Ok LED off, relay on
...
Well, I am glad that is settled.
So, IF we could find a SSR that didn't require a ton of current to turn on, then it could be connected between the output, DO, and common. The internal 10 kΩ pull-up resistor would supply current to the SSR when the output goes "high" and the open-collector transistor turns off. Unfortunately, even if the SSR input was a "dead short" when conducting, the current delivered through the 10 kΩ pull-up resistor would only be 0.00045 A or 0.45 mA with a 4.5 V DC supply. Not enough to operate any SSR input that I know about since these are usually an optical isolator with an LED and a series current-limiting resistor, about 440 Ω for the device under consideration.
The problem is, if you lower the value of the pull-up resistor to provide enough current to turn on the SSR when DO is high, it allows way too much current to flow through the open-collector output transistor when the DO is low. So... you need to insert a transistor in there to provide more current. Since output "high" turns the transistor on, we are back to the original NPN configuration with possibly a smaller "pull-up" resistor to provide sufficient base current drive to the transistor. I would start with 2 kΩ inserted between DO and Vcc, connect output DO directly to the base of the transistor, connect the NPN emitter to common, and insert the SSR or the original 3V relay coil between Vcc and the NPN collector. Connect a small 1N400x-series rectifier diode across the relay coil, cathode to Vcc and anode to the NPN collector. Use a small-signal transistor capable of carrying the coil current or the SSR control current. No heat sink necessary: the transistor is biased off when DO goes low and conducts in saturation when DO goes high.
This is basically the same circuit as the original post #1, except S1 is replaced with the circuit shown in post #7 and a 2 kΩ resistor is added from Vcc to DO to increase the base drive voltage. Try using a 2N3904 NPN transistor to operate the relay. Use a 4.5 V DC supply for Vcc. The 4.5 V DC supply will provide about 2 mA base current through the 2 kΩ resistor and the 2N3904 should be fully saturated and conducting when DO goes high.
And let's not get into any esoteric discussion as to whether the increased base current is responsible for driving the transistor into saturation and increasing the current available to operate the relay coil. We all know that it is the increased base-emitter forward voltage that decreases the base-emitter depletion layer width and allows more emitter electrons to drift across the depletion layer into the collector region to increase the collector current. The ratio of collector current to base current, called hfe or beta, is just a convenient observation sometimes used to incorrectly "explain" current "amplification" in a transistor amplifier circuit.
