highly recommended.

    :)
     
  10. alfacliff

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    resistors have resistance, diodes are different, they have a dynamic current depending on the voltage drop across them. same for led's. diode voltage drop depends on the current through them above their conduction point. non linear conduction, unlike resistors, that have a constant resistance. the reason resistance, voltage, and current are stressed is so you get the basics. reactance is just ac resistance, impedance is reactance and resistance.
     
  11. teesee

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    Jan 29, 2015
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  12. studiot

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    I find the new editor absolute rubbish.

    So I also often get the quotes in a twist .

    To start are you aware that points E, G and D are the same point from the point of view of circuit theory, as are
    points B, F H and C
     
  13. WBahn

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    Mar 31, 2012
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    DiodeSplit2.png

    You have six nodes. You get to arbitrarily assign whatever voltage you want to one of them. By convention, in this case, the negative side of the battery (the black node) is assigned a voltage of 0V.

    With that in mind, we can determine immediately the voltage on the brown node, namely 9V.

    We can also determine the voltage on the red, blue, and green nodes. So what are they?

    The next step is to apply Kirchhoff's Current Law (KCL) to the yellow node. To do that, write each of the currents through each of the resistors in terms of the voltages on each side of the resistor using Ohm's Law. Then sum them up according to KCL.
     
  14. tonyStewart

    New Member

    May 8, 2012
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    Ohm's Law still works for semiconductors and can be modelled as a fixed threshold voltage with a fixed linear resistance that you can add up.

    Since Resistors are linear they can be combined into one, so two 120 Ω resistors can be replaced with one 240 Ω value.

    We often call the diode series resistance Rs and the MOSFET series resistance RdsOn ( read... drain-source ON resistance or arrdeessON) or even Transistor Base Emitter and Collector Emitter resistance can be specified as a fixed value at rated current and assumed linear in that region for convenience.
    Even batteries have an Rs value that depends on size and state of charge.

    We most often use the same term to describe all or Effective Series Resistance or ESR.
    All Diodes (including light emitting ones) have a VI slope we call ESR.

    But for simplicity, I prefer to call each of them ESR and linearize them with rated forward current.

    Then Ohm's Law is very effective for estimating current in a loop.

    For example a 3V 2mm thick coin cell might have an ESR of 3 k Ohm but a 3.7 V LiPo cell might support 40 A short circuit current and thus have an ESR of 0.1 Ohm or even less. ( dont try that test more than a few milliseconds)

    If we put a 3 W LED that is rated at 3.5V @ 0.75A in parallel with a 3.7V LiPo battery without measuring or calculating the ESR from the datasheet, how will we know what the resulting current is? Well if you can calculate the slope and know the starting threshold is with no current you can calculate what the current might be or how many milliohms you need to add to limit the current just from the slope in the datasheet. The result is rather than use a 9V battery , I choose one to match the LED string (1) with slightly more voltage but less than one LED drop. This will always be most efficient usage of a fixed R current limiter.

    In this case 100 milliohms. with this Java simulation to illustrate.

    Ohm's Law is such a useful tool.
     
  15. teesee

    Thread Starter New Member

    Jan 29, 2015
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    Thanks again for taking the time to illustrate my circuit. I have been pondering the "next step" for a couple of days now and I need to ask for clarification ....
    If what you stated above as the "next step" is the same as performing the Junction and Loop Rules then, I'm on the correct track. Am I?
    The junction(s) are as follows
    I1=I2 + I3
    I2=I4 + I5
    I created three loops from top to bottom on my circuit A, B and C
    A) 9 - 120I1 - 600I3 - 1.89 = 0
    7.11 - 120I1 - 600I3 = 0
    B) 1.90 + 600I3 - 400I4 - 3.99 = 0
    -2.10 + 600I3 - 400I4 = 0
    C) 3.99 + 400I4 - 500I5 - 2.09 = 0
    1.09 + 400I4 - 500I5 = 0
    Having got this far I am having difficulty knowing how to calculate the currents. If you could clue me in I would really appreciate

    Your colourful illustration is my first experience with the "nodes" that you hi-lited. The only significance that I can see in them is that they make me aware that each colour is a different voltage at all points within the colour. Am I missing anything else?

    Why in this circuit does the Vt=ItRt formula not apply

    Many thanks again for the assistance.
     
  16. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    First let us recall the basics rules:

    I Kirchhoff's law (KCL)

    2.PNG

    In series circuit we have the same current but the voltage is divided in each individual elements
    http://forum.allaboutcircuits.com/attachments/17_1211642949-jpg.40685/

    II Kirchhoff's Law (KVL)

    [​IMG]
    In parallel circuit the voltage is the same but the current is splitting between the branches.
    So we have a current divider
    http://forum.allaboutcircuits.com/attachments/40_1211642965-jpg.40686/

    And be sure to read this
    http://www.ittc.ku.edu/~jstiles/312/handouts/312_Introduction_package.pdf (from page 3)

    Now let as back to your circuit but first notice that :
    Any point along the yellow wire (yellow node) is at the same voltage, and the same is true for any other color.
    So the KCL for is :
    I1 = I2 + I3 + I4
    Where :
    I1 = (Vbat - Vy)/R1 = (9V - Vy)/120Ω
    I2 = (Vy - Vr)/R2 = (Vy - 1.89V)/600Ω
    I3 = (Vy - Vb)/R3 = (Vy - 3.99V)/400Ω
    I4 = (Vy - Vg)/R3 = (Vy - 2.09V)/500Ω

    So we have
    (9V - Vy)/120Ω = (Vy - 1.89V)/600Ω + (Vy - 3.99V)/400Ω + (Vy - 2.09V)/500Ω
    Now you can try to solve this.
     
  17. WBahn

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    Mar 31, 2012
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    What are your currents? At first I assumed that Ix was the current through Rx in the direction indicated on the original diagram, so that R1 would be the current flowing left to right through R1, the 120Ω resistor on the top. But that can't be the case because you have an I5 and there is no R5 in the diagram. You need to clearly define your currents and voltages. Annotate the diagram with the location and direction of every voltage or current that you use in your work.

    You also need to track your units in your work from beginning to end. The term 120I1 is a current, namely a current that is 120 times the current I1. If you want the voltage across a 120Ω resistor then you need to multiply the current I2 by the resistance of a 120Ω resistor giving you (120Ω)I1. Now THAT is a voltage because it is the produce of resistance and current. Similarly, you can't subtract 120I1 from 9 because the first is a current and the second is just a number. Now, 9V-(120Ω)I1 is an operation that you can perform since both terms are voltages.

    Yes, the primary significance of a node is that the voltage is the same everyone along a given node. But the other significance is that KCL applies to the node so that the sum of the currents flowing onto the node must equal the sum of the currents flowing off of the node.

    I have no idea what you mean by Vt=ItRt since you don't define ANY of those three terms anywhere. What is Vt? What is It? What is Rt?
     
  18. teesee

    Thread Starter New Member

    Jan 29, 2015
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    Please ignore that last sentence. I got that totally wrong.

    now leaving town for about a month so would like to thank each of you sincerely for your advice and encouragement. I obviously have a lot of reading and re-reading to do in the mean time. Many thanks.

    Best regards
     
  19. WBahn

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    Mar 31, 2012
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    Good luck to you. Stop back by with any other questions you might have. We can continue with this whenever and to whatever degree you need/want.
     
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