12V relay keeps tripping in multisim 11

Discussion in 'The Projects Forum' started by bobon721, Mar 9, 2011.

  1. bobon721

    Thread Starter New Member

    Feb 8, 2011
    18
    0
    I'm using a 12V 400ohm spdt relay for my power supply schematic's short-circuit protection. Upon simulation, it keeps tripping for a while, then returns to its normally open status. I used this model to represent what I bought. Input voltage is 24V so I put a resistor in series before the relay coil to protect it. Expected output is 5V at 3A max. How do I make the relay work well? By the way, the screenshot I attached only has the 5V rail. I still need to attach the +/-15V and variable voltage rails. My schematics for each rail are already complete, ignoring the relay problem.
     
  2. wayne2056

    New Member

    Mar 9, 2011
    1
    0
    Hi,
    Maybe it's the photo but D3 and D4 anodes don't look like they are connected anywhere except to each other.

    Regards
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    If your input is 24v and your output is 5v, then the I-O different is 19v.
    3A x 19v = 57 Watts. Your TIP32 is rated for 2 Watts power dissipation. Thus, it will be fried to a crisp in a very short period of time; probably under 10 seconds.

    It's very inefficient to use a linear regulator with such an I-O differential. While power expended in the regulator would be 57 Watts, power in the load is only 15 Watts (3A x 5V)
     
  4. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    Maybe I'm reading this wrong, but it looks as if the TIP32 is also connected very oddly - base to the 5V regulator input, emitter to output?

    There also appears to be a 2N2222 with a base bias resistor of 25Ω fed from the 5V regulator output. 200mA base current ?

    This all looks very wrong. How exactly did you arrive at this circuit?

    Edit: And the Tip32 is PNP, but its collector goes to the rectifier positive, at least when the switch is closed.
     
    Last edited: Mar 9, 2011
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    It's similar to National Semiconducotors' application notes in the datasheet for the LM317 for increasing output current; but there they use a 22 Ohm resistor.

    I didn't want to point out too many problems at once. ;)

    It would be better if there was a small current sense resistor with an opamp or comparator measuring the current - but the relay would still clack on and off, burning the dickens out of the contacts.
     
  6. bobon721

    Thread Starter New Member

    Feb 8, 2011
    18
    0
    @wayne2056:
    The nodes should be more noticeable now.

    @SgtWookie:
    I can't use an opamp here. The components present here are the limitations. The TIP32 is supposed to increase the current output of the LM7805CT. And yes I'm aware that there's a large I-O difference. Reducing the supply and transformer's secondary coil voltage is out of the question though since this is supposed to be a multiple output power supply with +5V, +/-15V and a variable voltage output with a max of at least 20V. I'm using the +5V rail as a prototype of the other rails.

    @Adjuster:
    Haha. I've rotated TIP32 horizontally. Also, it seems I misunderstood the 200mA base current rating in the 2n2222's datasheet so I reduced it.
     
  7. bobon721

    Thread Starter New Member

    Feb 8, 2011
    18
    0
    Here's the new screenshot.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Then you will have a room heater/smoke generator with a life span measured in seconds once a load is applied.

    Yes, I realize that. It also has a power dissipation rating of 2W, and you will be dissipating ~57 Watts between the 7805 and the TIP32 with a 3A load on the output. Now if you can manage to keep the case of the TIP32 at 25°C, then maximum power dissipation will be 40 Watts. However, you'll need a VERY large heat sink - but even that probably won't be enough. Liquid nitrogen might work, but it's rather hazardous.

    You'll either need to use a separate transformer or a switching preregulator to avoid the power dissipation problem.

    Your variable voltage output regulator will have an even worse power dissipation problem when it's set to output a low voltage at high current levels. It's the Vin-Vout x current = power dissipation that's the killer. With linear supplies, after a certain point, you just can't get rid of the heat fast enough.

    Now it IS possible to build pretty powerful linear supplies, but it requires multiple output pass transistors and large fan cooled heat sinks. It's not particularly cheap, and certainly not efficient.
     
  9. bobon721

    Thread Starter New Member

    Feb 8, 2011
    18
    0
    That sucks. Now I have no idea how to make this work without going against every specification presented to us.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Well, one other thing you might do is to place a high-wattage fixed resistor between the filtered DC supply and the regulator/transistor inputs. That way the resistor will dissipate most of the power instead of the regulator and transistor.

    Let's say you have 20v in, and 5v out. The regulator itself needs at least 2v differential between the input and output; more at higher currents. Let's say 3v more than the output; total of 8v.

    20v in, 8v out, so 12v. Since R=E/I, a 4 Ohm resistor rated for ~72 Watts power dissipation would likely do the trick (12v*3A*2) or you could use four 1 Ohm, 20W power resistors in series, with plenty of room for air circulation.

    Then connect your relay coil across the resistors, with another resistor in series with the coil so that you can adjust the threshold for pullin/dropout.
     
  11. bobon721

    Thread Starter New Member

    Feb 8, 2011
    18
    0
    I wonder what size and price those resistors come in. The largest power dissipation I've seen on a resistor was 5W for 100Ohm and it was close to an inch in size. I don't wanna exceed 100sqin for my PCB's area if I put in all the rails, nor do I want gigantic components. My transformer's already quite bulky.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Radio Shack carries 1 Ohm 10W power resistors in a 2-pack for ~$2.
    http://www.radioshack.com/product/i...0&filterName=Type&filterValue=Power+resistors

    1A through 1 Ohm resistance will produce a 1v drop, and dissipate 1 Watt of power.
    3A will produce a 3V drop, and 9W of power dissipation. You generally double the power dissipation requirement to ensure long life. The resistors will get quite hot when your regulator is sourcing current, so you will need to provide adequate spacing and cooling to keep them from burning up.
     
  13. bobon721

    Thread Starter New Member

    Feb 8, 2011
    18
    0
    Another group's been able to fix their relay problem, which was similar to mine. I asked one of them and he said their lm7805's Vin was 36V at 1A as well and they didn't even do anything to reduce the regulator's Vin or use a heatsink on any component. Their on actual testing phase by the way.
     
  14. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Wait until they try putting a 3A load on the output. The regulator and TIP32 will most likely go up in smoke, unless the regulator just plain shuts down due to overheating.
     
  15. bobon721

    Thread Starter New Member

    Feb 8, 2011
    18
    0
    I see. So overall, the problem lies in the voltage difference between the voltage regulator's input and output. I've bought generic resistors. The helpers in the electronics shop made me choose between 1/4 and 1/2 something and I chose the 1/2 ones. Are these 1/2 and 1/4 the power ratings? There's nothing on the resistors to give me a clue.
     
  16. bobon721

    Thread Starter New Member

    Feb 8, 2011
    18
    0
    Oh yeah. I made some computations based on your suggestions and made some changes. I'm still not sure of this though. That last screenshot was based on the changes made by my instructor to the schematic after I consulted him about this. I'm actually thinking he's playing with me because one of the things he changed was actually originally his suggestion. He never even mentioned anything about the power dissipation of the components I'm using. I included screenshots with all the other rails attached. I repeat, I'm still not sure about this.
     
  17. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The 1/4 and 1/2 designation for the resistors is wattage. Use Ohm's Law to calculate the power requirement for each resistor. For reliability, double the result as I showed a few posts back. Generally, you can use higher wattage ratings than are called for, but they get physically larger and more expensive.

    Let's see about your 20 Ohm resistor power rating. The TIP32 will start conducting when Vbe (voltage on the base referenced to the emitter) is around 0.63v, and that might get up to around 1V-1.2v when conducting maximum collector current.

    I=E/R, so 1.2v/20 Ohms = 60mA max current through the regulator. P=EI, so 1.2v *.06A = 72mW, x2 = 144mW. 1/4 Ohm would be OK there.

    The way you're using the transistor & base resistor to drive the relay coil will provide an action opposite to what you want.

    When the voltage on the output of the regulator(s) starts coming up, the transistor will turn on, causing the relay coil to energize and the input current path to turn off. It'll never be able to get up to normal operating voltage.

    However, if the drain on the output of the regulator was really heavy, like a direct short, the transistor would not turn on.

    You really should be using the relays on the unregulated input side, developing the voltage across the power resistors as I mentioned previously.

    Determine what the pull-in (operate) voltage is for your relay coils. Just as a rough example, a 12v coil might operate at around 9.5v. Once energized, the contacts will stay open until the voltage across the coil drops to perhaps 1/2 to 1/3 the rated voltage.

    So, if you were using a 12v relay that operated at 9.5v, you'd need to drop that voltage across a resistor due to excessive current flow into the regulator.

    9.5v / 3A = 3.16 Ohms. 3 Ohms is pretty close to that.

    I'm really running out of time for projects on here.

    Do some calculations for power dissipation requirements like I've been showing you. Show the results to your instructor, and ask him about them. Print out datasheets for the components, and show them the power ratings for ambient temperature (like 2W for the TIP32) and when the case is kept at room temp (like 40W for the TIP32).

    You have to do the research and ask the questions.
     
  18. bobon721

    Thread Starter New Member

    Feb 8, 2011
    18
    0
    I don't quite get some parts but I'll read them over and over to get them. Thanks a bunch.
     
Loading...