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design darlington transistor interface problem

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azhar2035

New Member
i would like to design the simple IR detector circuit that can interface with relay 6v for my mini project.

my idea is when the IR detector detect the light(infrared), there will be more than 6 V voltage go to the relay out put

but if not detect any, the voltage go to the relay out put is 'zero'

But :confused:.....

As show in attachment,

i wonder, what the good value of resistor should i use (currently 470) for me to able to on the relay 6V

any idea or suggestion?:)
 

Attachments

EN0

Member
i would like to design the simple IR detector circuit that can interface with relay 6v for my mini project.

my idea is when the IR detector detect the light(infrared), there will be more than 6 V voltage go to the relay out put

but if not detect any, the voltage go to the relay out put is 'zero'

But :confused:.....

As show in attachment,

i wonder, what the good value of resistor should i use (currently 470) for me to able to on the relay 6V

any idea or suggestion?:)
Is the relay supposed to act like a switch? The way you have your circuit built is so that the darlington pair is always on. If it is supposed to act like a switch, you should have it above the 470Ω resistor.
 

azhar2035

New Member
as i know

"the small base current IB controls the large collector current Ic"

i would like to make only small current trough IB(current base) will on off the relay

emmmm...

so...this new attachment ok..?
 

Attachments

audioguru

Well-Known Member
Most Helpful Member
You show a photo-diode as the IR detector. It is forward-biased all the time so it conducts all the time. The darlington transistor will be turned on all the time.

An opamp is usually used an an amplifier for a photo-diode. The photo-diode can be with no bias voltage and behaves like a tiny solar cell generating a very small current that is amplified by the opamp. Or the photo-diode can have a reverse bias voltage then it leaks a small current when exposed to IR. the current is amplified by an opamp.
 

microtexan

New Member
Poll voting.

I voted #1 :p

PS how will you block ambient IR from turning on your relay?
#1, how could possibly chose #1. Nobody in there right mind would do that. The only real choice is obviously #7 or is it 4?:D
 

Electronworks

New Member
You want to connect an extra resistor from the base of the left transistor to ground. This will keep the first transistor off until you get sufficient conduction from the IR diode to bias the base on and switch on the Darlington. Experiment with values, but start with 10k and work up.

Not sure if the following will work, but try it - put a 3V Zener in line with the relay coil to drop the extra voltage down to 6V. You must also put a 1N4148 in parallel (but pointing upwards) with the relay coil to damp the back emf that will be generated when the relay de-energises
 

audioguru

Well-Known Member
Most Helpful Member
The reverse-biased photo-diode will leak current when it is exposed to IR light. The current will turn on the darlington transistor when it is high enough. It might never be high enough, which is why an opamp is usually used.
 

audioguru

Well-Known Member
Most Helpful Member
An opamp has about 30 transistors and will work in your circuit.
If you don't want an opamp but you have plenty of trsnsistors then keep adding transistors to your circuit until it is sensitive enough.
 

mneary

New Member
A 3V zener is unnecessary and unwanted for two reasons. First , presuming that 9V is coming from a "9V Battery" it isn't really 9V but is 7.2V after a few hours of use. Second, the Darlington combination shown will drop at least 1V most of the time making no more than 8V available to the relay under the best of conditions and more typically 6.2V (A 3V Zener diode will reduce these figures to 5.0 and 3.2 volts, not enough for a 6V relay).

The 1N4148 that was suggested is a good idea. The 470 resistor isn't needed unless you're using a regulated supply and not a battery. If it's a regulated supply, you can change the voltage to 7V or use 33 ohms instead of 470.

The experiment might "work". Build it and see. If you tell us what it does, and what you would like to improve, we will be happy to help you. (Well, most of us will help. Someone will say you can't use cheap parts from the wrong country but ignore him.)
 

Bob Scott

New Member
You need to know the resistance of the relay coil. Can you measure it? I cannot find a datasheet for the Westech SRD-06VDC-SL-C.

Measure the resistance of the coil and divide by 6. This gives you the coil's number of ohms per volt, or coil current in Amperes.
If the darlington pair has a voltage drop of 1 volt, and the coil needs 6, this leaves 2 volts out of the total of 9V to be dropped by the resistor. So using Ohm's law, the required resistor value should be V/I, or 2/coil current (two divided by the coil current [amps]).

Yes the photo diode is shown upside down. The drawing uses the library item for an LED, but that is minor. ;) (The schematic symbol for a photo diode has the tiny little radiation arrows pointing the other way; towards the diode.)

I agree with mcneary. Use the reversed bias rectifier diode across the relay coil. 1N4001 thru 1N4007, or 1N914, or 1N4148.....

Bob
 
Last edited:

mneary

New Member
I looked all around and found that the apparent "industry standard" coil power for similar relays is 400mW. This would make the current for a nominal 6V coil to be about 67mA, and of course the resistance about 90 ohms.
 
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