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Darlington pair not working?

HI,I've uploaded a schematic of a circuit I just built and hope everything is clear in it.I need the darlington pair(made using a couple of BC547) to allow flow of current through it to power the relay only when the SPST switch between the base of the first transistor and +6VDC line is closed.But here's the problem-Even before I close the SPST switch the relay's coil get power,as soon as power is served to the circuit.The situation can be compared to directly connecting a relay to a 6vdc source because all the other components in my circuit fail to perform their functions.FYI the LM 7806 regulator gives out a max.current of 1 ampere and I suspect the issue is related with the Darlington pair receiving too much current.I tried adding a couple of resistors to block the excess current,making little difference.I request for your assistance to correct the flaw.Thanks in advance for spending your precious time.

Picture of Darlington pair not working?
I'd guess you've blown the transistors with either the current that the relay takes, or the voltage generated by the release of the relay coil - you absolutely HAVE to have a diode across the coil, as I think I have already pointed out to you.....

Steve
Adarsh_tronix (author)  steveastrouk1 year ago
Will any ordinary diode like In4001 do the job?or is it a Zener diode?
DEPENDS on the size of the relay. What current does it need to pull in ?
Adarsh_tronix (author)  steveastrouk1 year ago
not sure,about 0.1 amps.
Too much. The transistor will fail with the surge current.
Adarsh_tronix (author)  steveastrouk1 year ago
Correction-It takes between 60- 70 mA.Is that within the limit?
Too close to the limit for my taste. It could pull easily more than double at the instant of switch on.
A normal diode. This is to short out the voltage produced by the relay coil when the magnetic field collapses when it is turned off.
Adarsh_tronix (author)  rickharris1 year ago
I believe the phenomenon is called flyback?In a circuit I had kept an LED in parallel to a relay,since both needed the same voltage.Both devices worked perfectly but the LED no longer glowed once I cut the power off and on again.It looked kinda burnt out.Only if I had known of this voltage produced due to magnetic field collapse then...
This back EMF can be quite big 100's of volts with a suitable relay.
Adarsh_tronix (author) 1 year ago
Putting into effect all of your suggestions,I've modified my schematic and uploaded it(with this comment).Do tell me if still any error exist.
new.jpg
Diode is backwards.
Adarsh_tronix (author)  steveastrouk1 year ago
Really?I thought the marked part in a diode(line) is the cathode(Negative) and that seems to be right in the schematic.I don't understand your statement.
Yes, that's right,and you have the marked NON marked end connected to the +ve supply, so the diode is conducting, and shorting out the relay....
Adarsh_tronix (author)  steveastrouk1 year ago
Oh!so our purpose here is to take advantage of a diode's resistance in blocking reverse current.Thank you Steve,you are the best!
No, the diode, when the switch is ON, isn't doing anything. When the switch goes OFF, the voltage across the coil THEN starts to reverse, and the voltage rises rapidly. Now the diode is FORWARD biased, and conducts, THEN the energy in the coil is converted to heat in the forward biased diode
Adarsh_tronix (author)  steveastrouk1 year ago
Alright.Thanks!
framistan1 year ago
I think a 100 ohm resistor is too small and maybe you burnt out the darlington pair. When transistors go BAD.... they almost always become SHORTED. Also, You might want to have a 100K ohm resistor from the BASE of the darlington to GROUND. This will prevent any stray voltage at the base from turning ON the darlington. With the BASE connected to just a resistor, it acts kind of like an ANTENNA. So any magnetic fields from fluorescent lights or other electricity can TRIGGER the darlington to turn ON. So I would put a 10K resistor or maybe a 5k ohm resistor at the base of your circuit, but certainly not a 100 ohm. That might overdrive the base current and burn out the daarlington. As steveastrouk says, you need a diode across the relay to prevent the pulse coming backwards out of the relay-coil which can be about 60 volts! This reverse pulse can also burn out the darlington.
Unlikely, with a BJT, MOSFET, yes.
Adarsh_tronix (author)  framistan1 year ago
Actually I've confined a part of the real circuit to a SPST switch in the schematic.The base meets positive voltage(6VCD) through water and not a switch.I assumed the resistance of water is high enough and an extra 100 ohms would be enough at the base.So,in that case do I still need that 10 k or 5 k?I once tried adding a 470 K resistor between the base and the ground but it still didn't work.I guess I might have already pretty much burnt out the darlington like you mentioned.I'll get a new one and start from the scratch.Also,could you suggest a common npn darlington(I mean the code and the number) suitable for my purpose as it would be better dealing with a single transistor than actually combining two separate transistors to perform the same function.
Adarsh_tronix (author)  Adarsh_tronix1 year ago
Do you think a BC 372 would do the job?It can handle high current(max.1A) and I believe is suitable for my purpose.
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