I have solid state relay (CPC1008N) Dataheet here. What is the resistor I need to use to connect it to a power supply with an output voltage of 4.5 volts?

  • \$\begingroup\$ Welcome to EE.SE. You can't assume that we all know the inner details of your SSR. Please hyperlink to the device datasheet in your question. \$\endgroup\$
    – Transistor
    Jun 17 '18 at 16:34
  • \$\begingroup\$ @Transistor - John D is ahead of me \$\endgroup\$
    – Wind
    Jun 17 '18 at 16:53
  • \$\begingroup\$ What is your control voltage? Also 4.5V? R = (4.5V - 1.5V)/5mA = 470R...1kOhm's \$\endgroup\$
    – G36
    Jun 17 '18 at 16:59
  • \$\begingroup\$ In Dataheet says: Input Control Current to Activate (Max) = 2 mA. Why 5 mA? \$\endgroup\$
    – Wind
    Jun 17 '18 at 17:32
  • 1
    \$\begingroup\$ @Wind The relay performs better at 5mA than 2mA, see speed curves. Also, with aging the output of LEDs decreases for a given current. Temperature may be an issue too, the data sheet values are at 25C. Also, you don't specify a tolerance on your 4.5V so 5mA gives some margin if there's any IR drop or tolerance on your control signal. \$\endgroup\$
    – John D
    Jun 17 '18 at 17:43

Assuming your control voltage is also 4.5V:

You need at least 2mA to guarantee activation of the relay.

The diode forward voltage is specified as 1.5V max at If = 5mA. So targeting 5mA to be conservative would mean you have 4.5V - 1.5V = 3V across your resistor.

Using Ohm's law V/I = R gives a resistor value of 600 ohms. Choose the next lower standard value, 590 ohms. There's plenty of margin to the 2mA minimum so you could go higher if you want to save power.

Since the abs max continuous forward current is 50mA there's no concern with exceeding that with the 590 ohm value.

  • \$\begingroup\$ For E24 series the 560R or 620R is the closest value. But I will use 470R or 680R \$\endgroup\$
    – G36
    Jun 17 '18 at 17:02
  • \$\begingroup\$ Sure, there's plenty of margin to go higher than 600 if you want, and you don't need 1% tolerance. \$\endgroup\$
    – John D
    Jun 17 '18 at 17:04
  • \$\begingroup\$ Input Voltage Drop: 0.9 <= Vf <= 1.5, Then: 4.5 - 1.2 = 3.3 V; R = V / I = 3.3 V / (0.45 + (2 - 0.45) / 2) mA = 3.3 / 0,001225 ~ 2694 Ohm \$\endgroup\$
    – Wind
    Jun 17 '18 at 17:28
  • 2
    \$\begingroup\$ @Wind if you're building a lot of these you'll not want to use the typical values, see my comment above about why the target current is 5mA not 2mA. You'll likely be fine with the typical values if you're building just one, and if you're not a single circuit is easily adjusted. \$\endgroup\$
    – John D
    Jun 17 '18 at 17:45

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