Now supply current should be 2.85 mA, but mu power supply is giving around 500mA, then how to step down it, and will the circuit diagram be same for this module as well?
This can be seen frequently on Internet. I will suggest Ohm Law for start.
Current in circuit is limited with resistance in that circuit.
Your module will use amount of current what he need, what is lower than your psu capabilities.
When you buy a car which can go 250km/h, does the seller said you must drive this car 250km/h ?
Best regards,
Peter
:wink:
Nice way of explaining :razz: so what exactly what you mean is that I need not worry about the current requirements of my module...as suppose my power supplu is giving an output current of 1 A, then also my module will take only what it requires...in our case it is 2.85 mA...... nyways i like your ways of explaining the concepts...Thanxs a lot...
Regards
as per my calculations if I want an output current of say 2.78mA, i will be requiring a resistance of value around 1.8K ohms... Am I right peter????
Regards
I am using atmega 16..
Dear Peter
When you purchase relays, you generally have control over several variables:
The voltage and current that is needed to activate the armature
The maximum voltage and current that can run through the armature and the armature contacts
The number of armatures (generally one or two)
The number of contacts for the armature (generally one or two -- the relay shown here has two, one of
which is unused)
Whether the contact (if only one contact is provided) is normally open (NO) or normally closed (NC)
What does this mean? I mean what will be the values of these parameters for my application. I want to convert the 5 Volt output of controller to 230 VAC, 20A.
Regards
Code C - [expand] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 #include<avr/io.h> #include<util/delay.h> void main() { DDRA = 0b00000000; DDRB = 0b00000001; While(1) { int x = PINA & 0b00000001; If (x = 0b00000001) { PORTB = 0b00000001; } else { PORTB = 0b00000000; } } }
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