0 -5 V DC Signal

Discussion in 'The Projects Forum' started by Oceanamie, Apr 17, 2014.

  1. Oceanamie

    Thread Starter New Member

    Apr 17, 2014
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    Hello, world!

    I have a general question.

    I have a 60 V DC power supply, and I want to get a 0 - 5 V DC signal out of it. So the 0 V would be 0 V on the power supply, and respectively, the 5 V DC would be 60 V DC on the power supply (of course).

    I know I need to create a voltage divider circuit, but my question is; which Ohms value should I have, and at what watt rated resistors? The amperage will be changing, but typically would be set at 2 A DC.

    Please advise.
     
  2. MaxHeadRoom

    Expert

    Jul 18, 2013
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    Is this a Linear (large power transformer) or switching supply?
    Max.
     
    Last edited: Apr 17, 2014
  3. Oceanamie

    Thread Starter New Member

    Apr 17, 2014
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    It's just a bench laboratory analog DC power supply with a constant current/constant voltage option.
     
  4. MaxHeadRoom

    Expert

    Jul 18, 2013
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    The reason I asked, is the first thing I would look at is if you could add a small overwind that would be capable of the 5v 2a supply, this might be the most efficient way.
    You should only need 12-15 turns.
    Max.
     
  5. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    The amperage of what would be changing??? That to me adds some confusion to your question..

    At first guess I'd think you simply have some sort of metering device but its full scale voltage is only 0-5V or some datalogger or whatever that you want to use to relate to the labs fully 0-60V output voltage.
    Is that what you are doing?
     
  6. Austin Clark

    Member

    Dec 28, 2011
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    A resistor divider in this case would be extremely inefficient, and would give you wildly varying voltages depending on your load (which, as you said, will be changing). A resistor divider is mostly used with very small and constant loads, like an Analog to digital converter, or an amplifier input.
     
  7. inwo

    Well-Known Member

    Nov 7, 2013
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    A pass transistor with base connected to a 12/1 V divider?

    "I want to get a 0 - 5 V DC signal out of it"

    Only a signal??
     
  8. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    There are a couple of ways to do this, giving you a regulated voltage output that does not change significantly with variations in load, and with an output voltage that tracks as a percentage of the input voltage. The hard part is getting it to regulate down to 0 V, for two reasons. First you must have a small negative voltage power supply so your circuit has a common mode range that truly extends to ground. But the real problem is that your regulator is powered from the source it is regulating, and when the source is turned down to zero there is nothing to power the regulator.

    One alternative is to leave the lab supply at 60 V and just turn down the control voltage to the regulator. If the 60 V is across a 100K pot, that's less then 0.1 W dissipated in the pot and you have a full 60 V adjustment range. The wiper of the pot goes to the control input of the regulator circuit that is powered from the constant 60 V even when the control is turned down to zero.

    Another way is to do what you stated in your post, combined with inwo's suggestion in #7. A fixed resistor divider followed by a complimentary darlington emitter follower (PNP followed by NPN) will work for an input voltage down to about 3 V, or 0.25V output.

    The problem with that approach is that you need a relative low value resistor to provide enough base current at low Vin, and that resistor will have 54 V across it at 60 Vin. At 60 Vin and 2 Aout, that's about 36 W in the resistor, plus 110 W in the output transistor.

    What is the output driving?

    ak
     
  9. crutschow

    Expert

    Mar 14, 2008
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    It would help greatly if you would explain exactly why you need to do this.
     
  10. Oceanamie

    Thread Starter New Member

    Apr 17, 2014
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    Apologies about the confusion, but I would be testing this at 2 A DC constant current and then after a few weeks, 3 A DC constant current.

    This is precisely what I'm trying to do. The data logger can only handle an analog signal from 0 - 5 V DC. We need to calculate the resistance of the load, therefore if we know what the voltage is and at constant current, we calculate via Ohm's Law.

    Only a signal because it's use for a data logger unit.

    Thanks for the information. The output isn't driving anything; it's meant for data logging purposes. We know if the data logger is at 5 V, then we say it's 60 V DC and so on and so forth down to 0 V (accordingly). A linear trend, obviously.

    See above.
     
  11. BobTPH

    Active Member

    Jun 5, 2013
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    In that case, you just need a simple voltage divider with a ratio of 1:11 to divide by 12. The values of the lower resistor should be < 1/10 of the input impedance of the data logger. 1K and 11K would probably work just fine, but would draw 5mA at 60V.

    Bob
     
  12. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    The amount of current whatever your lab supply is providing has NO bearing on the 0-5V part and as such caused all this confusion above..

    You simply need a 1/12 voltage divider and done..
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html
    start with like a 10K for R1 and go from there..
    adding a zener to overclamp isn't a bad idea either.


    My "reading between the lines" was strong that day.. :)
     
  13. Oceanamie

    Thread Starter New Member

    Apr 17, 2014
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    So, a 1/4 W resistor would be fine, right?
     
  14. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    depends on the values chosen for the resistors but yes a 1/4W should be just fine assuming your resistors are 1000 ohms or more.. even 1/8th should be fine
     
  15. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    If you know the input impedance of the data logger, you can build a simple 2-resistor voltage divider with a Thevenin equivalent impedance 10% or 1% of that, and connect directly. What is your error budget for the measurements?

    ak
     
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