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Common Emitter Transistor

Discussion in 'General Electronics Discussion' started by vick5821, Jul 27, 2012.

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  1. vick5821

    vick5821

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    Hey guys,

    How can I modify this circuit so it become a common emitter configuration and able to amplify ?

    [​IMG]

    According to the picture, the transistor is acting like a switch but does not have the amplify purpose.

    So, how can I modify it ?

    Anyway, the operation of this circuit is like this :
    When the LDR detect low light intensity, it will output a HIGH from the comparators which activates the another circuit(Blinking LEDs circuit). When in bright condition, the blinking LEDs wont blink. :)

    Thank you :)
     
    Last edited: Jul 27, 2012
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    What do you expect it to do?
     
  3. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Homework?

    As it is, the transistor will limit the supply voltage to your blinking circuit due to the base-emitter voltage drop.

    Change the NPN to a PNP. You can then easily make a common emitter configuration by making some changes in the way the PNP is connected to Vcc and the blinking circuit

    Since the PPN will be activated by a low voltage on the base, you will have to make another change to the comparator circuit. I leave this to you.

    A hint: Don't stick too much to this
    Remember that there is very often more than one way to go about a problem.

    It should not be too difficult.

    Harald
     
  4. CocaCola

    CocaCola

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    Apr 7, 2012
    On a side note you really should organize that toolbar, you are wasting a lot of screen space having them stacked like that vs just using a few lines and having them go the length of the screen...
     
  5. vick5821

    vick5821

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    Jan 22, 2012
    The 2nd comparator circuit ?
     
  6. vick5821

    vick5821

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    Jan 22, 2012
    which toolbar you mean ?
     
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Your circuit already works and does what you want, doesn't it?

    What are you trying to change? Why? Is this a homework question?

    At present, the transistor is being used as an emitter follower, aka common collector configuration. It amplifies current (the op-amp output can only supply, say, 20 mA maximum, but the transistor can supply a lot more), but it doesn't amplify voltage; in fact some voltage is lost across the base-emitter junction, so the second part of the circuit (the LED circuit) won't get the full supply voltage.

    As Harald said, you can change the transistor so it's operating in common emitter mode, which means you will get nearly the full supply voltage to your LED blinker circuit. You need to make the following changes.

    1. Change the transistor from NPN to PNP with a similar current rating.
    2. Exchange the collector and emitter. So the emitter will be connected to the positive supply, and the collector will feed the LED blinker circuit.
    3. Increase the base resistor (R2) to, say, 4.7 kilohms.
    4. Exchange the inputs of the first comparator.

    Exchanging the comparator inputs means that the comparator output will go LOW in darkness, instead of HIGH. This LOW voltage on R2 causes current to flow from the positive supply, through the emitter-base junction of the transistor, through R2, and into the comparator output, which is roughly at ground potential. This current biases the transistor ON, so it conducts. Then, current flows from the positive supply, through the emitter-collector path of the transistor, and into the LED blinker circuit.

    When the comparator output is LOW, most of the supply voltage appears across R2 (since the base of the transistor will be about 0.7V below VCC and the comparator output is close to 0V), so R2 determines the current that will flow into the base of the transistor. A value of 4K7 will give a base current of about 1 mA. A typical small-signal transistor has a current gain of 200~300 so with 1 mA base current, the transistor's collector-emitter path can supply as much as 200~300 mA. You want to keep the transistor saturated (conducting as hard as it can), so you can't draw that much current, but you can draw somewhat less current - a typical small-signal transistor with 1 mA base current can supply a collector current up to around 100 mA while still in saturation. The LED blinker doesn't use that much current, so you're safe - the transistor will stay saturated.

    Also, you are using the wrong symbol for the LDR. That symbol is for an LED. An LDR is shown as a RESISTOR, with arrows pointing TOWARDS it.
     
    Last edited: Jul 27, 2012
  8. CocaCola

    CocaCola

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    Apr 7, 2012
    Instead of this, note green highlighted area...
    [​IMG]

    Do something like this...
    [​IMG]
     

    Attached Files:

  9. vick5821

    vick5821

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    Jan 22, 2012
    Okay sure :) Thanks for the advice ^^
     
  10. BobK

    BobK

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    Switching to a PNP will not do the trick without other changes.

    The LM325 can only ouput as high as V+ - 1.5V, so the PNP would be on even when the opamp went high.

    I would switch the LM324 to a comparator with open collector output and a pullup to V+.

    Bob
     
  11. vick5821

    vick5821

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    Jan 22, 2012
    How was it ?
     
  12. BobK

    BobK

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    The max output voltage of the LM324 will be V+ - 1.5V, so in this case 3.5V. So you have the emitter of the PNP at 5V and the base at 3.5V, which is enough to turn the transistor on. There is no way to turn it off when the base is connected to the output of the LM324, the voltage at the base must be > about 4.5V to turn it off.

    Bob
     
  13. vick5821

    vick5821

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    Jan 22, 2012
    Then how can I make that the transistor acting like a switch for second circuit and at the same time, the blinking circuit can get the full 5V ?
     
  14. BobK

    BobK

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    1. Use a P channel MOSFET with at gate threshold around 2 or 3V
    2. Use a PNP and change to an open-collector comparator + pull up
    3. Use and NPN and switch ground instead of V+
    4. Use a PNP and use and NPN as a level shifter to drive it's base.

    Bob
     
  15. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    BobK, I think the LM324 output will swing quite close to the positive rail when the load is towards VCC. The specification for pulling to VCC-1.5V applies when the load current is in the other direction (it is specified with a resistive load to GND in the data sheet). But you would have to test it by experiment, and it could vary from one manufacturer to the next; this information isn't given in the data sheet.

    Yes, a P-channel MOSFET would be better than a transistor from a component count point of view. Also you would avoid wasting base current. The other option would be to add a base-emitter resistor on the transistor. I should have suggested that originally. But a P-channel MOSFET is a cleaner answer.

    vick, you can use my previous suggestion - you may need to add a second 4K7 resistor between the base and emitter of the PNP, or you can use a P-channel MOSFET instead of the transistor, with no resistors needed. Either way will give you the full supply voltage for your LED blinker.
     
  16. TedA

    TedA

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    Sep 26, 2011
    vick5821,

    There should be some fairly simple ways to improve your circuit.

    First, I think there's a typo in the connections between the output of U2A and the non-inverting input. There should be a resistor between these pins. The drawing seems to show a direct connection.

    The circuit might oscillate, as drawn, or might not. And the frequency would likely be much lower than intended, as well as unstable.

    As drawn, R4, R8 and R10 don't do much, as they all just inject some very small current into the output pin.

    Another possible problem with the circuit as-drawn, is that LED1 will have a much higher drive current, or voltage, or both, than LED2 will get. The different values for R5 and R9 increase the difference in drive caused by the fact that the LM324 output pulls down better than it pulls up. Is this indeed what you want?

    BTW, it is good practice to avoid making dots necessary to interpret connections on a schematic, where lines cross. Better to make all lines that cross ones that do not connect. The dots on this schematic are too darn small, anyway.

    I think you would want to use a single LM324 powered off the full 5V power supply, then use logic level or high level switching in the blinker part of the circuit to make both LEDs go off when light is detected. Don't power one LM324 from an output controlled by another one.

    There are several ways the light detector stage can control the LEDs.

    It could drive a couple of transistors that switch off the power to the LEDs.

    It could drive inputs on other sections of the same LM324, forcing the outputs to the LED off states.

    Are the other op-amp sections in the LM324s available for use?

    If the rest of your LM324 is available, two sections can be used as LED drivers. Each LED can then be driven when its own output goes low. The oscillator will be more predictable, since the load on the output can be much smaller. And the light detector circuit can force the driver sections off when light is detected. This will just take a handful of resistors. And you have eliminated one LM324 and all transistors.

    Ted
     
  17. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    The problem of complete turn off in PNP high side switching arises often. Diodes and or LEDs can be used to insure that Vbe goes below the base threshold voltage. A resistor between the base and emitter can also be added to further insure collector current turn off.
    Since this sounds like home work no values are included.
     

    Attached Files:

  18. vick5821

    vick5821

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    Jan 22, 2012
    Using one LM 324 ? and used the two comparator inside ?
     
  19. TedA

    TedA

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    Sep 26, 2011
    vick5821,

    Each LM324 package contains four independent op-amps.

    These can be used as comparators, but are optimized for the op-amp role. They are stable with 100% negative feedback.

    Your circuit could be built with actual comparators, such as those found in the LM339. Note that many comparators have open collector outputs, so pull up resistors may have to be added to a circuit originally intended for the LM324.

    Ted
     
  20. vick5821

    vick5821

    700
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    Jan 22, 2012
    [​IMG]

    I am trying to connect like in this example. Using two transsitor. NPN and PNP type transistor .

    I think it should work ?
     
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