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Sink current for TLP521 via 74HC595 or ULN2803A



I have a TLP521 driving a set of relays. The circuit is shown below.

To switch on the relay, I will need to sink the current to complete the P521 circuit, which will trigger the relay.

I am thinking to use a ULN2803A to sink the current as shown in the below diagram. This would mean that when I pull the Qa pin of shift register as HIGH, the 1C point (on ULN2803A) should be zero (grounded) and the circuit should be complete. Is this the correct way to do it? Please note that the common and gnd of ULN2803A are grounded and there is no +Vcc on the IC. Is this OK, or should I connect the COM to +5? (I don't want to source current to 1c by mistake or during initialization)


Another question: Can I get rid of ULN2803 and directly sink the current from 595? I dont think it can sink this current.

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I thought I had responded to this question back when you first asked it, but clearly this isn't the case; I apologize.

Looking at what you have, you could probably get rid of the ULN2803 and just use the shift register outputs instead; you would connect the ground portion of your TLP521 to the ground on the shift register and then have your shift register output a high signal to "activate" the TLP521. I don't know what sort of voltage it needs to turn on though; glancing at the datasheet for the TLP521 it looks like it wants a small voltageĀ (~1.15V typical) to turn on the LED that activates the transistor but I don't think the shift register or the ULN operate at that low of a voltage...

You could keep your current setup, but I would connect COM to your 5V. Maybe you don't have to (now that I looked through the datasheet some more) but I don't have the Darlington array on hand to test it out.

Let me know if you have any more questions.


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