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SSR relay blown or low amps ?

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harps

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hello : )

i receieved a nice little relay in the post today, im sure if i blew it up (due to misleading websitedata) or have a current problem??

on the website i bought it from,(see link below) the data section says input control voltage 1.7v

so i thought approx 1.7v would be fine. ( i think i may have given it 1.85v at max)

anyway the relay did not work , i thought oops : i then download the official data sheet and it says :

Input Forward VF (volts)
min1.1
typ1.3
max 1.65

so i now think,oh boy, did i over cook it?

the reason i send this message on the board is that i am not an expert and the way i get my voltage is via a 2.5v regulator and i drop the volts using approx 5 M ohm resistor. i am not near my breadboard to give the exact ohms rating which i used but via my multi meter the voltage i had when it was not connected to the relay was about 1.75v. Is this wise ? Do you think my amps could be too low due to the large M ohms i use?

anyway i have reordered the same relay again and shall try 1.3volts instead, but that means i will need much more resistance maybe 7>8M ohms to get 1.3v??

any thoughts would be great

here is the relay, it is a dual/ double relay in one package. Avago Technologies | Relays, Switches and Indicators | Relays | PCB Solid State Relays (SSR) | PCB Mounting SIL, DIL & SMT |ASSR-1420-002E
 
No, you need to limit the current through the input LED.

What a stupid data sheet! The input device is CURENT OPERATED; not VOLTAGE OPERATED. The current required to turn on the LED is not specified, other than to say it is "LOW".

Try an input control voltage of 3 to 5V, in series with a 100 Ohm resistor.
 
oh no , i really am confused now.

but the data sheet says control voltage 1.65v max?

i have a 5 volt regulator at hand i could try (7805 i think?) and a 100ohm resistor.

i always thought a component drew the current it needed from a psu so there was no need to worry, just need to get the voltage right?
 
Last edited:
The relay will work on any voltage from 3v to 12v.
Here is how to do it:
For 3v operation, add a 68R to 220R resistor to pin 1. Connect pin 2 to 0v rail and the other end of the resistor to 3v.
For 6v operation, add a 220R to 560R resistor to pin1 etc
For12v operation add a 680R to 2k2 pin 1 etc.
 
wow ,

o.k, i just re-tried my 2.5 regulator, but i took out the Mega ohms of resistance and first tried a small 220 ohm resistor!

It works!! : )

my multi meter still reads about 2.5 volts so i am way over the maximum voltage!


so i have no idea why it has not blown or quite what is going on!

i think i shall buy a 1.5 regulator and use that instead, just incase

thanks mike, i shall see if i can cancel this re-order now ( fingers crossed)

: )
 
The input device is an LED. As with any LED, you "turn it on" (make it emit light) by applying the rated current (typically ~20mA for a small LED) to it (from a voltage source in series with a resistor), and it will have a forward voltage drop of about 1.65V (measured across just the LED). The opto-coupler is no different. The guy that wrote the data sheet should be fired!
 
opto

Just to make sure...... Will your circuit look like this?
 

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Oops

Don't know where that schematic came from.
 

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Don't know where that schematic came from.

Why bother with the regulator? If you have 9V, and you want to turn on the LED, and if the Forward Drop of the LED is 1.65V, then just use a resistor = E/I = (9-1.65)/0.02 = 367Ω (use a standard value of 360Ω).
 
thanks everyone its working fine now.

hi ronv , i do electroincs by trial & error, as i was never taught well at school, so unfortunately i can not make head nor tails of the diagram you posted. but hey it all works now, i use a 2.5v regulator, with capacitors to make sure the power is smooth and a 220 ohm resistor. the relay switches very quick , it works fine. : P
 
Mike,

On the Web page:
"Control Voltage 1.7 V"
"Maximum Input Current 25 mA "

In the first paragraph of the datasheet:
"The relay turns on (contact closes) with a minimum input current of 3mA through the input LED. The relay turns of (contact opens) with
an input voltage of 0.8V or less."

And yes, it is poorly written.
 
... do electroincs by trial & error, as i was never taught well at school, so unfortunately i can not make head nor tails of the diagram you posted
Being able to read a schematic is fundamental to working with electronics. You might look at one of these articles.
 
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