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i'm a new bee and trying a DIY project with DC solenoids. 

I'm having some basic (i hope so) with powering this solenoid - please let can you me know what is wrong with my attempt.

  • Solenoid is : 12v  540mA DC (normally closed) 
  • Power Source :  12v 23A Alkaline 
  • Diode :  IN4007

I have tried the approaches described below and in both approaches, Solenoid was not moved !  

i have tried this solenoid on a commercial product and it works just fine. And battery is tested with a multimeter and it is working independently BUT not in my circuit. so, i don't understand what i'm doing wrong here - any help is much appreciated.

  1. Directly power solenoid with diode  (see image 1). I have also tried as explained here Can i use 12 v solenoid valve directly connect with the 12 v battery?. But mine did not work :(
    enter image description here

  2. Use arduino uno board to drive solenoid (see image2) - i have connected DC converter (12v 1A)  as power source plugged into Arduino board

enter image description here

int solenoidPin = 4;                    //This is the output pin on the Arduino

void setup() 
{
  pinMode(solenoidPin, OUTPUT);          //Sets that pin as an output
}

void loop() 
{
  digitalWrite(solenoidPin, HIGH);      //Switch Solenoid ON
  delay(1000);                          //Wait 1 Second
  digitalWrite(solenoidPin, LOW);       //Switch Solenoid OFF
  delay(1000);                          //Wait 1 Second
}

Here is image of Solenoid and Battery:

enter image description here. enter image description here

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  • \$\begingroup\$ The Fritzing cartoon doesn't show us the pinout of the transistor. (That's why we use schematics in which the symbols show the schema of the circuit.) Unplug the yellow wire from the microcontroller and connect it to the breadboard's red line. If the solenoid doesn't work the problem is on the breadboard. If it does the problem is on the microcontroller board. \$\endgroup\$
    – Transistor
    Commented Jan 7, 2021 at 22:06
  • 3
    \$\begingroup\$ 23A batteries are very weak, not suitable for driving a 0.5 amp solenoid. "23A" is the type of battery, it has nothing to do with current capacity. \$\endgroup\$
    – Mattman944
    Commented Jan 7, 2021 at 22:11
  • \$\begingroup\$ Read the datasheet for that battery. (And weep). \$\endgroup\$
    – user16324
    Commented Jan 7, 2021 at 22:22

1 Answer 1

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Typical A23-type batteries are rated for about 15mA max load. Their short circuit current could be about 500mA. Estimating from that, the battery has an internal resistance of roughly 24 ohms.

The solenoid is rated for 6.5W at 12V, or 540mA. That is about 22 ohms.

Connecting those, you get roughly 6V over the solenoid with roughly 260mA of current.

That battery has no chance of operating the solenoid. It is meant for light pulsed loads such as remote controls, or even lighter constant loads.

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  • \$\begingroup\$ @Venu. This. Try using your multimeter to measure the voltage across the solenoid when powered from your 23A battery (i.e between the two twisted wire connections in your photo). You should find it's nowhere near 12 V. \$\endgroup\$
    – Graham Nye
    Commented Jan 8, 2021 at 0:39

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