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Insufficient Power on my Load

Discussion in 'Power Electronics' started by DarkArrow29, Sep 22, 2018.

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  1. DarkArrow29


    Sep 22, 2018
    Hi! I have this simple circuit that activates a load, whenever the switch is pressed (shorted). When the switch is released from being pressed, the load will power down slowly. I have already put this up on a breadboard. When the load is an LED, it works! But the actual load that I will use is a 2W MP3 Decoder. The load only receives 17.4 mW (based on my schematic diagram below) but before reaching the LM317, the power is 4.11 W. How can I make the load receive at least 2W. Please help! Thank you!

  2. kellys_eye


    Jun 25, 2010
    The load itself determines how much power is drawn - you can't 'force' power into a load.

    When you state a '2W MP3 decoder' what is the actual stated current consumption? Is the '2W' derived from the audio output specification?
  3. Ylli


    Jun 19, 2018
    Posted on at least 3 forums...

    Someone might give you a circuit that would actually work, but what you really need to do is go back and learn how transistors work and the proper way to apply them to a given circuit.

    In this case, closing the switch will immediately blow the b-e junction of the NPN transistor. Even if it didn't and just turn the NPN on, the following PNP is improperly bias to generate any current from the emitter. It most likely would fail with a blown c-b junction.
  4. duke37


    Jan 9, 2011
    I do not see 4W anywhere.
    U1 is a regulator giving a constant voltage output so long as the input voltage is a couple of voltage above the output. You can calculate the output voltage since there will be 1.25V across R5. There will be negligible current going into the ADJ pin.

    There is no positive current to the base of Q4 so no output current. Perhaps Q2 collector is biased the wrong way to supply some base current for Q6.
    Ah, I see that the BC558 is pnp so you have fitted it upside down.
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    Excellent, then it will be a race to see which transistor fails first.

    Even in the current configuration both transistor are at mortal risk, but the BC558 has a somewhat unfair head start if the switch is open.

    Why not remove all the transistors and put the switch in series with the regulator?
  6. davenn

    davenn Moderator

    Sep 5, 2009

    none of which make any sense ! ….

    eg the yellow text box that points up to the V position

    Vrms = 1.34 kV ??/ What The ...? rms refers to an AC voltage
    Vfreq = 15.4 kHz ??? again What The ...?

    I don't know how you got that info but it isn't correct ..

    @Ylli gave good advice
  7. DarkArrow29


    Sep 22, 2018
    Just made a big change, sorry for ignorant use of devices. So my new simulation makes the LED light up at 4.17V. If I will replace the LED with an MP3 Decoder Module with 2W and 3.7 to 5V requirement, how can this circuit activate the said module? See attached file.

    Forgot to explain the 'switch'. So the 'switch' is composed of a bolt and a nut on which when bridged by a finger, the "switch" closes.

    Attached Files:

  8. Bluejets


    Oct 5, 2014
    Knew there was a nut in there some where.........o_O
    davenn likes this.
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    As mentioned previously the emitter and collector of Q3 are drawn back to front

    Once you fix that you will need a resistor in series with the base of Q3 or it will likely expire the first time you use the circuit.
  10. duke37


    Jan 9, 2011
    R2 and R3 are connected in series, one resistor can do the job.
    davenn likes this.
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