Optocoupled Relay LED indicator

Thread Starter

pcmerioles609

Joined Sep 8, 2015
1
Hi, i am trying to drive relays from a parallel port.
I put an optocoupler between the buffer register and the relays
(if double isolation proposes a problem, please tell me)

Optocoupler Relay LED.png

The problem is I want my LED indicator to indicate whether the coil is activated
which is the exact opposite of what my circuit does.

You can see that I attempted to put the indicator at the emitter of the 2N2222
but the relay wont activate so I guess that's not an option.

Any ideas?
Thanks!!
 

gerty

Joined Aug 30, 2007
1,269
I would put the led in series with the base of the transistor, your 330Ω resistor should be about right to give you about 10ma through the led.
 

ian field

Joined Oct 27, 2012
6,539
Hi, i am trying to drive relays from a parallel port.
I put an optocoupler between the buffer register and the relays
(if double isolation proposes a problem, please tell me)

View attachment 91266

The problem is I want my LED indicator to indicate whether the coil is activated
which is the exact opposite of what my circuit does.

You can see that I attempted to put the indicator at the emitter of the 2N2222
but the relay wont activate so I guess that's not an option.

Any ideas?
Thanks!!
The LED in series with the relay & transistor is going to cost you nearly 2V out of Vcc - and then there's the volt drop on its current limit resistor, I assume Vcc is 5V.

The LED probably has a maximum of about 20mA - probably not enough for the relay.

Move the LED complete with its 330R current limiting resistor to in parallel with the relay coil - you still need the back emf diode.
 

MaxHeadRoom

Joined Jul 18, 2013
19,705
Many relay manuf. supply relays with LED indicators across the coil using a series resistor for the LED.
Simply duplicate this.
Max.
 

AnalogKid

Joined Aug 1, 2013
8,251
The upper circuit should do what you want. Does it? The lower one might work if Vcc were 12 V or higher, but will not with Vcc of 5 V.

Also, unless the optocouplers have a current transfer ration (CTR) of greater than 100%, they are not getting you anything in this circuit. However, if you eliminate them and connect the base resistors directly to the LS373 two things will change. First, the logic polarity will change; it will take an output high to turn on the relay. Also, an LS373 can not source as much output current as it can sink. Both things can be remedied by changing the drive transistors from 2N2222 to 2N2907, 2N3906, 2N4403, etc.

ak
 
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