Basic Circuit Questions (Ohms Law)

Discussion in 'General Electronics Chat' started by beeson76, Dec 17, 2010.

  1. beeson76

    Thread Starter Member

    Apr 19, 2010
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    Here is just a basic question that it really isn't covered in the stuff I am reading.

    When your designing circuits, basic or complex, we should always design it by using Ohms law right. We also use ohms law in detemining individual components information, such as voltage etc.

    But how do the 2 relate?

    For example suppose you have a basic circuit with more than one component such as 2 LEDs and a current limiting resister. You have a basic power supply of 10 volts. Each LED takes volts 20 mA. From this I determine that I need a 200 ohm resister. But from my 2 LEDs I determine that each one has a resistance value of 100 ohm. So my circuit has a total resistance value of 400 ohms. Is that correct? More questions will follow if I can determine that this is correct.

    Thanks for any help. This problem has been bugging me for a while.
     
  2. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    Use the forward voltage drop rating of the LED in question. Example: Vf = 3.5V, so two LED's in series will drop 7.0 volts. Your supply is 10 volts, and 7 volts are used on the LED's. You have 3 volts to work with. A 100 Ohm resistor would let 30 mA flow, so that would be to much current.

    You have 3 volts and choose a 160 Ohm resistor. This allow 18.75 mA to flow through the circuit. Max for the LED's is usually 20 mA. Check the data sheet and see what yours are, then use that as the target.

    Do you follow the math needed to do this?
     
  3. beeson76

    Thread Starter Member

    Apr 19, 2010
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    Thanks Kermit2

    I was using figures in a more general use, but here are the exact specs for the LED's in question. Forward Voltage: 1.9 Forward Current:20 mA.

    So from my 10 volt power supply, I have 6.2 volts to work with. So this is where I am having my problem. For the specific circuit I take 6.2/.02 to get the resister value. (10-3.8)/.02 That would give me a resistor value of 310 correct.

    So from that I have for each LED 95 ohms. I get this simply from applying Ohms Law R= V/I. So the total resistance of my circuit would be 500 ohms correct?
     
  4. #12

    Expert

    Nov 30, 2010
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    That's one way to look at it, but as soon as temperature or supply voltage change, the equasion becomes useless. LEDs are "current driven". They do not have linear voltage changes as the current changes. Trying to call them a resistor has a very small usefulness.
     
  5. iONic

    AAC Fanatic!

    Nov 16, 2007
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    I would discontinue worrying about the theoretical resistance of the LED's and stick with the resistor you need to consume 3.8V

    If you are working with a Vf of 3.1V/LED (3.1V x 2 = 6.2V) and a If of 20mA you would need to drop 3.8V across an unknown resistance, but with a known current. Now use Ohms Law. 3.8V/.02A = 190 Ohms.
     
  6. beeson76

    Thread Starter Member

    Apr 19, 2010
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    I will just ask the question I am leading up. This circuit that we have been discussing:

    If there are 10 volts coming from the power supply, and total resistance is 500 ohms (2 LED's and 1 resistor), can I use Ohms Law to figure out how much mA I need to power this circuit by knowing Voltage of Power Supply and total resistance based on the totals for the Circuit instead of individual components? or Do I have to figure totals for each component?
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    Rlimit >= (Vsupply - (Vf_LED * Number_of_LEDs_in_series)) / Desired_Current
    Rlimit >= (10v - (1.9v * 2)) / 20mA
    Rlimit >= (10 - 3.8) / 0.02
    Rlimit >= 6.2 / 0.02
    Rlimit >= 310 Ohms

    A decade table of standard EIA resistance values resides here:
    http://www.logwell.com/tech/components/resistor_values.html
    Bookmark that page.
    Look at the E24 values (green columns), as that's what's commonly available.

    You'll see 300 and 330 Ohms as being the closest values to 310 Ohms.

    6.2v / 300 Ohms ~= 20.7mA ; a bit much.
    6.2v / 330 Ohms ~= 18.8mA ; comfortably under the maximum specifications at 94% rated current.

    Then you need to determine the power requirement for the resistor Rlimit.
    P=EI, so 6.2v * 19mA = 117.8mW. Double this for reliability's sake, 235.6mW. You can use a 1/4 Watt resistor.
     
  8. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    The leds are current dependend.
    To calculate the series resistor you use the following procedure:
    Rseries = (Vsupply - Vled) / Iled.
    Vled can be multiple leds.
    Example: Vsupply = 12 Volts, Vled = 2 X 1.9 Volts (2 X red leds), Iled = 20 mA, you will get:
    Rseries = (12 - 2X1.9) / 0.02 = 8.2 / 0.02 = 410 Ohms.
    The first higher value E12 series resistor is 470 Ohm.
    The current will be about 8.2 / 470 = 17.45 mA.

    Bertus
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    Bertus, your math appears to be a bit off. It seems to be the same kind of math that our politicians use.
     
  10. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    @ SgtWookie,
    I just gave the 12 Volts example, for the basics.
    For 10 Volts you are correct.

    Bertus
     
  11. beeson76

    Thread Starter Member

    Apr 19, 2010
    185
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    Thanks guys for the replies. I now understand clearly the concept of how to figure resistance values for components.

    I finally found the diagram that I was looking for when I read it. It is in Simple Series Circuits in the Electronics Book provided for this forum. Vol 1. DC. The box there is the box I am referring to about Totals for the entire circuit. I made a box similar to this and am filling it in for my simple circuit and just had some questions about it.

    I guess this may help come from a different angle.

    I will fill out the box and post it and see if you guys thinks it looks ok. I will add some components to it also.

    Thanks.
     
  12. beeson76

    Thread Starter Member

    Apr 19, 2010
    185
    1
    Here is my chart for my simple circuit. Does everything look right. Especially look in the Totals Column at the far right. I left the Amp box empty but I think it should be 20 mA (9/450).

    This will help a lot when I get your responses.

    Thanks.
     
  13. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    You can not see the leds as plain resistors.
    The "resistance" of the led will change with the applied current.

    Bertus
     
  14. beeson76

    Thread Starter Member

    Apr 19, 2010
    185
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    Oh wow. So I cannot use this chart for planning a circuit. Im glad we are having this discussion then.

    So if the circuit stays the same, as far a components should current change at all? My power supply is regulated. It is 9 volt with .660mA. So I should always go with the component that has the highest mA as far as planning how much my power supply should supply as far as current. For example my LED needs 20mA but if I have a component that needs 40mA. I go with at least a 40 mA power supply and just control current with resistors to anything else that is lower?
     
  15. iONic

    AAC Fanatic!

    Nov 16, 2007
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    I think that using your method will get you into trouble. It will not hold true in every circuit you apply this method to.

    The method offered (above) and several online calculators all point to 190 - 220 ohms, for 20mA and Vf of 3.1V/LED. It really comes down to what the typical, and max forward voltage and current is. Many LED's I've been seeing the typical is 20mA and the max current is 30mA. some voltages can be between 2.9V - 3.6V. It all depends on determining the safe numbers to start with. Ideally, be safe and under estimate the current a bit as you will never see the difference visually.

    Online Calcs.
    ------------
    Linear
    Quickar
    ledcalculator/

    I'm not sure, but I believe your power supply is 660mA and not 660mA? Yes?
    You power supply must be rated greater than both the Voltage and current needed for the circuit. Your circuit could probably run fie with a 6.5V, 30mA supply. Of course it would be difficult to find one with those specs. A 9V supply with a current rating of 100mA - 500mA would be easier to find for a good price.
     
    Last edited: Dec 17, 2010
  16. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Yes, In general that is correct.
     
  17. beeson76

    Thread Starter Member

    Apr 19, 2010
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    So in this circuit, the only thing I need to worry about is:

    1. Find the correct value for the resistor.
    2. Make sure the voltage is sufficient for the circuit.
    3. Make sure the current (in this case the mA) is more current than the highest rated component.

    Other than that I shouldn't worry about the "specs" for the whole circuit such as doing Ohms Law to find the voltage, amps, and resistence.
     
  18. iONic

    AAC Fanatic!

    Nov 16, 2007
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    You would also need to determine that the power requirements of the Resistor chosen are sufficient. (IsqR), (.02 x.02)/R.
     
  19. beeson76

    Thread Starter Member

    Apr 19, 2010
    185
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    Thanks iONic. Could you give more details. When you refer to Power Requirements, are you referring to the Watts. I see that there are several different wattages available starting at 1/4 watt and going up to 1 watt. What does IsqR stand for and is that the formula after that (.02x.02)/R.

    Thanks.
     
  20. Wendy

    Moderator

    Mar 24, 2008
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    LEDs and diodes (and many other components) drop a constant voltage. In this case Ohm's Law does not apply. Instead, you use Ohm's law outside the constant voltages to set up a predicted current. Ohm's Law is used to set what is basically a quantum effect.
     
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