Who can explain this page in more detail to me?

Discussion in 'General Electronics Chat' started by gte, Sep 18, 2009.

  1. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    This page says that we can calculate the voltage drop across multiple resistors in series with each other, by using a total current calculation figure, but the total current calculation figure seems like you just pick and arbitrary number, so I'm not sure I understand.


    If you wanted to offer a different example, that applies to a circuit I am currently working on, I am trying to calculate the voltage drop across different resistors on a 5vdc regulated circuit. Does the math work out to (with a 20k total resistance) you'd lose .250vdc for each 1k of resistance?



    http://www.allaboutcircuits.com/vol_1/chpt_5/2.html



     
  2. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
    230
    You've answered your own question within your post, so I'm not sure if you still have a question or you've simply stated a learnig path.
     
  3. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    Ok, I guess I simply stated a learning path, haha :)

    Scenario - I have a 5v regulator and need to output 1.75v and 1.5v . I've been using potentiometers, but they are getting wiped somehow and not offering any resistance.

    So If I wanted to use less than (20) 1k resistors in series, I could use a 12k resistor to drop to 2 vdc and then to step it down to 1.75vdc and 1.5vdc, using (2) 1k resistors respectively?

    How do I know what wattage to use, for this?




     
  4. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
    230
    say because .25 volts is your smallest increment, then the 5v supply / .25 = 20. I suspect thats how you came out with the 20 1k resistors.

    So starting from common (or ground as some would say), working back towards the supply.

    1.5 volts require 1.5/.25 = 6 1Kohm. Because your next level is .25 volts higher, we'll add another 1K in series to obtain that voltage. Your next step is the 5v supply, which is a difference of (5-1.75)/.25 = 13 1kohm. So now add them up and you get your 20 1K resistors.

    Calculating this example, the resistors in series is a total of 20Kohm. At the supply of 5v that will deliver 5/20K = .25mA, so a total of .25mA*1K = .25v drop across each resistor.

    Wattage = VA, so your 13Kohm resistor drops 3.25v at .25mA, your dispatted wattage being .8 milliwatt.

    So now you have your refernce voltages, but they are only that. Once you add any load, you will need to recalculate based on the effects that the load contribute to the circuit.
     
    Last edited: Sep 18, 2009
  5. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    Hi,

    Thanks for the reply, so am I understanding you correctly, per my diagram?

    (Keeping the multimeters negative probe on the negative side of the battery and moving the positive probe to the different locations in between the resistors that I have marked?)


    [​IMG]




     
  6. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    I guess I did something wrong, because when I tried to do this in practical application, it does not work out to even close to the right voltages?
     
  7. bertus

    Administrator

    Apr 5, 2008
    15,641
    2,344
    Hello,

    What kind of multimeter did you use?
    The total current is only 0.25 mA.
    Any load will lead to a misreading of the voltages.

    Greetings,
    Bertus
     
  8. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    I used an auto ranging digital multimeter, why, should I use something different?


     
  9. bertus

    Administrator

    Apr 5, 2008
    15,641
    2,344
    Hello,

    Do you know the input resistance of your meter?
    This resistance is parallel to the resistors when you measure the voltage.

    Greetings,
    Bertus
     
  10. Wendy

    Moderator

    Mar 24, 2008
    20,764
    2,534
    Most decent DVMs have a 10MΩ input impedance. VOMs are considerably lower, around 100KΩ is typical (or less).
     
  11. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Question - how close is "not even close"? What is the tolerance of those resistors? How close to 5 VDC is your source - how many decimal places does your meter give? What are all the actual voltages measured?
     
    Last edited: Sep 19, 2009
  12. Wendy

    Moderator

    Mar 24, 2008
    20,764
    2,534
    Which leads me to another thought. If you are touching the meter leads your body's resistance could be pretty low, throwing off the readings. People in general are pretty variable as far as their internal resistance goes.
     
  13. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    Can't I calibrate my DMM and be ok as far as resistance from the leads and wire?

    My voltage values do not have to be exact, but I believe my hurdle is understanding different ways I could have 1.5v and 1.75v and 2.0v in my arrangement of resistors?

    Do I have to have (20) 1k resistors, or could I have (1) 10k resistor and (10) 1k resistors? And in the second scenario, does the large resistor go closer to the positive voltage or ground side of the resistor arrangement?
     
  14. Wendy

    Moderator

    Mar 24, 2008
    20,764
    2,534
    The DVM input resistance is fixed, defined by design. Most cases this is 10MΩ, which is pretty invisable for resistance measurements. Since we don't know the meter you're using we can only speculate, but it is an educated guess.

    I haven't been following the rest of this thread too closely, so I'll let someone else pick up the slack.

    I will say this, figure total resisance, then resisance to the common point, and you can calculate your voltage pretty simply.
     
  15. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    I'm using an inexpensive radio shack multimeter

    I just went out and bought (20) 1k resistors, and I'm able to get close to the voltage resolution I need, but I'm sure this is an inefficient way of doing this, and I could use larger value resistors in place of some of the resitors (to replace the areas that I don't need to any voltage value from) to get the same resolution ... but I don't know how to calculate for this?




    [​IMG]
     
    Last edited: Sep 19, 2009
  16. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    Ok,

    This seems to easy, but maybe this is correct

    Take the dc voltage and divide it by the number of total resistance and you have your resolution.

    Then find the window of that voltage that you want and you can use large resistor values until you get down do the voltage window that you need small increments in?

    (E/R) * % of total voltage that window resolution is

    ??
     
  17. Wendy

    Moderator

    Mar 24, 2008
    20,764
    2,534
    There are several problems in previous posts to choose from. Which one are you referring to? Or are you trying another example?
     
  18. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Let's go back to your:
    First, let's decide on a range of current that you need in this scenario. Let's just say you simply needed the voltages present, and you weren't going to power anything. Somewhere around 1mA should be enough.

    So the question now is, how much resistance is needed to limit current to 1mA with a 5v source?
    R = E/I, or Resistance(Ohms) = Voltage/Current(Amperes)
    Since E=5v, and I=1mA = 0.001 Amperes...
    R = 5,000 Ohms.

    So to get 1.5v, we know that R = E/I, or Resistance(Ohms) = Voltage / Current(Amperes).
    Since E=1.5v and I = 1mA, R= 1.5/1mA = 1.5/0.001 = 1,500 Ohms (1.5k)

    Now you need 1.75v - but that's just 0.25v higher than the 1.5v you already have.
    So to go 0.25 higher than the 1.5v to get the 1.75v, you'll use R=E/I again.
    R = 0.25v/1mA = 250 Ohms.

    We're up to 1.75v, and have to drop the rest of the 5v. 5v-1.75v=3.25v.
    R = 3.25v/1mA = 3250 Ohms.

    Another way to look at it would be since 5v/5,000 Ohms = 1mA, and 1mV/1 Ohm = 1mA, for each Ohm resistance there is 1mV dropped in this particular circuit that has a total resistance of 5,000 Ohms.
     
  19. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    I'm trying to see if I understand the calculations correctly by presenting a different example.

    I was able to raise my resolution by raising my resistance to 40k, and then the 5v signal was decreasing by .125 for each 1k of resistance. Since I wanted to key off of the voltage values of 2v, 1.75v and 1.5v, I used (2) 10k resistors close to the voltage regulator, then (10) 1k resistors in series following that and then (1) more 10k resistor after that to finalize the 40k of total resistance. This gave me 2vdc @ 24k of resistance, 1.75vdc at 26k of resistance and 1.5vdc at 28k of resistance.

    If I had a 24k resistor and a 12k resistor, I'd only have to use (6) total resistors, but unfortunately I don't have those values, so I had to use more resistors in series.

    Am I understanding what I did correctly?



     
  20. Wendy

    Moderator

    Mar 24, 2008
    20,764
    2,534
    You had a schematic for the previous example. Schematics are the language of electronics, they have no ambiguity. So lets start from this example:

    [​IMG]

    You tell us the values of Vcc and R1-6. We'll show you how to solve for them.
     
Loading...