Will Raspberry Pi burn out when connecting a relay via GPIO?

Question

Electronics

After having a lot of blinking on / off the LED, I decided to go further and make the relay control.
Based on this article .
Bought from a local radio store:
Relay Fangke JZC-23F (5A/220VAC) DC 5V
Transistor BC546B
Diode and resistor found at home.

I assembled it all on the circuit board according to the diagram on the site.
The article says that the raspberry will burn out if more than 20 mA flows through the leg. Therefore, as a control signal, I connected two finger batteries together and got 3V. And to power the coil, I took a 5V power supply. I have put their cons together.
The scheme is working. To measure the current, I connected an ammeter in series between leg #25 on the diagram and the minus common for the batteries and the power supply.
The ammeter showed 60 mA . It turns out that if I connect this circuit to raspberry, then it will burn? Where did I go wrong?

Answer 1

It's hard to understand from the photo, but in my opinion, you need to measure the current in the wire break that goes from the plus of the battery to the resistor.

Answer 2

You don't have a ground.

Answer 3

scheme in the studio!
what is the value of the resistor? (for a 3V battery, the resistor to the base should be about 470-1000 ohms, then the base current will be 3-5 mA)
Where 60 mA comes from is not clear. If the resistor is 47 k, maybe there is 60 uA ? (very similar). But this current may not be enough for the relay - the transistor will be in active mode.

Answer 4

Thank you all, everything worked!)
Do I think correctly that for 220V and 40 watts such thin wires will be too weak?

Answer 5

In general, the current did not kill me) Already good. Made a web control of a light bulb over the Internet. Who cares, they can wink at her in my room)