How to understand relay data.....
ex. relay is...Data sheet :Single-pole 10-A Power Relay
View attachment 67258
Dear Friend
Hi
No . it just mean that your relay can support 5 amperes at the out put contact . and your input current will given by : 12v/DC impedance of it's coil .( if that is DC)
Best Wishes
Goldsmith
Sorry so confused here.
I attached a file. Where did that 5A came from?
I thought 5A is max amps required to activate the relay coming from the input where the transistor is.
Where did that 5A came from?
I thought 5A is max amps required to activate the relay coming from the input where the transistor is.
Ans: it is given from an external circuitWhere did that 5A came from?
the answer maybe pretty much clear now.......I thought 5A is max amps required to activate the relay coming from the input where the transistor is
View attachment 67280
Our relay is activated. when the current is passed through the relay coil.. Now again have a look in the schematic. Here the positive voltage applied to the base of the T1 (since NPN) will helps to create the path for the current to pass through the coil and activate the relay. Note since the other end of the relay is connected to the battery array, the current flows from the positve terminal of the battery, through relay coil, then to the transistor T1 and then to ground. (have a look in the schematic to have a proper understanding).. If you understood up to this point, then continue reading. Otherwise understand it first....
Now we connected 5A to the Normally Open point and now it is connected to the coomon point. thus 5A is reaching th common point. Now look, according to your schematic, the common point is again connected to the battery array..( I can't explain why it is connected so.Because you didn't mention the purpose of this circuit anywhere). Now the 5A (either ac/dc -since you didn't mentioned it) will be reaching that point.
(note this circuit may be used for battery charging controlled by another source through the port pin).
The Diode in Red boundary is connected in wrong direction as it is in forward bias connection and as soon as the Relay Contact is closed the diode will conduct in parallel to the relay contact and will provide additional load to the battery for nothing. Abbreviations are as follows:
NC = Normally Closed Contact
NO = Normally Open Contact
COM = Common Contact
There are two things are mentioned normally for a relay,
1. The coil operating voltage + current (some times)
2. The Contact current which can be switched for maximum.
Yes, I have understood this part.
So a relay is activated by a given current. How do I determine the minimum required current for a 5A 12v relay to activate? given that 12v DC is the given voltage.
I searched and is the V = IR the right solution? Or as we say the Ohm's Law.
Actually Yes, But since the current sources are not readily available, we use the voltage sources........ The rating 12v 5A means the switching is occured at 12VSo a relay is activated by a given current
How do I determine the minimum required current for a 5A 12v relay to activate? given that 12v DC is the given voltage.
I searched and is the V = IR the right solution? Or as we say the Ohm's Law.
Currently I'm using this to control a coin slot to power it on or off. At first I didn't mind what values the relay have as long as it is 12v not until I was questioned why I chose a 5A 12v DC relay and it started all of my confusions.
Thus the current flows through the relay coil when T1 is ON......... Now look, according to your schematic, the common point is again connected to the battery array..
This is as I mentioned is flyback or free wheeling diode) Gold smith on post #17 gave why it is connected so........So I will just reverse the diode to make it right? right?
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