low current relay circuit question

Thread Starter

darylrobert

Joined Sep 30, 2020
9
I have a 5v dc output connected to an indicator LED that is active when i need it de-activated. visa-versa it turns off (deactivated) when i want the LED to light. I need the current draw to be as low as a LED draw, i think a relay with draw too much current?.. i was looking at this relay, the control would be input but what is pin 3 N/C (normally closed) ....thankyou
 

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Martin_R

Joined Aug 28, 2019
137
I have a 5v dc output connected to an indicator LED that is active when i need it de-activated. visa-versa it turns off (deactivated) when i want the LED to light. I need the current draw to be as low as a LED draw, i think a relay with draw too much current?.. i was looking at this relay, the control would be input but what is pin 3 N/C (normally closed) ....thankyou
Easiest solution is to use a small signal PNP transistor as shown in the attached circuit. It's function is to invert the input signal.
1601624829732.png
 

PhilTilson

Joined Nov 29, 2009
131
You don't specify what this 5V output is - is it a microcontroller output pin, for example? If so, then why not simply connect the LED to the other rail? For instance, if the LED is presently connected + to 5V rail and - to output pin, connect instead + to output pin and - to ground. This will 'reverse' the operation as you require.
 

MrChips

Joined Oct 2, 2009
30,706
We don't know how the LED is being driven or what is driving it. But as @PhilTilson said, you may be able to drive the LED through one resistor, depending on whether the LED needs to be on the high side or the low side.

Hi-side low-side LED.jpg
 

MisterBill2

Joined Jan 23, 2018
18,167
Mr Chips offers the simple solution in post #4, but that solution is at best a guess because we have not a hint as to if the output is sinking or sourcing current, nor how much current it can provide in the other mode. The suggestion about simply reversing the LED is inherently wrong, also.
So now the question goes to the TS to describe the driving circuit details as asked.
 

Thread Starter

darylrobert

Joined Sep 30, 2020
9
thankyou all, i have sourced a ready made 5v relay module thats trigger current is 5ma. I have tested it and its more complicated than i wanted but it works.
 
thankyou all, i have sourced a ready made 5v relay module thats trigger current is 5ma. I have tested it and its more complicated than i wanted but it works.
I have a 5v dc output connected to an indicator LED that is active when i need it de-activated. visa-versa it turns off (deactivated) when i want the LED to light.
 

MisterBill2

Joined Jan 23, 2018
18,167
I have a 5v dc output connected to an indicator LED that is active when i need it de-activated. visa-versa it turns off (deactivated) when i want the LED to light.
Use one of the circuits posted by Mr. C in post #6. Presently your circuit probably looks a lot like one of them, so use the opposite one to reverse the logic. Fairly simple and really cheap.
 

MisterBill2

Joined Jan 23, 2018
18,167
If your low=on output can sink 20 mA then the upper circuit shown in post #6 will work quite well. If not, your circuit with R2=470 ohms and R1 =2.7K ohms should work.
 

MisterBill2

Joined Jan 23, 2018
18,167
NO, I don't think so. My answer, as the others, have been in response to a question about an output able to light an LED, requesting that the solution use no more current than that.
If the output is an open collector, or an open source, that is an unstated possibility, and in that case the answers need to be a bit different.
 

BobTPH

Joined Jun 5, 2013
8,804
If you want help, you need to answer our questions. Solutions have been proposed that May or may not work depending on what is driving the LED.

So, once again, what is driving the LED? If you don’t understand the question, please ask for clarification.

Bob
 
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