Darlington pair 10A 12VDC transistor spec?

Discussion in 'General Electronics Chat' started by hobbybuilder, Oct 26, 2007.

  1. hobbybuilder

    Thread Starter Member

    Mar 24, 2007
    16
    0
    I attach a circuit diagram.

    darlington6.jpg

    I'd like to know, can one simply replace TIP31C with 2N3055 and expect this to work?

    Also, regarding the diode, is there such a thing as a 10A diode? Can I take 10 1A diodes in parallel as a substitude?

    Reference to a resource where one can look this sort of thing up would be much appreciated.
     
  2. lightingman

    Senior Member

    Apr 19, 2007
    374
    22
    Hi,This is one inefficient way of controling motor speed....Rite, although the 2N3055 has a rated Ic of 15 amps, it would be pushing it to control that 10 amp motor......For continus opperation from this circuit, you wolud need a couple (maybe 3) 2N3055's in parallel, with an emiter resistor (0.2 ohms) on each device, along with a LARGE heatsink.....At slower speeds, the transistors will dissipate a lot of heat.... If you want to control a motor of this size, you would be better of looking into PWM (pulse width modulation) control, as in this type of control the output device is on or off, therefore keeping dissipation down to a minimum....Daniel.
     
  3. Gadget

    Distinguished Member

    Jan 10, 2006
    613
    0
    not sure of the specs of the driver transistor, but it must be capable of supplying the extra base current required for the 3055's as well.

    That 1N4148 will be toast as well
     
  4. monu

    New Member

    Oct 26, 2007
    1
    0
    what is the wattage of1K&10K
     
  5. lightingman

    Senior Member

    Apr 19, 2007
    374
    22
    Yes that's rite.... You will need to drive the 3055's with a TIP41, and then drive that with a BFY51.... Don't foreget the voltage drop through the devices....Diodes in parallel...."not a good idea" they will never be matched exactly, and thay will have very slightly differing forward voltages, causing extra dissipation from some of the diodes...Daniel.
     
  6. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    12V / 10,000\Omega = 1.2mA. 12V * 1.2mA = 14.4mW

    Hobbybuilder, have you thought about using pulse width modulation (PWM)?
     
  7. Raam

    New Member

    Apr 22, 2008
    6
    0
    Diodes are typically rated to peak inverse voltages rather than current(10A diode). However, u need to look at the peak power the diode can dissipate- (current squared times the bulk resistance of the diode).Say, if the bulk resistance of the diode is 7 ohm,then for 10A, power diss = 10*10*7= 700W. A peek into the datasheet of the diode wud give u the above rating. If the diode icannot handle that power, it is likely to break down.
     
  8. Caveman

    Active Member

    Apr 15, 2008
    471
    0

    Actually, it's way worse than that. As one diode receives more current, its voltage drop will reduce which will cause it to pull more of the current. The other one will just sit there pulling less and less current until the first blows. Then the second one is all alone, and will blow soon after.
     
  9. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The max saturation voltage loss of a 2N3055 transistor when it conducts 10A is 3.0V when its base current is a whopping 3.3A. Then the TIP41 driver transistor has a max saturation voltage loss of about 1V when its base current is 300mA, then the BFY51 pre-driver transistor has a max saturation voltage loss of 1V when its base current is 20ma.
    You still need an additional transistor to supply the 20mA.

    The max voltage to the motor will be only a few volts.
     
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