I am learning electronics so this is learners question. I wired up the LM339N quad voltage comparator (powered using a 5v power supply through a current limiting resistor of 40 ohms)-in a straightforward standard way- like at
but for the potential divider I used two 4.7k resistors-not a pot- and I only used one comparator. I also had an output LED too.I added the standard 0.1uf capacitor near the power supply. So the circuit worked as it should (for the input voltages to the comparator I used AA batteries that went through a current limiting resistor of 500 ohm into the comparator input) at 1.5 volts or less the LED turned off. At higher input voltages the LED turned on.
I want to replace the LED with a pc123 optoisolator, so the optoisolator behaves like a switch (just on or off, with a resistance of between 1 and 60 ohms when on, or zero ohms when off)-so what is the circuit to do this? So when the opto is off there is no resistance .
There maybe other components (im not sure of this) required in addition to the pc123 for answering this question . I replaced the led with a pc123 optoisolator but smallest resistance I could get (directly from the phototransistor) out of it was between 5k ohms and 6k ohms and I used a 40 ohm resistor for limiting the current into the opto. If its not possible to get between 1-60 ohms , then whats the the best that can be done to control the output resistance?
here is data sheet for the opto
Also, for a different and related circuit, I want to replace the LED with a bc547 transistor to turn on a 5v relay -so what is the circuit to do this?
I dont need a fancy answer. I want an answer that show clearly how to answer the question using those components in a minimal way.
Come on folks, try and answer the question!