Guys,
Struggling with this seemingly easy circuit to function. The input signal at J2 is an arduino output and ground. Power at J1 is 12V, required by the solenoid. When J2 is disconnected, I've verified the arduino correctly outputs a logic high (4.9V). However, when my arduino is connect to J2, the output pin only achieves .7 volts. Any help would be appreciated.
Thanks -D
the output pin only achieves .7 volts.
What are you considering the output pin?
Common error is missing pinMode(x, OUTPUT) declaration in the setup. Voltmeter shows pull-up resistor at digital input
Hi, have you got a solenoid or some sort of load connected to J3, if not then the Q2 has no load, so no current flow.
The output will go to 0.7V if the input at J2 is HIGH, then current will flow through the transistor and the LOAD.
Tom..... 
I hope you have gnd of the arduino connected to pin 2 of J2, and the arduino output connected to pin 1 of J2.
Hi,
J2 is the output of the arduino and j3 is the output to the solenoid. Pin one of j2 only reaches a high of .7v when the cable from my proto board is connected to this circuit. When the cable is not connected, the output pin of the arduino reaches 4.9v.
With my meter connected at J3 I see 12v when the arduino sets j2 - pin 1 high, but when the solenoid is attached it is never activated. Clearly no current is flowing. Depressing the switch does activated the solenoid's coil. I hope this is clear.
D
To OP: Have you decided to ignore reply #2, or I'm "invisible ghost" on this forum?
Sorry, I was not in front of my project code to reply earlier. At the time I was thinking 'of course I declared all of my I/O pin modes'. Well, in checking, I declared every other pin EXCEPT that output pin. Once declared, the circuit works as expected.
Strange that the build process did not flag it and allowed me to write (high/low) to the pin without being declared an output.
I humbly say thank you for your help and I wish you a good day.
D
Strange that the build process did not flag it and allowed me to write ](high/low) to the pin without being declared an output.
Not at all - pins are INPUT by default if not declared as OUTPUT. Writing HIGH to an INPUT pin enables the internal pullup resistor, and LOW disables it.
Interesting, I did not know that. Clearly I'm new to the arduino/atmel project group. I guess that explains the voltage swing I was seeing on an unloaded pin.
Thanks,
D
Glad to hear the issue is solved. Yes, it's confusing when pin declared as input provides a voltage, at least if it's not so , someone 'd "intuitively" look into software
The output is 0.7V because its a darlington output, which saturate at a diode-drop.
The zener is doing nothing, the darlington base will not rise above about 1.5V anyway.