Help me understand this circuit.. me newbie ==

Discussion in 'General Electronics Chat' started by syaf921, Jan 4, 2013.

  1. syaf921

    Thread Starter New Member

    Dec 21, 2012
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    Could you tell me what is the function of each component in this circuit and how it related to each other. This is an automatic dark sensor circuit.

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  2. SPQR

    Member

    Nov 4, 2011
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    I'll take a shot at this, and then let the experts chime in.

    1. Whenever you see an LED and a resistor in series, the resistor is there to limit current flow so you don't torch the LED when it comes ON.

    2. Whenever you see something connected in series to the collector of an NPN resistor, it is there to "turn on" when there is collector-emitter current (the transistor must be "turned on" before that current flows)

    3. The arrow on the emitter points in the direction of current flow

    4. If you ever see a resistor connected to the base of a transistor, it is probably there to control the base current or voltage, and prepare it for "another signal". When that "other signal" comes, it will turn on the transistor and allow collector emitter current to flow

    5. That LDR1 looks like it might be "another signal". (maybe a sound or light detector?)

    6. A battery gives you "energy" - just like caffeine

    So I'll bet you can now tell us what the circuit does.
     
  3. JMac3108

    Active Member

    Aug 16, 2010
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    LDR1 is a light sensitive resistor. It probably increases its resistance as it gets dark. This turns on the transistor and the LED when it gets dark.
     
  4. ScottWang

    Moderator

    Aug 23, 2012
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    Last edited: Jan 4, 2013
  5. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    100% correct. An LED has little to no internal resistance, so if you connect it directly to a battery with a voltage higher than the LED forward voltage drop, you're theoretically allowing infinite current to pass through the LED. This will destroy it. For that reason, you use an external resistor to limit the current.

    Overall this is correct. When the base voltage reaches a certain level, it will connect, allowing current to pass between the collector and emitter. It acts as a switch, allowing the LED to turn on.

    Close enough, though there's a bit more to it than that :D

    In this case the resistor between the base and +V, along with the LDR, creates a voltage divider bias. This means that as the resistance of the LDR increases or decreases, so does the voltage on the base. Looking at the circuit, I would say that when the LDR has a low resistance, there will be little to no voltage on the base, and the LED will turn off.

    "LDR" stands for "Light Dependent Resistor". The more light it receives, the more resistive it becomes.

    Though this may be true in a sense, I'd watch the terminology a bit. A battery supplies a voltage, which causes current to flow.

    Hope this helps!
    Regards,
    Matt
     
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  6. syaf921

    Thread Starter New Member

    Dec 21, 2012
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    If i change the transistor into darlington pair, will the current increased at the output? The output nodes are at above R2 and below D2.
     
  7. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Changing it to a Darlington Pair will increase the gain of the circuit, which means the value of the current limiting resistor on the LED would have to be recalculated. There really wouldn't be any advantage of doing this. I'd just stick with a single NPN transistor, such as the 2N3904.
     
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  8. Brownout

    Well-Known Member

    Jan 10, 2012
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    Changing to a darlington does nothing, as the transistor operates in cutoff/saturation. Darlingtons aren't meant to operate that way.
     
  9. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Thank you Brownout, I failed to mention that. I was going to expand on my statement about increasing gain, and I guess I forgot :D

    Darlingtons are intended to operate in a linear mode, for amplifying circuits. The amount of gain determines the level of amplification.

    Switch circuits operate, as Brownout said, in the cutoff/saturation mode, meaning switching between fully off and fully on. Gain has absolutely no use in switching circuits. It only has an effect between the cutoff and saturation points.

    EDIT: Looking back, I did not say what I thought I said. My statement about changing the current limiting resistor on the LED was incorrect. Please disregard that part, and only refer to Brownout's post (quoted above) and this one. Thank you :p
     
  10. P-MONKE

    Member

    Mar 14, 2012
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    Sorry to be really pedantic, but surely here the LDR has less resistance as the light increases? Otherwise this circuit would be a "light" detector rather than a "dark" detector.
     
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Many available Darlingtons are intended to be used as switches. ULN2803A is an example. TIP12x is another.
     
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  12. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Hmm, you learn something new every day.... :cool:
     
  13. Brownout

    Well-Known Member

    Jan 10, 2012
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    Those darlington switches operate in the linear region, a disadvantage in power usage.
     
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    It's true that MOSFETs are usually a better choice than Darlingtons for switching. The big advantage of ULN2803A is density. You get 8 switches in a package. There a a few MOSFET arrays available, but I don't know of any with 8 to a package, except for an 8 bit shift register/latch/open drain driver combo.
     
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