In the event of transistor failure ?

Hey,

I am building an Arduino circuit to drive a solenoid door lock. I have been following this http://www.arduino.cc/playground/uploads/Learning/solenoid_driver.pdf schematic on the tutorials page, and I have a few questions.

  1. If the transistor was to fail for some reason, due to a short or over heating, would the collector-emitter continue to conduct, keeping the solenoid powered permanently ? This is an important design consideration as this scenario could potentially leave the door unlocked constantly - I would rather the door remained locked until the circuit could be fixed.

  2. The solenoid lock I'm using is much lower power than the one in the above schematic - 250mA 12V - and will probably only be active for 60 seconds, once or twice a day, would a heat sink on the transistor still be advisable? I was of the belief that if the correct resistors were used, the voltage drop across C-E would be minimal, and so the transistor would not dissipate that much heat.

  3. I'm not too hot on transistors. I have a BC489 http://uk.rs-online.com/web/p/general-purpose-transistor/5450040/ , would that be an acceptable substitution for the TIP102.

Thanks in advance,

Rich

  1. There are different kind of failure, so it's quite possible. Put a safety watch, resistor as current sensor + timer, if current flow more than 60 sec = do something: ring a warning bell , shutdown power etc. 2.& 3. Your transistor BC489 not good as TIP102, but would be o'k for 250 mA w/o heat-think with proper base resistor (220 Ohm).
  1. If the transistor was to fail for some reason, due to a short or over heating, would the collector-emitter continue to conduct, keeping the solenoid powered permanently ? This is an important design consideration as this scenario could potentially leave the door unlocked constantly - I would rather the door remained locked until the circuit could be fixed.

I used to work in the security industry and the two situations you are describing are known as "fail safe" and "fail secure". The fail secure situation, while you might think is desirable, opens up a whole load of legal issues that might have you prosecuted in the event of a fire or other injuries.

A transistor might fail short but it is more likely to fail open.

I was of the belief that if the correct resistors were used, the voltage drop across C-E would be minimal,

No, this voltage is known as the saturation voltage, and can be significant with a large current, in the order of 1 to 2V. That is why FETs are so much better.

  1. If the transistor was to fail for some reason, due to a short or over heating, would the collector-emitter continue to conduct, keeping the solenoid powered permanently ? This is an important design consideration as this scenario could potentially leave the door unlocked constantly - I would rather the door remained locked until the circuit could be fixed.

The transistor could short and the lock will remain unlocked. An important consideration in your design should be something like Magician suggested because not all locks are rated to be continuous duty. If you have an intermittent duty is lock powered up continuously, you will burn out the solenoid.