aaah.. how to tackle with these resistors....????

Discussion in 'The Projects Forum' started by onlyvinod56, Aug 16, 2013.

  1. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
    362
    1
    Hello....

    Every time when i work with transistors, fets, LED circuits, i am facing problems with current limiting resistors or biasing resistors.

    This time i want to get it clear.
    How to select a proper resistor when using with these components.??

    I am attaching some pictures.

    when using LEDs: for 12v supply R=1K; for 5v supply R=330 ohm.

    When using transistors: from 12v analog IC, base resistor =1K
    from 5V microcontroller, resistor = 330 or 220 ohm

    when using optocouplers: diode side 1K for 12v, 330 for 5v
    transistor side 1K for 12v for any mosfet.

    99 percent i'll get positive results.
    but sometimes not. Here is an example.

    Consider the LDR circuit shown in the attachment. the output leds are responding to the LDR status. but not driving the relay. Tested with 741 and 358 also.

    can anybody give me some suggestions.??
     
  2. bertus

    Administrator

    Apr 5, 2008
    15,645
    2,344
    Hello,

    What happens with the relays?
    Is it always powered?
    What current does the relays need?

    The base current of the transistor should be about 1/10 of the collector current,
    for full saturation of the transistor.

    Bertus
     
  3. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
    362
    1
    Hello bertus,

    The relay is at normally open condition. Whenever the circuit powered up, the contacts get closed. When the LDR status is changed, the LEDs are responding but the relay continues to be in closed position.

    The current rating of the relay is not mentioned. But the coil resistance is 400 ohm. So, the relay takes 30mA. According to your statement, the base requires 1/10th of Ic. So, base requires 3mA. Right..?.

    How can I pump 3mA into the transistor?

    Does the 1/10th logic valid for all the transistors?. If not how can I get that ratio info?
     
  4. bertus

    Administrator

    Apr 5, 2008
    15,645
    2,344
    Hello,

    Do you mean that the relays always activated?
    If that is the case, the voltage on the base does not go low enough.

    Bertus
     
    onlyvinod56 likes this.
  5. wayneh

    Expert

    Sep 9, 2010
    12,089
    3,027
    It's valid, because it's sort of preparing for the worst case scenario. Just about any functioning transistor (BJT, anyway) will be fully on when the base has 10% of the collector current. This will be certain to saturate it and turn it fully on. Often this could be nearly accomplished with far less current, but you cannot count on that in a design. Hence the 10% rule of thumb.

    The base current is calculated with Ohm's law just like you would for a resistor. You choose the resistor by assuming essentially zero ohms from base to emitter. The only wrinkle is that a drop of ~0.7V will occur from base to emitter regardless of current. So with 5V on the base, the ∆V for your ohm's law calculation is 4.3V.
     
    onlyvinod56 likes this.
  6. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    As Bertus stated, the transistor is not turning off.
    Measure the voltage between ground and the op-amp output when the top LED is on (op-amp output low, the transistor should be off). If it's not less than 0.5 volts or so the transistor will begin to turn on.
    In this state the op-amp is sinking about 10ma which is probably pulling it's output up enough to keep the transistor on.

    EDIT: I think the 741 output (with single supply) can't get that close to ground any way.

    A simple fix would be a pull-down resistor on the base and/or a diode or two in series with the base to drop the excess voltage.
     
    Last edited: Aug 16, 2013
    onlyvinod56 likes this.
  7. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
    362
    1
    yes!

    The opamp's output voltage is 1v(with light) and 9.5V(without light)

    Voltage after 1K (@ base of the transistor) is 0.7v(with light) and 0.8v(without light)
     
  8. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
    362
    1
    That means, for 12v, considering Vbe as 0.7, the ΔV = 11.3.
    accoriding to 1/10th thumb rule, the base current required is 3mA. so, the required resistance is 11.3/3m = 3.7K.

    am i correct?
     
  9. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
    362
    1
    its 1V.

    Now I am using 358.

    I"ll try and update. thanks.
     
  10. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
    362
    1
    Hi tubeguy.

    as you said,

    1) i used a pull down resistor.the transistor base voltage is 0.4v(with light) and 0.83v(without light).

    2) used a diode to drop the voltage. the transistor base voltage is 0.63v(with light) and 0.83v(without light).

    Its working.

    Thankyou Tubeguy.
     
  11. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    Glad you got it working!

    Could post a drawing of the working circuit ?
     
  12. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
    362
    1
    YES:)

    sure. check out the attachments.
     
  13. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    Nice drawings.
     
Loading...