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Newby tired of burning component

Discussion in 'General Electronics Discussion' started by bibiz999, Feb 3, 2011.

  1. bibiz999

    bibiz999

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    Feb 2, 2011
    Hi,

    I'm having fun trying to learn electronic on myself but I need help here.
    I've attached part of my schematic !
    All I need is to connect a 12V DC .28A Fan on a 36V CT 72VA transformer without burning it or busting another fuse!
    I have 53V DC on the circuit (due to C1 I assume) so I've divided the voltage in half between R5 (47 Ohms) and R6 (47 Ohms) to 26.5V and around 0.55A (V/Rt = 53 / 94).
    I used a 7812 Positive Voltage Regulator (rated at 35V in and 1.5A) to insured the Fan receive 12V using 0,26A.
    Any reasong why something should eexplode in the desing?? R5 seems to have a hard time surviving my experiment!

    Thanks in advance!

    Hugo...
     

    Attached Files:

  2. davenn

    davenn Moderator

    13,246
    1,745
    Sep 5, 2009
    2 questions for a start :)

    1... what wattage ratings are the 2 x 47 Ohm resistors ? 1/4W, 1W etc
    2... what is on the rest of the cct to the right of the transistor T1, how much current is being drawn by the rest of that circuitry ?

    The resistor is getting hot cuz its dissapating a lot of wattage

    voltage drop across R5 = 26.5V with a current of .55A W = V x A = 26.5 x .55 = 14.575 Watts

    The fan prob doesnt need a regulator to supply it. how about just choosing the appropriate value resistors for R5 and R6 to produce 12V at the centre tap point.
    the fan is likely to be ok with a volt or 2 either side of 12V

    Dave
     
    Last edited: Feb 3, 2011
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,174
    2,690
    Jan 21, 2010
    You don't need great regulation, but you do need the regulator to handle significantly more than the 35 to 40 V they are usually designed to handle.

    Have a look at this. It is simpler, and some advice is given on selecting components.

    Given that your fan is rated at less than 1/3 of the DC input voltage, the regulator will dissipate over twice the power that your fan consumes. The transistor will need to have a heatsink, and will preferably be rated at around 60 to 80V (Vceo) and perhaps 1A.

    You would be better off with a lower input voltage though. Are there any other options? Have you considered a 24 or even 48V fan? (the latter would require far less regulation -- maybe even none at all)
     
  4. Resqueline

    Resqueline

    2,848
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    Jul 31, 2009
    In addition to the above; you have a center tap transformer there. Feeding the regulator from the center tap would do away with the power resistors, just add a small cap..
    Oh, and calculating the exact power loss in R5 is done like this:
    E=53V/2=26.5V Z=47ohms/2=23.5ohms UR5=E+0.28A*Z=26.5V+6.6V=33.1V IR5=33.1V/47ohms=0.7A PR5=33.1V*0.7A=23.3W
    (UR6=19.9V IR6=19.9V/47ohms=0.42A PR6=19.9V*0.42A=8.36W) (total power waste=23.3+8.4+2.2=33.9W) (efficiency=3.36W/37.3W=9%)
     
    Last edited: Feb 4, 2011
  5. bibiz999

    bibiz999

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    0
    Feb 2, 2011
    Thanks for the answer.. I've breach 1 watts and 2 watts resistor on R5 and I now see why!
    The rest of the cct is here and consume nothing except for the LED, but it will feed a TV Flyback for Hight Voltage output (I want 30kV)via a NE555 timer and a NPN transistor.
     

    Attached Files:

  6. bibiz999

    bibiz999

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    0
    Feb 2, 2011
    I need the hight voltage but I will check carefully the link you provides... thanks a lot for the help!
     
  7. bibiz999

    bibiz999

    43
    0
    Feb 2, 2011
    Hi,

    Yes, I suppose I could use the center tap (Yellow/Green) and route a cct (with it's own bridge and all) using only 18V ... the regulator would manage the power isue and I could put the correct resistor value to have around 0.3A... very good idea!
    Regardin the power calculation... I'll check it all!

    Thanks a lot, I may save a resistor or 2!
     
  8. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    You don't need an extra bridge, just route it directly to the regulator input, having the minimum capacitance there (1uF or so). You need to heatsink for 4W of dissipation.
     
  9. bibiz999

    bibiz999

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    0
    Feb 2, 2011
    Ok I suppose it is obvious for everybody except for me but, do you mean I can feed the 7812 with AC current, because the specs seems to indicate DC current and I've seen what a reverse bias 7812 exploding looks like... it's not pretty, it's noisy and it smell bad :rolleyes:
    In this case the - voltage is already clamp by the bridge so it won't reverse but will it works smoothly? Would it be a good idea to add a Zener on the center tap also?

    I'm considering merging your center tap solution with the "Resistor - Zener diode - NPN transistor" solution provided in a link above by steve.
    I don't really need a fan, I'm putting one there to see if I can do it. I found out the best way to learn is by try and error so I'm making a lot of both :)

    Thanks to all, very usefull!
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,174
    2,690
    Jan 21, 2010
    Aaaaah! So your intent is *not* to drive a fan, but to make a 12V power supply and use it (at first) to operate a fan.

    You're probably best off with a bridge rectifier, filter capacitor, and 7812 as you have now (but without the voltage divider).

    All Resqueline is suggesting is that instead of using the two "outside" secondary terminals (3 and 6 in your diagram) and connecting that to your bridge rectifier, use one "outside" terminal and the centre tap (3 and 4, for example). This will halve the AC voltage and bring the resulting DC voltage into an appropriate range for the 7812.

    There is nothing wrong with using what is effectively half the secondary winding and you don't have to do anything special.

    You don't show it on your circuit diagram, but I would assume you currently have terminals 4 and 5 connected together (or they're internally connected).
     
  11. bibiz999

    bibiz999

    43
    0
    Feb 2, 2011
    Well a want to use the cct as a 53V/12V dual power supply (see new cct attached) to drive a High Voltage 30KV output (see cct attached). And yes the center tap are internally connected (didn't fine the exact transformer model in Eagle library).

    I just wanted to connect the fan for better heat dissipation since everything is in the same box.

    I've finnaly use Resqueline center tap solution, so simple! I'll see when everything is connected together but it seems to work just fine!

    Thanks again...:D
     

    Attached Files:

  12. davenn

    davenn Moderator

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    1,745
    Sep 5, 2009
    you cant feed AC into the IC2 the 7812 it will kill it... you have to have a bridge rectifier in there

    the 7812 and anything following it is expecting to see a rectified DC voltage !! :)

    Dave
     
  13. davenn

    davenn Moderator

    13,246
    1,745
    Sep 5, 2009
    so you want to put AC into the 7812 ?? ;)

    Dave
     
  14. Resqueline

    Resqueline

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    Jul 31, 2009
    The center tap of a transformer is not at AC relative to the bridge outputs, it's "always" at exactly half the rectified DC from the bridge.
    If you think of it as a dual (centertapped) +/- supply I'm sure you'll recognize it from somewhere else.
    The 7812 will live quite happily with this solution, but how much of a capacitance is needed to fill in the gaps at zero crossover is open for discussion however.
    Thinking about it 1uF might be too small, 220-330uF might be more appropriate.
     
  15. davenn

    davenn Moderator

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    1,745
    Sep 5, 2009
    I need to learn some thing here ... I dont quite understand :rolleyes:
    you really have me stumped, I cant see how its (the regulator) getting a +DC voltage

    Dave
     
    Last edited: Feb 5, 2011
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,174
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    It's a trick ;-)

    Have a look at how you use a centre tapped transformer with 2 diodes to produce a full wave rectified output that is +ve with respect to the centre tap.

    Then add a similar circuit with the diodes the other way around to produce a full wave rectified output that is negative with respect to the centre tap.

    Then check out the arrangement of the 4 diodes.

    Typically you then have capacitors from +ve to 0V and -ve to 0V. In this case there's a capacitor between +v and -v (which have other labels because all things are relative).

    My concern is that without a significant capacitance (as Resqueline has just suggested) before the regulator that the existing filter capacitor will be relatively ineffective at filtering the new mid-rail.
     
  17. davenn

    davenn Moderator

    13,246
    1,745
    Sep 5, 2009
    yup well familiar with that ... each diode conducts on each half cycle :)

    so that automatically makes the CT a +DC rail with respect to the more -V rail

    interesting ... have never thought of it like that b4, ya know I'm going to have to get an old CT'ed xformer out of the junkbox and experiment now huh hahaha

    OT... MAN ... I'm so sick of this heat Steve !!! this is the 8th day in a row over 35C and its over 40C here again today in inner west Sydney :( brain isnt working too well

    Dave
     
  18. bibiz999

    bibiz999

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    0
    Feb 2, 2011
    Yep it work fine...Daven if you check the above circuit diagram you'll see I already use a full brige rectifier so the current fedding the 7812 is never reverse!

    Thanks to all again for helping a newby like me!
     
  19. bibiz999

    bibiz999

    43
    0
    Feb 2, 2011
    I have successfully blown a 2N3055 NPN TO3 Power Transistor (Base pin and Collector pin now have 0 Ohms resistance)... I'm now re checking everything before puting it back together but I suppose putting the multimeter between the HV plug and the 0V was not a good idea after all! Fortunatly I had a fuse there :) I'll re paste the whole circuit when I'm sure of it and ask for advice then!!!!
     
  20. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Are you saying that you tried to measure current across a voltage source? It's kinda amazing your meter survived.

    Anyway, the transistor tried to do the right thing by sacrificing itself to try to protect the fuse. (That's simply an amusing way of saying that semiconductors can die a lot faster than fuses can blow.)

    If you want to limit the current then you're going to need something to do that, it won't just happen.

    A simple way involves a low value resistor in series with your load that turns on a transistor when the voltage across it hits about 0,6 volts. As the transistor turns on, it shunts current away from the base of the pass transistor leading to a lower voltage output (and thus hopefully a lower current requirement)

    When you post the circuit I'll try to show you how to add current limiting to it.

    Note that current limiting is no panacea. You may have to cope with the entire power being dissipated in your pass transistor (for a dead short). And that's likely to make it get very hot, very quickly.
     
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