Dear all, for my project I have a Power Supply (PC) - having max power of 500W.
Connected to the power supply is:
OPAMP AMPLIFIER - TC7650CPA -
LCD
LOAD CELL
VOLT REG - 7905
SOLENOID VALVE - 20W
RELAY -SOLID STATE
Before i connected the solenoid valve and relay to the circuit everything was working fine....
Having the solenoid connected, the voltage regulator became damaged....
The voltage regulator 7905 has a limited power of P <= 15W.
Is this why it got damaged?? since the sol valve dissipates power greater than 15W?
What can I do to not damage the regulator or the circuit???
The 7905 is a negative voltage regulator. Why are using that? Also I don't know where you get a rating of <15W. When properly heat sinked, the 7xxx series regulators generally only provide 1amp. They don't generally have a power rating.
What is the solenoid connected to?
What are you using to load the ATX power supply? (ATX supplies generally require between 500mA and 1A of constant load on each of its voltage rails to keep the voltages in spec.)
Neg Volt reg since opamp requires negative and pos voltage.
Ok I dont know what happened first but I soldered another volt reg and it worked....I dont understand the concept of power....
The ATX has a 10ohm 10W res connected, it gives volt in spec
If for example my sol (20W) is connected to my +12V and gnd supply, and from that same 12V and gnd I power the arduino wil the arduino burn???
The opamp is used to amplify the loadcell o/p voltage, since the loadcell, when giving it a full load (25kg) it gives mV and i amplified it to 5V having a gain of approx 1600.
The voltage regulator 7905 has a limited power of P <= 15W.
Is this why it got damaged?? since the sol valve dissipates power greater than 15W?
The power dissipated in a load is NOT the same as the power dissipated in the regulator or switching circuit.
Power is the product of voltage and current, but is not the only limit on a component.
A regulator dropping from 12V to 5V has 12-5 = 7V dropped across it. If there is 1A flowing through it then the power being burned off in the regulator is 7 * 1 = 7W.
If the 5V is being used to power a 5 ohm resistor it will have 5 / 5 = 1A through it. However the resistor is only burning 5 * 1 = 5W.