How does the Arduino 5VDC relay module work? Using SRD-05VDC-SL-C

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I was wondering how this 5V relay module works exactly.

I understand how the actual relay circuit (SRD-05VDC-SL-C) works but I’m not totally sure how the signal and 5V inputs work.

My guess would be that the 5V is somehow dropped down to, lets say 4V, and then is combined with the signal voltage (let’s say 3.3V from a 3.3V pin from an Arduino.) If the signal input is HIGH then the relay (SRD-05VDC-SL-C) receives a voltage greater than 5V and closes, but if the signal input is LOW then the relay is receiving just the 4V from the VCC and stays open.

If that is the case, how does the relay module do this using diodes, resistors, etc.

If that isn’t the case then does anyone know how this thing works?

arduinovoltagerelay

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edited 6 hours ago

JRE

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asked 6 hours ago

thomas

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• This might help: How to properly use a relay module with JD-VCC from Arduino/Raspberry? Asked 1 year, 11 months ago Modified 24 days ago Viewed 23k times electronics.stackexchange.com/questions/505318/… – tlfong01 6 hours ago  
• @tlfong01 that relay module is different than this one. That one has opto isolation and independent logic and relay voltages. – Passerby 6 hours ago
• Part A of my TLDR answer says this: Part A – Simple High and Low level trigger relay modules without optical isolation: To explain the idea of Low and High trigger relays, let us start with the no optical method and look the the respective schematics below, … The OP’s relay is perhaps like this: forums.raspberrypi.com/viewtopic.php?f=37&t=77158#p1323061. – tlfong01 1 hour ago   
• And the schematic is perhaps like this: imgur.com/gallery/RRjyeNn. – tlfong01 1 hour ago  

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The arduino controls a npn transistor. The transistor controls the relay. There is no mixing of voltage as you describe. The transistor is current controlled. The relay will get 5V across it, or be disconnected like a switch, if we only look at on/off operation.

The basic schematic is basically this 

The transistor acts like a switch.

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edited 6 hours ago

answered 6 hours ago

Passerby

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• Great! Thanks a lot, that makes total sense. – thomas 6 hours ago

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I would guess the schematic is like this:

^{simulate this circuit – Schematic created using CircuitLab}

The relay coil requires about 90mA so the base current should ideally be roughly 5mA+, meaning for a red LED the input voltage should be 3.3V to 5V for “ON” and more like <1V for “OFF” (there’s no emitter-base resistor so leakage should be minimized).

It’s a pretty minimalist circuit with a shared supply and no isolation beyond what is provided by the relay.

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answered 6 hours ago

Spehro Pefhany

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