Measuring current used by Relay Module switching operation

Other Hardware General Electronics
1

I have small knowledge about electronics in general, my question is how to measure the current used by switching operation on relay ?

where should I put my (+) probe and (-) probe ?

Correct me if I'm wrong,
I break the 5V to VCC connection then,
I connect my (+) probe to the 5V of arduino then,
I connect my (-) probe to the VCC pin of Relay then,
I jump ground to the IN pin of Relay

2

Please post a link to the relay module, a wiring diagram and state which Arduino your are using.

The 5V output of most Arduinos cannot supply much current. You should avoid drawing more than about 100 mA from it, depending on how the Arduino is powered.

Check the specifications of the relay module to learn what the current draw should be.

3

So in general, it's not safe to powering the relay directly from 5v of arduino ? Should I use external power supply instead ?

4

Some relays may not be powered from the 5V output. Yours may be fine.

Post a link to the data sheet for your relay.

5

don't have the datasheet, but my relay is exactly this one :

https://components101.com/switches/5v-single-channel-relay-module-pinout-features-applications-working-datasheet

6

Thanks for posting the link to the product page. These are the important data as far as the Arduino is concerned. It is OK to power one of these modules from the 5V output.

  • Supply voltage – 3.75V to 6V
  • Quiescent current: 2mA
  • Current when the relay is active: ~70mA
7

but isn't it exceed the current limitation of arduino pin which is 40 mA ? in that explanation it was +- 70 mA

8

The 70 mA is the current drawn by the relay coil from the 5V power source Vcc.

The current drawn from the port pin by the "Input" terminal is limited to about 4 mA by the series 1K base resistor in the schematic below, so that is safe too.

750×500 58 KB

9

Ahhh... I see... so the 40mA limit is for the I/O pins right? question : so If I want to measure the current signal,

I should break the connection between arduino pin - IN pin of relay, and then connect my (+) probe to the arduino pin, connect my (-) pin to the IN pin.
Is this the right way to do it ?

10

To measure current with a meter, you have to break the connection. The meter adds some resistance, so the measured current will be less than the original.

11

Do you want to measure the current needed to switch the relay?
Or the current going via the switch of your relay?

Measuring a current means always:

  • break the wire
  • put an AmpMeter in series into the broken wire (to close it again, now via the AmpMeter)
  • set a Multi-device to Amp mode (not voltage mode)!
  • watch carefully how large the current could be for the AmpMeter (it has a fuse to protect it),
    which connectors on AmpMeter to use and the value range to set

Measuring amperage is measuring the current "going through" (therefore AmpMeter is in series).

But you could also use a "shunt resistor":

  • you put a 1 Ohm resistor in series (break wire, put it in, in series to connect the "wire" again)
  • measure the voltage over the 1 Ohm resistor (as voltage)
  • Ohm's Law tells you: 1V measured over a 1Ohm - the current is 1A
12

Thank you for the explanation. I want to measure the signal current (the current that going through I/O Pin), so should I measure on the IN (input) side ?

13

You can only measure a current coming out from an OUTPUT pin.
But this goes via a resistor, e.g. this 1K to the base of a transistor. There will not be really a remarkable current.

On a GPIO input you cannot measure any current: it is CMOS logic which has a very high internal resistance (not taking "any" current).

It does not make sense to measure the current out from an OUTPUT and going to a transistor, it should be really low.
You can measure the current on the other side of transistor, e.g. from VCC, going through the relay coil and transistor (the current into collector of transistor, e.g. on 2).

It means: measure the current on the load side. The driver (your GPIO output), going to a transistor (as a driver) should take very small current. You would be interested in the "switched load current", the current through the relay or the current through the switch in relay (the final load).

14

For those relays, the typical logic signal current is 2 ~ 4mA, the coil activation current about 72 ~ 75mA (including the transistor base current, 2 ~ 3mA).
OptoRelayChannelData-575

15

I agree, makes sense.

16

But OP posted a link to a different relay module, one without opto coupler.
The one OP linked to has a base drive current of 4.3mA, and a relay/LED current of 75mA.
Leo..