Finding values

Discussion in 'General Electronics Chat' started by Robin Mitchell, Dec 13, 2010.

  1. Robin Mitchell

    Thread Starter Well-Known Member

    Oct 25, 2009
    734
    200
    Hi everyone,

    Having a few problems

    [​IMG]

    Uploaded with ImageShack.us

    Lets say we have the circuit above, how do i find suitable values for the resistors? (9V supply)

    Do i start by deciding what currents i have in different areas and the voltages and then choose resistor values that give those. And which bits can i treat seperatly?

    Thanks :)
     
  2. iONic

    AAC Fanatic!

    Nov 16, 2007
    1,420
    68
    I think all you would be concerned with is the voltage source, the resistor and the specs of the LED (ex: Vf typical of 3.0V, If typical of 20mA)

    Single LED resistor Calculator
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    You don't need any resistor in series with gate (AND).
    And resistor in base of the BJT is also unnecessary. Honestly you don't need this whole circuit. Simply connect LED + resistor to power supply.
     
  4. Robin Mitchell

    Thread Starter Well-Known Member

    Oct 25, 2009
    734
    200
    it was just an example, obviously i would never use this circuit. But what i want to do was just find values.

    So lets say i want 13mA going in and out. Determine what current i want in the different wires, then select resistors so that it works....is that how you do it?
     
  5. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Only the transistor has any current. Its collector to emitter has the current for the LED and its current-limiting resistor. Its base has up to 1/10th the collector current.

    The inputs of the gate do not have any current (Cmos gate) so its input resistors are not needed and its output current is the low current to the base of the transistor so the gate's output resistor is not needed.

    So the only thing to be calculated is the calue of the resistor that limits the current of the LED.
     
  6. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Many times it doesn't matter, a wide range of resistors will work fine. In that case I lean towards the higher resistances for power efficiency. Other times the constraints are rigid, like with powering an LED. You need to be able to pick between the two.

    That schematic the base resistor might be needed on other configurations. I answered a question two days ago on exactly that issue. The resistors on the gate inputs are not needed. AG has just explained how the base resistance might be selected.

    [​IMG]

    While this is an OR gate the principle is the same. Rotate the diodes 180° and it is a AND gate.
     
  7. Robin Mitchell

    Thread Starter Well-Known Member

    Oct 25, 2009
    734
    200
    OMG!!!
    CMOS gates dont have any current? how does that work?
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    Well, the input current is equal almost 0A thanks to CMOS
     
  9. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    CMOS is a highly efficient logic family. Their typical current draw while in a static mode is in the µA, sometime in the picoAMP range. Only when they are switching do they draw current in any quantity. Besides their wide voltage range this is another reason they are so popular.

    Modern CPUs use CMOS, but they have billions of transistors and switch extremely fast.
     
  10. Robin Mitchell

    Thread Starter Well-Known Member

    Oct 25, 2009
    734
    200
    So messed up XD
    Thanks a milion
     
  11. iONic

    AAC Fanatic!

    Nov 16, 2007
    1,420
    68
    with 9V, a forward voltage of 3V, and a desired forward current of 13mA, a 470 ohm resistor will do.
     
  12. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
  13. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Yes.
    Cmos gates have no input current.
    The gate of a Mosfet is insulated and has no DC current.
    Read all about it in Google.
     
Loading...