PNP transistor issues

Thread Starter

abcinc

Joined Apr 29, 2009
26
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
 

Thread Starter

abcinc

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

Thanks

Dave
 
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.
 

Thread Starter

abcinc

Joined Apr 29, 2009
26
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

 

Ron H

Joined Apr 14, 2005
7,012
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?
 

Audioguru

Joined Dec 20, 2007
11,251
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.
 

Thread Starter

abcinc

Joined Apr 29, 2009
26
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?
 
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?
 

Ron H

Joined Apr 14, 2005
7,012
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?
Which output do you have connected to the 4.7k base resistor? This is the Range connection.
 

Thread Starter

abcinc

Joined Apr 29, 2009
26
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.
 
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.
 

Ron H

Joined Apr 14, 2005
7,012
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.
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.
 

Thread Starter

abcinc

Joined Apr 29, 2009
26
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?
 
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!
 
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