Hi all,
I plan to control a Solid State Relay RA4850D12- see datasheet , with 15 V dc.




My question is:

RA4850D12 datasheet: 40 mA at 32 V dc So R '' inside '' SSR 800 Ohm?
Control voltage 3 32 Vdc

I thought of the usual way for Led devices, Vf, If max, and not found in this data...

According to 10b-SSR application datasheet, they add - for their products - Rx with Vf Led, and max Input current.

So say 6 V dc for the SSR, and 10 mA for the external Led.

Trying to find RX value: 15 - 2.4 - 6 / ( 10 mA) = 660 Ohm ?????

Or with '' Inside SSR Resistance " of 800 Ohm:

Always for 10 mA:
V SSR input: 800 * 10 mA = 8V
Vf External Led: 2.4 V

So 15 - 8 -2.4 = Rx * 10mA

Rx = 4.6 / 10 mA
Rx = 460 Ohm ????

I'd like to know the right calculus for this external resistance, before Led.

Thanks a lot for your reply.

Jm

 

Not sure of the question?
There is normally no resistor needed for these SSR's.
It is all taken care of internally.
Max.

 

My question is:

RA4850D12 datasheet: 40 mA at 32 V dc So R '' inside '' SSR 800 Ohm?
Control voltage 3 32 Vdc

I thought of the usual way for Led devices, Vf, If max, and not found in this data...

The data sheet says the control voltage is 3 to 32 VDC, which is typical of solid-state relay. No external resistor required.

For a raw opto-isolator, with only an LED on its input, your concerns would be appropriate; but SSRs generally have a resistor built-in, making it unnecessary for you to concern your self with providing one.

Just apply an input voltage in compliance with the specified input voltage range, and your job is done.

 

It mentions input impedance 15k?
Max.

 

It mentions input impedance 15k?

I see 1.5 kΩ:

 

Hi,

Thanks a lot for your fast reply.
Nice day guys.

Jm