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I am trying to use a simple Arduino circuit to turn an AC lamp on and off. This is the circuit I was about to try which I got from a website.
I don't understand why a transistor is required there. I thought we already have a 5 V output which we can directly connect to the relay.
The average microcontroller like on your Arduino has only limited ‘strength’ on the port pins. Typically it might be 2..20mA. Your relay might require 60mA at a guess so you require some means of amplifying the port pin. A transistor or mosfet is commonly used.
Note your diagram is missing a resistor in series with the transistor as the transistor only requires around 0.7V to turn it on. The resistor drops the port pin voltage from 5V down to 0.7V.
Ok, here's (almost) everything you need:
simulate this circuit – Schematic created using CircuitLab
The Arduino uses an ATMEGA MCU (most of the AVRs have similar specs so it's not critical) so the datasheet has the relevant informations:
At 5V supply the output port HIGH voltage is at least 4.2V, with 20mA load
Still a 5V, the output port LOW voltage is at most 0.7V, with 20mA load
Since they quote the current at 20mA it's safe to assume that the current is available (up to 400mA total for the chip, as in the absolute maximums).
From the BC547 datasheet you can see that:
Maximum current capability is 100mA (the size of your relay coil)
DC current gain (hFE) at 2mA is at least 110
Base emitter saturation is typically 0.7V
Base emitter on voltage could go low at 0.55V (this is a beginner's trap)
Now, for the calculations: you want to put some mA in the base with the GPIO high: so you have a 4.2V source, the base resistore and the B-E junction (which is substantially a diode for this purpose, and drops 0.7V)
You then leave with at lease 3.5V on the resistor to drop. The proposed 1k resistor limits the current to about 3.5mA (minimum). That, multiplied by the hFA gives a potential 350mA of collector current which is enough (as in, if you try to pull more than 100mA the BJT dies). So even a slightly larger resistor will work (not too much however)
Now, for the above mentioned trap: if you are really unlucky you could have an AVR with a very high low output (0.7V) and a BJT with a very low VBE (0.55V). So the relay could actually trigger when the GPIO is low!
For that reason it's useful to add another resistor between base and emitter to shunt the excess current. The calculation is not immediate (think a resistor in parallel to the B-E junction diode). However a 10k resistor is useful and traditional for many decades.
Let me try to simplify:
If B>A then you don't need a transistor, otherwise to source extra current required by the relay you need transistor.
There are relays which don't need a transistor to be operated: for instance the HE3300 reed relays need as little as 10 mA and can be driven directly by a microcontroller pin. Note that such relays are in turn rather limited w.r.t the power their contacts can switch (for HE3300 that would be 10W max).
In addition to limited current capacities of outputs: Sinking significant current, even within specification, will create more heating of the microcontroller (not good for reliability); also, there might be a limit to total output (sink) current for the whole microcontroller, limiting the amount of relays that can be directly controlled by it.
Also, driving significant current spikes through the ground connection of the MCU can potentially create signal integrity problems (forcing a bus ground in a place where it might not be desirable), especially if the MCU has some analog I/O and no separate AGND.
Also, relays and motors are considered devices that create back EMF spikes and other effects which, while you have to mitigate them with devices like freewheeling diodes and/or snubbers in any case, you want to keep somewhat isolated from a sensitive LSI part like a MCU.
