What LED do I use?

Thread Starter

mbxs3

Joined Oct 14, 2009
170
Hi,

I recently built out the circuit depicted in the schematic attached. I inadvertently purchased and soldered in an LED that has a forward voltage rating of 12V. While I was certainly not paying attention during many steps of that process, I can't say I would have ended up with the right LED for the circuit had I noticed the 12V rating and immediately known that wasn't correct. Basically I am just trying to control this solid state relay via a digital output and would like to have the LED to indicate that the relay is active. I am limited to a 5VDC relay input and 3.3vdc signal input. Any guidance would be greatly appreciated.

upload_2018-10-27_0-8-1.png
 

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ebeowulf17

Joined Aug 12, 2014
3,307
Hi,

I recently built out the circuit depicted in the schematic attached. I inadvertently purchased and soldered in an LED that has a forward voltage rating of 12V. While I was certainly not paying attention during many steps of that process, I can't say I would have ended up with the right LED for the circuit had I noticed the 12V rating and immediately known that wasn't correct. Basically I am just trying to control this solid state relay via a digital output and would like to have the LED to indicate that the relay is active. I am limited to a 5VDC relay input and 3.3vdc signal input. Any guidance would be greatly appreciated.

View attachment 162502
It looks like an essentially good setup. Replace the LED with a basic red or green one (which means Vf around 1.8-2.2V typically.)

Also, I could be wrong, but I think I'd reduce R1 to 1k, providing more base current and ensuring the transistor saturates.
 

AlbertHall

Joined Jun 4, 2014
12,338
And you need a diode, a 1N4148 would do, across the relay coil, stripey end to relay pin 3.
This will catch the back emf from the coil and stop it damaging the transistor.
 

Thread Starter

mbxs3

Joined Oct 14, 2009
170
It looks like an essentially good setup. Replace the LED with a basic red or green one (which means Vf around 1.8-2.2V typically.)

Also, I could be wrong, but I think I'd reduce R1 to 1k, providing more base current and ensuring the transistor saturates.
Thanks for the info!
 

Thread Starter

mbxs3

Joined Oct 14, 2009
170
It looks like an essentially good setup. Replace the LED with a basic red or green one (which means Vf around 1.8-2.2V typically.)

Also, I could be wrong, but I think I'd reduce R1 to 1k, providing more base current and ensuring the transistor saturates.
I followed your suggestions, the circuit works to activate the relay and illuminate the LED.

When I activate (high) GPIO3_16, the LED illuminates and the relay activates.'
When I deactivate (low) GPIO3_16:
1) If I do not have my "load" connected to the relay output, the LED turns off and the relay output stops supplying 5vdc
2) If I do have my "load" connected to the relay output, the LED turns off but the relay continues to supply 5vdc. (My "load" is the Vcc input on an arduino mega ISP pin)
a) If I remove the wire going from my relay output to the Vcc input on the arduino then my relay deactivates and does not activate again until I take my GPIO input high. I can also deactivate the relay by removing the ground wire on the arduino ISP connector.

Any thoughts?
 

ebeowulf17

Joined Aug 12, 2014
3,307
I followed your suggestions, the circuit works to activate the relay and illuminate the LED.

When I activate (high) GPIO3_16, the LED illuminates and the relay activates.'
When I deactivate (low) GPIO3_16:
1) If I do not have my "load" connected to the relay output, the LED turns off and the relay output stops supplying 5vdc
2) If I do have my "load" connected to the relay output, the LED turns off but the relay continues to supply 5vdc. (My "load" is the Vcc input on an arduino mega ISP pin)
a) If I remove the wire going from my relay output to the Vcc input on the arduino then my relay deactivates and does not activate again until I take my GPIO input high. I can also deactivate the relay by removing the ground wire on the arduino ISP connector.

Any thoughts?
Doh!

I hadn't looked at what you were switching before. I only looked at the control side, because that's where you were having trouble.

That SSR is not suitable for switching DC. It uses a phototriac for isolation. Once a TRIAC is triggered, it's essentially "latched" on until current stops flowing through it. When you're controlling AC, the operation is more or less transparent, because alternating current changes directions regularly, crossing through zero-current along the way. The TRIAC turns off at these zero-crossings. When you try to control DC, there are no zero-crossings, so it never turns off.

You'll need a different SSR which is rated for DC loads, instead of (or in addition to) AC loads. Here's the spec that shows your problem:

9872F279-7E99-4B8F-8AF2-BA02EF4D68CC.jpeg

Sorry to be the bearer of bad news.
 

ebeowulf17

Joined Aug 12, 2014
3,307
Ah well that stinks! Thanks for pointing that out.

Do you think this would be a suitable replacement? https://www.alliedelec.com/m/d/3ade70f3255083b5b8c77de16c708fe1.pdf
Looks promising. I'd go with the 282 model rated for 500mA and 60V - it offers the lowest on resistance.

Beware, it looks like there's no built in current limiting resistor on the input. So, unless I'm misreading something, I think you'll need to add a resistor on the input, either between D1 and pin 3 or between Q1 and pin 4. If I've done the math right, you'd need a resistor of 750 ohms or less. I'd probably go lower, like maybe 470. That would provide a little over 5mA to the opto.
 

Thread Starter

mbxs3

Joined Oct 14, 2009
170
Looks promising. I'd go with the 282 model rated for 500mA and 60V - it offers the lowest on resistance.

Beware, it looks like there's no built in current limiting resistor on the input. So, unless I'm misreading something, I think you'll need to add a resistor on the input, either between D1 and pin 3 or between Q1 and pin 4. If I've done the math right, you'd need a resistor of 750 ohms or less. I'd probably go lower, like maybe 470. That would provide a little over 5mA to the opto.
Thanks for the explanation and guidance! I will give this a go.
 
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