short or open circuit detector

Discussion in 'The Projects Forum' started by sengand, Jul 8, 2008.

  1. sengand

    Thread Starter New Member

    Jul 8, 2008
    2
    0
    Hi
    i am try to building the 5V base short and open circuit detector. it mean that if detection of more than 4.5V is open or less than 0.8V is short circuit.
    i try to use the LM393N but i fail to built the circuit. anyone who is expect to built this kind of circuit? or i should use other Ops Amp to built.
    the purpose for this circuit is for the Automotive engine sensor open and short circuit simulation..

    Adrian
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Can you share the circuit used with your LM393N? Are you using half the chip to sense for 4.5V or higher and the other half to sense for 0.8V or lower? What are you using to create the reference levels?
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You may want the low threshold to be a bit lower, say 0.55v, unless the transistor base you're checking will always be either saturated or off.

    Comparators are the proper device to use for your circuit.

    See the attached schematic, made with LM339's. The outputs really aren't quite proper, as LM339's can only sink up to 10mA, so R4 is much too small. With red or green LED's, R4 should be 1k Ohm and 1/4W.

    D1 lights if the input is less than 0.6v. D2 lights if the input is greater than 4.5v.
     
    Last edited: Jul 8, 2008
  4. sengand

    Thread Starter New Member

    Jul 8, 2008
    2
    0
    thank you sgtwookie
    i will try your circuit. by the way attachment is my own circuit, do you think that any improper connection? i want to use only single LED to made the light on when lower or more than the voltage threshold
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Ok, the LM393 can sink up to 20mA current with it's output, and the outputs are open-collector. You can tie pins 1 and 7 together (the two outputs) and use that on the cathode of the LED to sink the LED current. You don't need the transistors at all.

    The 4.7k resistor to ground from the LED's cathode isn't correct. Please remove it.

    Now you need to calculate the current limiting resistor for the anode of your LED. You use the average voltage specification, not the maximum voltage specification. The LM393 can sink 20mA. If the LED's maximum current is less than that, use the LED's current, otherwise use 20mA.

    Rlimit = (SupplyVoltage - Vled) / 20mA
    Use the closest equal or higher value that you wind up with. For example, let's say you're using a red LED that has a typical Vf of 1.8v at 20mA, and you're powering the circuit from a 9v battery.
    Rlimit = (9-1.8) / 0.02A
    Rlimit = 7.2/.02
    Rlimit = 360 Ohms.
    That happens to be a standard E24 value. Connect the anode of the LED to the + supply, and you should be OK.

    Now, I notice that you are using 10k pots to set your reference levels. That would be OK if you were using a regulated voltage supply, however you are apparently running from a battery. This means that as the battery wears down, your reference levels will constantly shift, and will require constant adjustment.

    Notice that in my schematic I used a Zener diode and a 1N4002 diode; the voltage drop across those two will remain far more stable from when a battery is fully charged to when it is nearly spent than a resistor divider network will.
     
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