PNP transistor issues

Discussion in 'General Electronics Chat' started by abcinc, Apr 29, 2009.

  1. abcinc

    Thread Starter Member

    Apr 29, 2009
    26
    0
    Hi, I am trying to use a 2n3906 PNP transister to swhich 12 V DC to a relay. The relay has a diode accross it also. The issue I have is when the circuit is turned on, E has 12.28 V DC, C has 3.4 V DC and B has 11.68V DC. This is even if I have B hooked to nothing. I have C hooked to the input coil on the relay. The coil is 500 ohm. The other end of the coil is hooked to ground. Pleas help.

    Thanks
    Dave
     
  2. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    If you connect B to the same point as E, do you still get current through the transistor?
     
  3. abcinc

    Thread Starter Member

    Apr 29, 2009
    26
    0
    Yes, but the relay does not activate. If I connect e to c, then the relay activates.

    Thanks

    Dave
     
  4. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    A simple schematic would help out here to make sure we are talking about the same thing.

    Is your diode that is across the relay coil in the correct way? It should not be conducting unless there is a spike in voltage during the relay switching.

    If you connect e to c you are simply putting the relay coil between 12V and GND so it should activate. Connecting b to e should not activate the relay and the transistor should be off.
     
  5. abcinc

    Thread Starter Member

    Apr 29, 2009
    26
    0
    The doide has the black line on the C connection side. A circuit descripition is below. The relay I have is NTE R56-5D.5-12

    [​IMG]
     
  6. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    Did you verify the pinout of the 2n3906? Not all transistors are pinned out the same.
     
  7. abcinc

    Thread Starter Member

    Apr 29, 2009
    26
    0
    Yes I did with a datasheet.

    Thanks

    dave
     
  8. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    You are trying to drive the base through 4.7Meg. That will not provide enough base current to turn the transistor on.
    Where did you get that schematic?
     
  9. abcinc

    Thread Starter Member

    Apr 29, 2009
    26
    0
    Last edited: Apr 29, 2009
  10. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The jumper from the 4.7k base resistor to the RANGE output of the CD4060 IC provides enough base current for the transistor and stops the oscillator through the diode.

    The transistor is faulty since it leaks current. A base to emitter resistor probably will not fix it.
     
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    A thousand pardons. I missed the Range connection.
     
  12. abcinc

    Thread Starter Member

    Apr 29, 2009
    26
    0
    It is a new transistor. I tried other ones with the same result. I calculated .0026 A to the base through the 4.7meg at 12.24 volts. The transistor needs under 2200 ohm to get over the .005A for the transistor. I put at 780 ohm resistor in and still get the same issue. The 4060B puts out the input voltage at 0.2 A at each pin when latched. Any ideas?
     
  13. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    I was going to second Audioguru's comment on the transistor being bad.

    12.24 volts and 4.7Meg resistance give 0.0000026 A. The 4.7k resistor is between the output of the 4060 chip and the transistor to provide the 0.0026 A.

    If you remove the range connection an put it to +12V, does the transistor turn off, ie. relay in default positioin? If you put the range connection instead on the GND, does the relay go to energized position?

    Maybe a picture of the circuit would help? Is that possible?
     
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    Which output do you have connected to the 4.7k base resistor? This is the Range connection.
     
  15. abcinc

    Thread Starter Member

    Apr 29, 2009
    26
    0
    Pin 13 which gives me a 15 minute timer. Shouldnt thre transistor turn on with 12V on the base side? This is a PNP type.
     
  16. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    The PNP type transistor will turn off if you apply 12V to the base since the emitter is also at 12V. The base needs to be approximately 0.7 volts below the emitter to turn the transistor.

    For an NPN type, the base voltage needs to be approximately 0.7 volts higher than the emitter for the transistor to turn on.
     
  17. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    Just to clarify: In both cases, the current should be limited by a resistor. 0.7V is the approximate base-emitter voltage when current is flowing in the transistor.
     
  18. abcinc

    Thread Starter Member

    Apr 29, 2009
    26
    0
    This needs a transister to latch the Vin of 12.24 volts to the coil on the relay when the 4060B latches the output pin. The diagram shows a pnp transister to do this and that is what I thought a PNP transister will do is turn on when the base get voltage and current to latch the emmiter and collector together. Am I wrong?
     
  19. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    That is correct. The part that I think you are missing is there is only current flows out of the base of the PNP transistor when the 4060B outputs a low value.

    When the output of the 4060B is at 12.24Volts the emitter and the base of the transistor are at the same voltage and therefore no current flows.

    Thanks Ron H for the clarification!
     
  20. abcinc

    Thread Starter Member

    Apr 29, 2009
    26
    0
    OK, how do I get the relay to turn off then?
     
Loading...