power supply short circuit protection, overcurrent draw

Discussion in 'General Electronics Chat' started by relicmarks, Nov 2, 2008.

  1. relicmarks

    relicmarks Thread Starter Active Member

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    I have a 12 volt power supply , but what can i use "inbetween" the power supply and the circuit under test that has indicator lights to tell me if there is a short on the circuit under test?

    I know they make variacs with a built in amp meter

    But is there something cheap to put inbetween a external power supply or wallwart that has indicator lights or something to tell me there is a short on the circuit being test

    Cause i don't want to turn the circuit under test on for the first time and it burns the PC board or burn some IC chips
  2. relicmarks

    relicmarks Thread Starter Active Member

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    I tried using a Variac with the current meter, but if the board/circuit is drawing alot of current at a low voltage than its hard to troubleshoot where the short is at

    I start the Variac at O zero volts and increase it up very slowly, but mostly around 3 or 4 volts if there shorts on the board i will see the current meter jump very high trying to draw alot of current

    Than i try to use my DVM meter set to voltage , checking various grounds to see i read any voltages on grounds.

    If there is voltages on the grounds then there is a SHORT

    A short circuit protection circuit i need will still supply the 12 volts and "normal rating current draw" even tho its trying to draw alot of current since there is shorts on the circuit under test

    It like it will Clamp or regulate the current/voltage from increase under test
    but will have a indication lights tell me its trying to draw alot of current cause there is shorts on the board

    example: circuit board under test 12volts at 1mA rating

    Shorts on board under test:

    1.) If i use a Variac with current meter is will be at 3 volts with 1 Amp
    2.) Than i have to tests the grounds with a DVM set to volts
    3.) Try to find which grounds have 3 volts on the ground nodes

    Example: circuit board under test 12 volts at 1mA rating

    1.) If i use a short circuit protection circuit
    2.) the circuit board is trying to draw 1 amps
    3.) The short circuit protection circuit lets me test the circuit under test still at 12 volts at 1mA rating, Clamps regulated the current so i can try to find the shorts

    It will be easy to troubleshoot having the correct normal rating voltage and current while trying to find shorts
  3. mik3

    mik3 Senior Member

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    If the power supply 'sees' a short circuit directly on its terminals caused by the circuit then you can detect it simply with a continuity meter. Apply your multimeter probes on the circuit power lines and see if the buzzer sounds. However, if your circuit contains active components you cant do it because they may cause a short when a voltage is applied. In this case use a resistor in series with the circuit, which in case of a short cant cause a current of more than 10 amps to flow (calculate its value depending on the supply voltage), and use your multimeter set on 10 amps range to measure the current flowing to the circuit. If it much higher than expected then a problem occurs. Be careful, because some power supplies, when a short circuits occurs, they limit the current to a safe value so you wont be able to detect if there is a short circuit.
  4. relicmarks

    relicmarks Thread Starter Active Member

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    1.) If the power supply 'sees' a short circuit directly on its terminals caused by the circuit then you can detect it simply with a continuity meter. Apply your multimeter probes on the circuit power lines and see if the buzzer sounds.

    Yes i'm not talking about this


    2.) However, if your circuit contains active components you cant do it because they may cause a short when a voltage is applied.

    Yes this is what i'm talking about only when voltage is applied

    3.) In this case use a resistor in series with the circuit, which in case of a short cant cause a current of more than 10 amps to flow (calculate its value depending on the supply voltage), and use your multimeter set on 10 amps range to measure the current flowing to the circuit. If it much higher than expected then a problem occurs. Be careful, because some power supplies, when a short circuits occurs, they limit the current to a safe value so you wont be able to detect if there is a short circuit.

    Is there another way than using a resistor in series because if i get the resistors value wrong than it will smoke the board or components

    I need a circuit board protection network , so if any voltage or current increases it will just CLAMP and regulate it back to the normal rating

    4.) How do you guys find SHORT when the circuit board is Active , having power ON ?

    Continuity checks RULES::

    a.) Can't use a DVM meter on continuity check mode when the applied voltage is ON only when the circuit board is off under test

    b.) Continuity check with a Analog volt meter set to OHM's doesn't help either ON or off when the applied voltage is ON because there is series/parallel circuit resistance

    c.) So its hard to Continity check with a Analog or DVM set to OHM's when trying to hind a short

    d.) Maybe using a 12 volt LAMP with low current to do continuity check when power is applied to the circuit board under test

    e.) Or Set your DVM meter to volts, apply power to the circuit under test and than probe around grounds,inputs,outputs,VCC , if you don't get any voltage drops than u might find a short but its still confussing to try to find no voltage drops around IC chips because if there is ONE 12 volt or 6 volt short if makes all the IC chips look like they all have 12 volt shorts because there is no voltage drop
  5. scubasteve_911

    scubasteve_911 Senior Member

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    Why not have the return path go through a small sense resistor, then use a differential amplifier off of the resistor, then run it into a comparator (with hysteresis)? The comparator is set to disconnect through a SSR or MOSFET. You can also run a circuit(LED?) to indicate when the SSR/MOSFET has switched.

    Steve
  6. Bailey45

    Bailey45 Member

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    The attached circuit limits when there is .7 volts acroos the emitter resistor independent of supply voltage.

    The transistor will require a heat sink.

    Attached Files:

  7. SgtWookie

    SgtWookie Expert

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    You might consider simply using an automotive light bulb. Light bulbs are unusual in that they have a relatively low resistance when they are cold, but increase in resistance considerably as the filament grows hot.

    If your circuit under test was not shorted, your lightbulb would glow dimly. If it were shorted, it would glow brightly.

    You would need to select the appropriate bulb wattage rating for the unit under test.
  8. relicmarks

    relicmarks Thread Starter Active Member

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    If i use a automotive light bulb on the output of a 12 volt power supply in "series" with the + red positive lead going to the DUI circuit under test
    and the circuit under test had a bunch of shorts or just one short its going to draw alot of current like 1Amp or so its going to still burn out the IC chips , PCB and other components

    Yes the automotive light bulb will give me a indication that its drawing alot of current

    But i still need a Current Limiter or something to STOP the circuit under test from or trying to draw alot of current since it has shorts on the board

    I don't want to put a fuse , because a fuse would just blow out and i can't test voltages or find the shorts when power is applied if it keeps blowing fuses

    example

    DUI circuit under test normal rating is 12volts at 40mA

    if the board has shorts its going to want to draw 1A, how can i current limit even if the board has shorts on it?
  9. scubasteve_911

    scubasteve_911 Senior Member

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    I still fail to see why my current sense solution isn't feasible. It will not limit current, but it certainly will be able to disconnect the load.

    Steve
  10. relicmarks

    relicmarks Thread Starter Active Member

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    If i use a current sense resistor i would need a comparitor circuit or detecting
    circuit plus it disconnects the load or circuit under test if any voltage or current goes "above" the threshold

    I don't want to Turn OFF the power if the current raises really high or draws alot of current cause there is shorts on the board

    Current sense resistor will turn the circuit under test OFF not keep it on

    Some type of "Current Divider circuit" is all i can think of to keep the current constant or a current regulator?

    A current regulator circuit that will always keep the source current out of a power supply at 100mA even if its trying to draw 1A or 800mA the current regulator circuit will not change
  11. scubasteve_911

    scubasteve_911 Senior Member

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    So, how does the constant regulator work then? It is impossible to control both voltage and current. I have seen overcurrent protection done with switching off and on the circuit, but how would you do that linearily while maintaining voltage?

    Steve
  12. relicmarks

    relicmarks Thread Starter Active Member

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    u know how a regulator works , if the LINE voltage drifts up or down the regulators output will not go up or down it stays constant

    If the LOAD voltage drifts up or down the regulator will keep a constant voltage at all time

    Current regulator will not supply more current even if the LOAD changes or draws more current
  13. SgtWookie

    SgtWookie Expert

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    If you have a current limited output on a voltage regulator, and the load suddenly starts drawing more current, the current limiter will reduce the voltage output until the current is within the specified limit.

    That's the way current limiters work.
    I=E/R
    E=IR
    R=E/I
    You're not going to be able to change Ohm's Law.
  14. relicmarks

    relicmarks Thread Starter Active Member

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    So using a High Value resistor in series and a series low current lamp bulb for indication with the power supply would "protect" and stop the DRAWING of current?
  15. SgtWookie

    SgtWookie Expert

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    A low-wattage 12V bulb should be sufficient to limit current, if the circuit under test was designed to run on 12v.

    Since P=EI, a 1.2W bulb would allow 100mA maximum current to pass through a dead short in the unit under test. If the circuit was not shorted, and required much less than 100mA, the bulb would glow very dimly, if at all.

    #1112 Miniature bulb, glass wedge base are 12V, 1.2W. They're sometimes used as side marker lamps.

    Here's a website that shows a bunch of different types of bulbs with their wattage/current ratings.
    http://www.bulbtown.com/12_Volt_Light_Bulbs_s/365.htm
    I am not suggesting that you purchase anything from them; I've never used them before. You can probably get what you need at your local auto parts store, and save on the shipping costs.
    Last edited: Nov 4, 2008
  16. relicmarks

    relicmarks Thread Starter Active Member

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    A low-wattage 12V bulb should be sufficient to limit current, if the circuit under test was designed to run on 12v.

    1.) How is it going to limit the current when the circuit under test is going to draw 1Amp of current?



    a 1.2W bulb would allow 100mA maximum current to pass through a dead short in the unit under test.

    2.) isn't that bad to pass 100mA of current through a Dead short? or what currrent rating is bad to pass thought a dead short?
  17. relicmarks

    relicmarks Thread Starter Active Member

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    Wouldn't the lamp bulb just PASS the 1 amp drawing current through the lamp bulb and not limit the current cause its to much current its drawing from the power supply?
  18. scubasteve_911

    scubasteve_911 Senior Member

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    I was merely stating that it is impossible to maintain a constant current and voltage. I am well aware, as should have been indicated by my post, of how these circuits operate. If one had an understanding each of each of these are, then they would know that both cannot co-exist.

    I have seen current limiters that work on a switching action. They will shutoff the load, then after a fixed off time, they will try to power the load once again. If overcurrent condition is reproduced, then it will repeat the cycle.

    Other limiters will linearly reduce voltage to the circuit to keep it fixed at a particular current; hence acting as a current source/regulator.

    What you are describing is two different circuit functions that need to be combined to produce the result that you are trying to achieve. #1) A voltage regulator at 12V #2) Upon overcurrent condition, apply a current source to the load. Implement this however you wish.

    Steve
    Last edited: Nov 4, 2008
  19. kubeek

    kubeek AAC Fanatic!

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    Look here http://forum.allaboutcircuits.com/showpost.php?p=95034&postcount=11
    This circuit is an overcurrent protection, voltage on the load R is full supply voltage until the current exceeds the set current, then it acts like a current source.

    The only missed components in that are zeners on the inputs of the most right opamp, and between gate and drain of the mosfet
  20. SgtWookie

    SgtWookie Expert

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    As the filament in the bulb heats up, it increases in resistance. If the circuit under test has a low resistance (12V across 12 Ohms=1 Ampere), the bulb will limit the current available to the circuit; it's resistance will increase to approximately 110 Ohms due to the filament heating up. Since the bulb is in series with the 12 Ohm "short", total circuit resistance will be 110+12=122 Ohms. Since I=E/R, and E=12, R=122, circuit current will be about 98mA.


    Letting 100mA pass through a dead short is far better than allowing 10A pass through a dead short! What you have to realize here is that the high-resistance light bulb will act as the primary load, instead of the low-resistance short acting as the primary load.

    You COULD use a semiconductor-based current limiting circuit instead. However, you would have to build it, and there would be fixed losses in such a circuit; for example an LM317-based current limiter has a 3v minimum dropout voltage that you would have to compensate for.

    Light bulbs are cheap, and they work.
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