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24V 2A Variable Power Supply

Discussion in 'Power Electronics' started by abuhafss, Nov 27, 2014.

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  1. abuhafss

    abuhafss

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    Aug 3, 2010
    Hi

    I found this schematic the other day and assembled using 2N3772 (instead of 2N3055), TIP31 (instead of BD135) and 24V 2A transformer. The transistor is mounted on a heatsink.

    Reg Power Supply.png

    I have two issues:

    a) When powered on with the POT at minimum, the output goes to about 12V. It takes a minute or so to come down to its minimum voltage.

    b) When used for only 5 minutes or so, 2N3772 got hot enough to burn my fingers. Is it normal? Do I need a larger heatsink and/or cooling fan?
     
  2. duke37

    duke37

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    Jan 9, 2011
    What do you mean by used?
    T2 should not get hot unless you have a considerable load current.

    Have you connected the output stage correctly? With 0V on the base of T1, the output should be 0V.
     
  3. abuhafss

    abuhafss

    348
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    Aug 3, 2010
    When operated for only 5 minutes.
     
  4. abuhafss

    abuhafss

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    Aug 3, 2010
    Yes, everything is connected correctly.
    Can T2 be the culprit?
     
  5. davenn

    davenn Moderator

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    show us some photos of your construction, sharp enough that we can read component values
     
    KrisBlueNZ likes this.
  6. abuhafss

    abuhafss

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    Aug 3, 2010
    Will get back soon
     
  7. duke37

    duke37

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    You still havn't said what you mean by used. Was it driving an led at 2mA or charging a battery at 14V 5A?
    Measure the voltages on T2 and measure output current.

    Much more information is required.
     
    Arouse1973 and KrisBlueNZ like this.
  8. abuhafss

    abuhafss

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    Aug 3, 2010
    I just used to supply 15V 100mA to a circuit.
     
  9. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
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    Nov 28, 2011
    So you're drawing 100 mA out of the regulator, which is set for an output voltage of 15V.

    What's the voltage on the collector of T2?

    T2 is being used in its linear mode. You can calculate the amount of power it will dissipate by multiplying the current flowing through it (0.1A) by the voltage across it (that is, the collector voltage minus the emitter voltage). This gives a figure in watts.

    How big is the heatsink? Can you upload a photo?
     
  10. abuhafss

    abuhafss

    348
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    Aug 3, 2010
    To rule out my doubt, I removed T2 and found its base & emitter were short. Luckily I had a 2N3773 in my power transistor box, replaced it and got the circuit working. Here are the photos of the circuit and T2 on the heatsink.

    IMG_20141128_150155.jpg
    IMG_20141128_162036.jpg

    However, there is one annoying issue. When the POT is adjusted to a lower voltage (without any load) the output won't go down until C5 (I assume) gets discharged. Shall I replace it with a smaller capacity, like 10µF?
     
  11. duke37

    duke37

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    Jan 9, 2011
    You will need a much bigger heat sink if you are going to utilise the full capability of the circuit.

    T2 must not have a large negative voltage base/emitter. A higher output voltage than the base is not a good thing. Reducing C5 would allow the output to drop faster but I would not do this, instead I would put a diode in series with the base lead so that only a turn on current is allowed.

    Edit. The maximum permitted reverse b/e voltage is 7V.
     
    Last edited: Nov 28, 2014
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  12. abuhafss

    abuhafss

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    Aug 3, 2010
    Okay, I shall change the heatsink.

    Do you mean connecting a diode with its cathode clamped to the base of T2 and anode to junction of R3 & Emitter of T1?
    Doing that only protects the Vbe of T2 from going negative but, the output voltage would not drop faster without load. How about connecting a 10k resistor across the output?
     
  13. duke37

    duke37

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    Jan 9, 2011
    A rough calculation suggests that T2 could dissipate up to 70W. A BIG heat sink would be required for this.

    The added diode is there to protect the transistor, you do not want to pop off another.

    A 10k load on the output would drop the voltage quicker as you suggest. The time constant will be R * C and the voltage will drop to about a third in this time.
     
  14. abuhafss

    abuhafss

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    Aug 3, 2010
    Please confirm if I understood the correct placement of the added diode.......post #12
     
  15. duke37

    duke37

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    Confirmed.
     
  16. abuhafss

    abuhafss

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    Aug 3, 2010
    Thanks everybody, for their inputs....:)
     
  17. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Yes you can do that, as well as adding a resistor across the output as previously discussed.
    Yes, and the best way to prevent damage is to reverse-connect a diode across the transistor's base-emitter junction, not to connect a diode in series with the base (that will reduce the maximum output voltage unnecessarily). Just add a 1N400x with its anode to T2's emitter and its cathode to T2's base. And I would do the same for T1 to protect it, and add a resistor of around 1k between the wiper of the voltage setting potentiometer and T2's base, to protect the potentiometer!
     
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  18. duke37

    duke37

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    Yes, if a parallel diode is used, then both transistors need to be protected.
     
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  19. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    I always like to have a bleed resistor across the output to ensure the capacitor is discharged. Adding the two diodes is a very good idea but there could be a residual voltage on the output when the diodes stop conducting. It won't be much and maybe of no concern but I thought I would mention it just in case the value of capacitance is changed to a much larger value in the future.
    Adam
     
    KrisBlueNZ likes this.
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