Cheap Ebay Relays

R

rmtucker

Member
Just bought A cheap dual relay off Ebay and they sent me the following Schematic.
They reckon it Needs about 15-20ma on each input to drive it.
I am connecting this to a 28x2 shield and would like to know how i check the current with a multimeter.
I would like to connect 8-10 of these but the current is going to be a problem.
 

Attachments

G

Goeytex

Senior Member
I am connecting this to a 28x2 shield and would like to know how i check the current with a multimeter.
Click to expand...
If your meter has a current setting ( many cheapos don't) you put the meter in series with the load and take a reading.

Alternatively you can add a 1 ohm "current sense" resistor in series with the load to be tested and then measure the voltage across that resistor. Using Ohm's Law, the voltage across the resistor will be such that 1 mv = 1ma. So 10 mv will represent 10ma and 100mv will represent 100 ma .... and so on....

They reckon it Needs about 15-20ma on each input to drive it.
Click to expand...
According to the schematic, the current load on the Picaxe will be ~5ma (or less) per input not 20ma. The only way the current is 20ma per input is if the schematic is bogus and the Picaxe must also drive the indicator LED's ( I have seen this on some boards where the indicator LED is on the input side ).

Many of these relay board vendors do not have a clue how the board operates or what the real specs are. Remember they also sell hair clips and baby socks.
 
Last edited:
R

rmtucker

Member
When you say connect it in series with the load,would that mean "+" lead to picaxe output pin and "-" lead to inp1 on the Relay?
The meter has Current neasurement.
I am not worried about the coil Current as it will be running from an external 12v supply.
Which brings another question to mind,The Relays are 12v,Will they be ok at Car voltages ie 13.5v
 
I

inglewoodpete

Senior Member
I presume the circuit describes what is one the circuit board that the relays come mounted on.

You would connect an output from your PICAXE to one of the inputs of the board, which connects to the base of a transistor via a 1000ohm resistor. By the look of it, the relay needs a 12 volt supply. Songle's datasheet says the coil resistance is 320ohms for the 12 Volt model.

Assuming the PICAXE part of the circuit is supplied by 3 alkaline cells, or about 4.5 volts, the 1000ohm resistor would have about 4 volts across it. Using ohm's law I=E/R means that the PICAXE would have to supply about 4mA into the base of the transistor.
 
R

rmtucker

Member
Inglewoodpete

The picaxe is a 28x2 shield so it has on board power supply.
The relays are 12v.
Looking at your calcs 4ma is great,The picaxe would handle 10off no problem,but where do they get 15-20ma to drive the input?
I did not think with a transistor in there it would draw anywhere near 15-20ma?
 
R

rmtucker

Member
Goeytex

We seem to be getting crossed wires.
I am worried about the current at the picaxe output pins that are driving the relay boards,not the current to power the coils.
 
B

BeanieBots

Moderator
rmtucker said:
....They reckon it Needs about 15-20ma on each input to drive it....
Click to expand...
This is what is causing the confusion.
15-20mA sounds about right for the coil current (where did that come from?) but the calculation of ~4mA (based on the circuit diagram) for the input on the circuit board is correct and is what the PICAXE output will need to drive. (which will not be a problem for 8 - 10 off)
 
G

Goeytex

Senior Member
I am worried about the current at the picaxe output pins that are driving the relay boards,not the current to power the coils.
Click to expand...
I got that.

Make sure that the supplied schematic is correct. As I said before, many of these boards have the LED's on the input side increasing the drive current requirement. You've gotta ask yourself why they would say 20ma, and then supply a schematic that indicates 5ma or less. A visual inspection of the board should show where the LED's are connected.

Please let us know.
 
B

BeanieBots

Moderator
Technically, you are correct that 13.2v would be out of spec according to that datasheet. However, I would not lose much sleep about running any relay with a 12v nominal coil as high as 15v unless it was in a life critical circuit in which case I probably wouldn't be using a relay anyway.
If you are very worried about it, fit a couple of diodes in series with the supply to drop it by a volt or so.
Of far greater concern is that you are clearly a novice electrically and you are planning on fitting something in a car.
Not sure what country you are in but if the UK, then your insurance will be void if you do not tell them about it and when you do, they will require an engineering certificate from a certified engineer.
 
hippy

hippy

Ex-Staff (retired)
rmtucker said:
but where do they get 15-20ma to drive the input?
Click to expand...
Perhaps they were expecting a higher voltage driving the input ?

It seems poorly phrased to me but at least with the circuit it's easy to calculate what the current draw for each input will likely be.
 
S

SD70M

Senior Member
Isn't the 15-20 mA the saturation of the transistor?

According to the schematic, the LED and relay are powered directly via VCC, then through the transistor to groud. The transistors are switched via the In1 and In2 from either a switch or, in your case, a PICAXE. Surely you would use a darlington pair to connect to the In1 and In2 to allow the source +V to the base of the transistors on the relay board to be 12V too?

Is this wrong?

Angie
 
P

Paix

Senior Member
From the diagram in post #1 I would think that the PICAXE with a 4.5V PSU/Battery would be more than capable of driving IN1 or IN2.
 
B

BeanieBots

Moderator
@ SD70M,
In short, no.
No need for a darlington pair.
The maximum collector current in a transistor is the base CURENT * hfe. (hfe is the transistor gain).
The VOLTAGE at the base will be one semconductor drop (typ 0.6v)
The base current will therefore be (the drive voltage - 0.6) / the base resistor.
For a darlington, there are two drops and the current to the second transistor is fed from the first.
Hence, there are TWO drops (~1.2v) and the gain is product of both (very high).

Assuming the diagram is correct, that board will be fine for a PICAXE output as has been described in detail earlier.
 
G

Goeytex

Senior Member
Isn't the 15-20 mA the saturation of the transistor?
Click to expand...
No. The 15-20 ma is a misprint from the supplier and is likely the specification for a different board.

Surely you would use a darlington pair to connect to the In1 and In2 to allow the source +V to the base of the transistors on the relay board to be 12V too? Is this wrong?
Click to expand...
Yes, that is wrong.

The transistor is an NPN and does not require 12v on its base. The transistor will fully turn on (saturate) with about 730mv on its base. Adding a Darlington pair would be a waste of components and is completely unnecessary.

ONLY .... if the Transistor were used as an Emitter Follower would there need to be 12v on the base .. where the emitter voltage "follows" the base.
 
You must log in or register to reply here.
Top