24V to 5V conversion (output has current limit)

Discussion in 'General Electronics Chat' started by vijaybala85, Jan 27, 2010.

  1. vijaybala85

    Thread Starter Member

    Jan 7, 2010
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    hi Guys,

    just a basic question but I need to know if I am going in the right track. Please give me your opinion which circuit is better.

    @sgtwookie .. this is different from the earlier question. I looked at LM2675 and will be using that for later. For now, I would like your opinion on the 3 circuits I have attached in this thread.

    Thanks!

    V
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    The LM341T circuit:
    1) You have no input or output capacitors. The lack of an input cap can negatively affect the stability of the regulator.

    2) The regulator will be dissipating most of the power in the circuit, as it's a linear regulator - even though it won't be much power.

    3) Your 10mA fuse, with it's 3k resistance will allow only 1.666...mA current to flow from a 5v supply to ground, even if the other end of the fuse is shorted to ground.
    4) You have a diode in the current path. The diode will have a Vf, (forward voltage drop) depending upon the current flow through it.

    Using an opamp...
    Why?

    The resistive divider and diode - well, the voltage you get out of it will depend on your load current, and the Vf of the diode. With 24v in, you'll get about 7.333...V at the junction of the resistors, with no load. However, even if the junction of the two resistors were shorted to ground, there would be no more than 2.4mA current flow.

    24v / 10,000 Ohms = 2.4mA

    Maybe it would be easier to simply state what you are trying to do, rather than posting a whole bunch of circuits?
     
  3. vijaybala85

    Thread Starter Member

    Jan 7, 2010
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    thanks sgtwookie.... The reason I am apprehensive is that the device is expensive and I do not want to use a trial and error method regarding current specs as it may damage the device.

    I have a pressure monitor which is a digital switch. It works on 24V power and if it exceeds a certain power, It will send a 0V.

    I am trying to display that on NI's LabVIEW using DAQ card USB 6008 which requires Digital input to be within 5V and current within 5mA. otherwise the DAQ card will burn out due to excessive current.

    The fuse I am using is to prevent any excess flow of current because the DAQ is very sensitive to current specs.

    So, I am trying to use one of these 3 circuits, The previous person has used the LM341T without any capacitors and it seems to work fine. I am not sure if that resulted in the fault.

    REgarding diode, I am using it only for digital cases, If 5V is seen the diode is forward biased and 0V means reverse biased so that there are no floating cases

    Thanks!

    VJ
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    OK, then do something like this:

    [​IMG]

    R1 limits maximum circuit current to 24mA, even if there is a dead short to ground somewhere. It should be a 1W resistor. It is optional, but it will keep your regulator's temperature lower (thus more stable regulation) than if you omit it.

    C1 is mandatory. I have a couple of 7805 regulators that will oscillate at 3MHz and 8MHz without capacitors. You do not want that to happen.

    C2 is mandatory. Do not omit it. If you don't use it, you may experience high-frequency oscillations, and you may damage your expensive test equipment.

    R2 provides a constant 5mA load on the output of the regulator. This is mandatory to ensure guaranteed regulation. You can use a somewhat lower value if you wish, like 750 Ohms to 910 Ohms, but not much lower than that.

    R3 limits the maximum current through the DAC to 5mA, even if it is a dead short to ground. You might increase this slightly, but the higher input impedance you provide to the DAC, the less accurate your readings will be.

    Notice that there are no diodes in the circuit; you don't need them, and you shouldn't add them.

    There is nothing "extra" in the circuit that you don't absolutely need.
     
  5. vijaybala85

    Thread Starter Member

    Jan 7, 2010
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    thank you. I am not able to get a 0.33uF capacitor (330 nF) ... I have 2 0.1uF and i can maybe get 0.22uF. Ordering parts will take a couple days and I have to get this done by tomorrow. Thanks for all the help sgtwookie. I am really getting back in touch with electronics.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    You could use the 0.1uF on the output, and a 0.22uF and 0.1uF in parallel on the input. You would then be just fine.

    You would probably be OK with just one 0.1uF on the input and output. It's just that you shouldn't operate those regulators with NO capacitors on either input or output; they may very well oscillate.

    I am not guaranteeing that they WILL oscillate without the capacitors; that depends upon various factors as to how you've wired it up. But, if you follow the manufacturer's recommendations for capacitors on the input and output, it won't oscillate.
     
    Last edited: Jan 28, 2010
  7. vijaybala85

    Thread Starter Member

    Jan 7, 2010
    92
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    alrite then... This would work out perfectly. I just have one last question.

    In your circuit, how different is a condition when there is no R3 and when there is R3? And how different is the condition when there is R2 and no R2. What I am trying to understand is how we can manage current as per requirement so that in future, I will not have a doubt.

    1) with R2 and no R3
    2) with R2 and with R3
    3) no R3 and with R3
    4) no R2 and no R3

    what would happen in these for situations.?

    My guesses are:
    1) voltage will be 5V but current will be higher
    2) voltage will be 5V and current will be limited to maximum of 5mA
    3) voltage will be 5V ??? and current will be 5mA??
    4) voltage 5V and no idea in this case....

    Thanks!

    VeeJay
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    Why don't you model it in SPICE, or build the actual circuit, and see what happens?

    I have already explained why I feel that each component is necessary, and what it does.
     
  9. vijaybala85

    Thread Starter Member

    Jan 7, 2010
    92
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    oh cool! I just thought that your explanation in words would make more sense to me thats all.. No problem. Thanks a bunch for you help :)
     
  10. vijaybala85

    Thread Starter Member

    Jan 7, 2010
    92
    0
    oh by the way 78L05 is IC 7805 or do I use LM341T-5.0?

    V
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    The 78L05 is a low-power version of the 7805; it is capable of 5mA-100mA output.

    The LM341T-5.0 is a 500mA max output version, equivalent to a 78M05. You can use it as well.

    You could also use a 7805, which is a 1A version.
     
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