How to get supply output down to zero volts

Discussion in 'General Electronics Chat' started by blah2222, Aug 25, 2014.

  1. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Hello,

    Like thousands before me I am moving from the life of batteries and wall-warts and onto the excitement of building my own variable bench supply. I am interested in the design process so I figured I could give this a try.

    It seems as though the LM317 seems to be a common choice, but I have some discrete transistors and zeners lying around and would like to make use of them.

    It seems as though discrete designs using only diodes and transistors (sometimes op-amps) are fairly complex and take up a fair amount of space on a schematic.

    Might this be because the circuitry to get the output down to zero volts is that much more complex and also relies on temperature regulation as well, increasing the parts count?

    JP
     
  2. alfacliff

    Well-Known Member

    Dec 13, 2013
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    if you posted a circuit diagram, we could tell you where to put in some diodes to offset the output to zero. basicly, you offset the ground point of the supply to make the output go all the way to zero, since the regulator will not go down that far.
     
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Use an opamp powered with split-supplies in the voltage feedback loop to drive the Ref Pin of the regulator chip to -1.25V.

    Read the app notes supplied with the various regulators at TI's web site...
     
  4. eetech00

    Active Member

    Jun 8, 2013
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    Hi

    What is output voltage range desired?

    What is maximum load current desired?

    eT
     
  5. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    For my purposes a 0-25V supply would work well. At some point I'd like to figure out a current limit circuit but I don't really need more than 0.5 - 1A.

    What are thoughts on the attached circuit. The potentiometer R6 connected to the negative feedback of the op-amp through R5 allows Vout to go down to zero. Would need to get a centre-tapped transformer to get the negative rail for the op-amp. Or just a rail-to-rail op-amp. Some kind of output transistor for load current would need to be setup as well.

    <br />
V_{out}(x) = 5[1+R_{1}(\frac{1}{R_{4}} + \frac{(1-2x)}{R_{5}+x(1-x)R_{6}})]<br />

    If R6 << R5

    <br />
V_{out}(x) ~= 5[1+R_{1}(\frac{1}{R_{4}} + \frac{(1-2x)}{R_{5}})]<br />

    <br />
V_{out}(x=0) = 5[1+R_{1}(\frac{1}{R_{4}}+\frac{1}{R_{5}})]<br />

    <br />
V_{out}(x=1) = 5[1+R_{1}(\frac{1}{R_{4}}-\frac{1}{R_{5}})]<br />

    Where 'x' is the turn ratio of the potentiometer R6.
     
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    Last edited: Aug 25, 2014
  6. ScottWang

    Moderator

    Aug 23, 2012
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    In series the bjt to the output and in parallel the pot with the bjt, and then you can adjust the current.
     
  7. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Something like this? This technique seems feasible but I feel like fluctuations in the rails would causes the diode drops to change quite a bit, giving a poor reference.

    I understand that pulling the ADJ pin to -1.25 would work with a precision -1.25V regulator, but I am looking into feedback-driven regulation down to zero if that makes sense.

    Is there a reference diagram you can post that goes along with your post?
     
  8. #12

    Expert

    Nov 30, 2010
    16,270
    6,787
    Look at the VI curves for the 1N4001 series. In the 10 ma range, it takes a 5 to 1 current difference to produce a tenth of a volt of change. Are your rails going to change from 30 volts to 150 volts?

    PS, look at the LM723 chip for some good design ideas.
     
  9. ScottWang

    Moderator

    Aug 23, 2012
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    Here are four kinds of circuits, they are translate from google, you can choose like this one.

    [​IMG]
     
  10. Lestraveled

    Well-Known Member

    May 19, 2014
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    1,215
    ......what??.......no......That don't work.....

    Q3/Q4, Q1/Q2 form constant current sources.........Sure they will absorb some voltage but how does this make an adjustable power supply that will regulate to zero???
     
  11. ScottWang

    Moderator

    Aug 23, 2012
    4,853
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    Yes, it won't work, because this wasn't designed for zero Vout.
    I just want to suggest the current control, if he want to adjust the Vout to zero, it needs some other works.
     
  12. ScottWang

    Moderator

    Aug 23, 2012
    4,853
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    The following two circuits will be output less then zero voltage.

    [​IMG]

    [​IMG]
     
  13. Lestraveled

    Well-Known Member

    May 19, 2014
    1,957
    1,215
    . . . . . . .
     
  14. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
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    A simple LM317 circuit will do 1.25 to 25v no problem, and there is very little need to every go below 1.25v. Even when powering battery devices with your bench supply nothing uses LESS voltage than a 1.5v cell!

    Your need to go to 0v causes a lot of issues and you simply may never need that ability.

    Regarding LM317 and current limiting, I have had good success putting another LM317 as a current limiter BEFORE the voltage regulator LM317. That way you get independent control of current limit and the regulated voltage. Regulated voltage remains exact until the current limit is reached.
    :)
     
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  15. crutschow

    Expert

    Mar 14, 2008
    13,001
    3,229
    Below is the simulation of a discrete power supply using a single-supply op amp with the pot going from 0 to 100% and RLoad stepped from 15Ω to 30Ω in 5Ω steps. It goes to from 0V to 25V output from a 30V rectified supply and is current limited to a little over an amp. Q1 needs to be on a healthy heat-sink of course.

    But I agree with RB that an LM317 based power supply should meet most general bench power supply requirements, is more bullet proof, and is much simpler to implement.

    Pwr Supply.gif
     
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  16. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Could the same current limit approach you used with R7/Q3 be used to proceed and bypass an LM317, i.e. PNP collector attached (through a resistor) to Vout with emitter attached to unregulated supply and base attached to input of LM317 with R7 between base and emitter?

    Also, would there be any benefit to making the regulator with differential discrete transistors vs. an op-amp?

    Thank you
     
  17. crutschow

    Expert

    Mar 14, 2008
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    It could work but you can't "bypass" the LM317, you need to limit the current at its input. For that you'd have to add another transistor in series with the input, preferably a P-MOSFET (or use another LM317 as RB suggested). Also note the LM317 is inherently current-limited at its maximum current rating.
    None that I know of.
     
  18. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    I dont care much for the circuits posted here. Here is my version. Simulated only, not actually built, but I have high confidence in this LTSpice simulation.

    The way I offset the Adj terminal of the LM317 to -1.25V is by using a TL431 shunt regulator, which intrinsically creates a stable -2.495V. The current that flows in R1, R2(the adjustment Pot) and R3 is essentially constant, so to effectively subtract half the -2.495V reference voltage, I just added R3, which must be the same value as R1 because the internal reference inside the '317 is 1.25V.

    With R1=R3=205Ω, a 5K Pot adjusts the output voltage from 0V to 30V.

    The simulation includes a transformer, rectifiers, capacitors. Note that with a Center Tapped Transformer, the required unregulated negative voltage for the TL431 reference is developed.

    In the first simulation plot, I show the output voltages vs time as a function of the pot value. Note that with the transformer shown, the unregulated + voltage is ~34V with ripple. With the Pot set to 5K, the output voltage is ~31V, which is right at the point where the LM317 drops out of regulation, as evidenced by the ripple on the output...

    There is also a start-up glitch (when AC is first turned on) which is not trivial to get rid of. It is caused by a difference in time when the Unreg+ first appears vs when Unreg- appears.

    The second plot shows the dissipation in the LM317 as a function of driving a 25Ω load resistor and as a function of the output voltage, Obviously it would be much worse to ask the power supply to deliver a few mV at 1A where the dissipation in the 317 increases to ~35W.
     
    Last edited: Aug 26, 2014
    blah2222 likes this.
  19. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    In your note (bottom left) do you mean disregard R6 and R7 instead of R4?
     
  20. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Yes. Glad you caught that. LTSpice sims of transformers where there is no resistive shunt path ahead of the rectifier diode take forever.
     
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