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Base resistor from PIR

Discussion in 'General Electronics Discussion' started by KTW, Sep 27, 2016.

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  1. KTW

    KTW

    273
    15
    Feb 22, 2015
    Hi;
    I've made a similar circuit and omitted the base resistor prior to finding this schematic.
    The circuit will operate an LED string around 14 watts, the power source is 2 amps and the transistor that requires the base resistor is a bc547, the voltage is 9 volts DC.
    The output for the PIR is 3.2 volts.
    I've searched for formulas and calculators but my calculations aren't matching up.
    Can someone tell me the basic formula to nail down a base resistor for this project?
    It works fine without the base resistor but I want to do it right.
    Thanks.
    Kevin.

    PIR schematic.JPG
     
  2. KTW

    KTW

    273
    15
    Feb 22, 2015
    Found my answer on Youtube.
    For the benefit of others..
     
  3. Harald Kapp

    Harald Kapp Moderator Moderator

    11,655
    2,696
    Nov 17, 2011
    The circuit in the video differs from your circuit with respect to R1.
    Let's look at it in two ways:
    1. Assuming the transistor is turned fully on (saturated, Vce=0V for simplicity), and further using Vled=2 V (typical value from the datasheet), the voltage across R1 is 9V-2V=7V.
      This in turn means Vbe of the transistor is negative: Vbe = Vbase-Vemitter = 3.3 V - 7 V = -3.7 V. This is a contradiction because with Vbe < 0.7 V (typical value) the transistor is off.
    2. Assuming the transistor is on, this requires Vbe >= 0.7 V.
      With Vbe = 0.7 V and neglecting the voltage drop across R2 (for the sake of simplicity only!), we get V(R1) = 3.3 V - 0.7 V = 2.6 V. This drives a current of I = V/R = 2.6 V / 50 Ω = 52 mA trough R1. Certainly mor ethan the max. allowed 30 mA (datasheet) for the LED.
      The gain of a 2N3904 at 50 mA collector current is hfe >= 60. Base current is therefore Ibase = IC/hfe = 50 mA/60 = 0.8 mA. This currrent creates a voltage drop across R2: V(R2) = 0.8 mA * 100 Ω = 80 mV which justifies my prior neglecting the voltage drop across R2 (see above).
    Imho this circuit is not very good as it operates the transistor in the linear region which causes the transistor to dissipate lots of energy which in turn creates stress for the transistor. It is preferable to put the current limiting resistor R1 into the collector path of the circuit.
    Read these two ressources:
     
    (*steve*) likes this.
  4. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    A 2N3904 is rated at 200mA continuous current. You cannot use it to switch 14W at 9V which is 1.5A.

    Bob
     
  5. KTW

    KTW

    273
    15
    Feb 22, 2015
    Thanks for the responses.
    I have put together a schematic of exactly what I have.
    The 12 volt led strip operates at 9 volts making it slightly dimmer but exactly what I wanted to achieve.
    104 ma is close to half the transistors maximum of 200ma. PIR.LED schematic.JPG
     
  6. Harald Kapp

    Harald Kapp Moderator Moderator

    11,655
    2,696
    Nov 17, 2011
    Not good. Thsi will not work as the LED strip will raise the emitter voltage of Q1 higher than the base voltage, thus shutting Q1 off.:(
    Read the links I gave in post #3:rolleyes:. The LED strip needs to go from the collector of Q1 to the battery's "+" pole.
     
  7. KTW

    KTW

    273
    15
    Feb 22, 2015
    Thanks Harold.
    I'll have another look, it may very well be that way because when I initially hooked it up it didn't work and when I interchanged the wires at the collector and emitter it did work and I remember wondering why?
    Thanks for the explanation.
     
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