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LM723 PSU with 0V lowest voltage

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by earckens, Aug 26, 2016.

  1. spec

    spec Well-Known Member Most Helpful Member

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    POST ISSUE 2 of 2016_12_12

    Hi earckens,

    Below is my version of a power supply, current mode indicator (a touch of input offset voltage and hysteresis will be required for a practical implementation).

    spec

    2016_12_11_Iss1_ETO_LM723_PSU_CURRENT_MODE_INDICATOR.jpg
     
    Last edited: Dec 12, 2016
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  2. spec

    spec Well-Known Member Most Helpful Member

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  3. earckens

    earckens Member

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    LM723-labPSU-0V-46V-v2.png View attachment 103080 View attachment 103052 View attachment 103029
    ok spec, now I did this: I made a hybrid from your drawing 2016_12_08_LM723_POWER_SUPPLY_V4_MOD1.png and my idea of using a series voltage regulator with a Darlington TIP120. I also introduced your substitute values for the voltage and current feedback circuit (18k to 47k and 82k to 230k, twice).
    The value for R1 seems unchanged but I still have to update this after your post this morning about SOA.

    Would you mind having a critical look at it please?

    Thks, Erik

    PS: somehow this drawing now appears on top of this post??? ..big enough? o_O

    Edit 13/12: corrected errors; replace 470R between Vc pin 11 and emitter 2SC1815 with a short, and add R11 (re. post 135) View attachment 103079

    Edit 6/01/2017: replaced the series voltage regulator with a more stable version based on http://www.skillbank.co.uk/psu/vr2.htm, in the process introduced Q1 in the regulator.
     
    Last edited: Jan 6, 2017
  4. dave

    Dave New Member

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  5. spec

    spec Well-Known Member Most Helpful Member

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    Hi earckens,

    In terms of circuit function, your new circuit and your previous circuit are the same.

    The LM723 only needs around 14mA so the Darlington transistor is just adding to complexity and reducing performance- sorry.

    Your previous schematic was OK.:confused:

    spec
     
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  6. earckens

    earckens Member

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    spec, I think I was confused, I'll be back to the drawing board and review our last weeks' posts;
    Erik

    Edit: I forgot to mention, I also have to power a MCU and LCD display from the regulator, those are not shown here because they are beyond the scope of this thread.
     
  7. spec

    spec Well-Known Member Most Helpful Member

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    Yes, best to get the fundamental power supply working and then you can add on as many bells and whistles as you like.:cool:

    spec
     
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  8. earckens

    earckens Member

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    Hi spec, somehow I had missed this post. Where do IN- and IN+ come from? -comp and +comp form LM723? And is this with a LM393 (not very legible)?

    Erik
     
  9. spec

    spec Well-Known Member Most Helpful Member

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    Hi earckens,

    Yes, that is correct.

    Yes, an LM393 dual comparator. Sorry about the confusing sketch. I was not feeling too good yesterday- too much amber nectar the night before.:D

    The principle of the constant current mode indicator relies on the fact that if ever the IN+ (+comp) input of the LM723 is more negative than the IN- (-comp) input, the power supply must be in constant current mode.

    spec

    PS: I have made the sketch a bit clearer- I hope:)

    DATASHEETS
    http://www.ti.com/lit/ds/symlink/lm393-n.pdf
     
    Last edited: Dec 12, 2016
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  10. earckens

    earckens Member

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    Great! I will implement this one. Simple, effective and more economical than a software solution (time needed for programming, testing, calibrating, wiring.. ).

    You need some more of this amber stuff, really enhances your thinking :hilarious:
    Thks,
    Erik
     
  11. spec

    spec Well-Known Member Most Helpful Member

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    Hi Eic,

    The circuit is only an outline and, like I said, needs a bit of hysteresis, but that is no big deal.

    I went to town today and got some more amber nectar.:)

    spec

    PS: By the way you need a 100nF ceramic capacitor directly across the supply pins of the LM723 in your power supply schematic.
     
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  12. earckens

    earckens Member

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    Hysteresis on LM393: R1 from output to +IN, R2 from +IN to GND?

    Say Vth (threshold, output voltage)=Vcc (18Vdc)*R2/(R1+R2), so if R1=18k and R2=82k (so that I can recuperate my previously discarded resistors from your calculations of my voltage- and current feedback circuits) then Vth=3.25V. Correct?

    ..by the truckload I guess? :joyful:
     
  13. spec

    spec Well-Known Member Most Helpful Member

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    Nice logic and calculations earckens.:)

    I will give the hysteresis some thought- there are quite a few factors to take into account. The main factor is that the current limit indicator function must not degrade the fundamental voltage stabilization function.

    The most important thing is that we have what appears to be a workable architecture for the current limt indicator.

    Just a 20 pak... for today.

    spec
     
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  14. earckens

    earckens Member

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    Which previous schematic do you refer to spec?
    I could use your straight zener regulator but I need a bit more current for a MCU and LCD; I toyed with your idea of keeping it simple but I decided to go for the darlington regulator because it still is a straightforward circuit and I would need it anyway.

    I did some homework and verified the previous weeks' posts: my drawing of post 163 differs from yours', spec, in that I did not follow your use of a series zener for the voltage regulation (I use a darlington), and the voltage and current feedback resistor values have been corrected per your posts 113 ansd 119. Instead of a BC146 (Q6) I use 2SC1815 which is on hand here (Vce=50V, a bit too low, will be changed later to a better transistor).

    Started up at 46Vdc and boem, the voltage regulator bust in smoke. Then tried the transfo separately but the primary fuse (1A-f) kept blowing, put in a 5A-f but then all house lighting down :nailbiting:. I suspected an earth fault, removed the transfo and tried again with just AC and no rectifier and all ok. Checked the rectifier (50A beast) and AC to "minus" diode bust.
    Did find a wiring error though :banghead:; too embarassed to say where.

    Waiting for a new rectifier to arrive, but Friday we leave on holiday so the works will be for after the 28th.
    Erik
     
  15. spec

    spec Well-Known Member Most Helpful Member

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    Hi earckens,

    The previous circuit (and best) is shown by the .png file attached to your post #135 as opposed to the later circuit of post #163.

    The two VBEs of a Darlington will seriously degrade the regulation of the 18V Zener diode.

    Not only that, but the extra current will be flowing through the current sense resistor and will mess up the current sense accuracy. Also, the processor and LCD, being digital, will introduce a whole load of hash into the precision voltage control and current control sensing.

    There are three axiom in engineering:
    make it modular
    keep it simple
    never compromise the core function

    So it would be much wiser to keep the core voltage regulation functions isolated and separate from the processor and LCD.

    That is a good omen because it means that god is in a playful mood and just generating obvious faults, rather than more arcane problems.:D

    By the way, we have all done similar things- you are not a true power supply man if you haven't had a few explosions/fires.:p

    It would be too embarrassing to say what I have done in the past.:arghh:

    Another holiday- seems like you have a good life in Belgium.:)

    spec
     
    Last edited: Dec 13, 2016
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  16. earckens

    earckens Member

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    spec, could you send me a copy of that one, i am afraid not to have it anymore?


    ...o_O.. spec, can you please explain?

    ..I could also use a straight TIP41C..? ..or this (http://www.skillbank.co.uk/psu/vr2.htm)

    [​IMG]
    Edit: example Vz1=7.5V, R1=2.2k, R2=56k, R3=47k, Vin=46V, Vout=about 18V, TR1=TIP41C

    Ok; but now I am out of ideas: how would you then supply the extra 250mA max (10V to 20V or so: I use a LM4040-5.0V for the last stage)?

    Thks, Erik
     
    Last edited: Dec 13, 2016
  17. spec

    spec Well-Known Member Most Helpful Member

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    Hi earckens,

    I never did have a copy of that schematic; it is a schematic that you drew in EAGLE and attached to your post 123. It is still there- just click on the shortcut.

    I would use a separate mains power module, like this: http://www.befr.ebay.be/itm/5V-600m...808143?hash=item4ae58d144f:g:qAYAAOSw-0xYSCqw. There are many different types on Belgium EBay.

    You can then connect the 0V output of the auxiliary mains power supply without worrying about current going through the current sense resistor.:)

    spec
     
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  18. earckens

    earckens Member

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    Hi spec, thanks for your answer; however: post 123 is by Les?

    And did you have some thought about my calculations for the hysteresis on the LM393 (your post 172)?

    You also mention that the 2 Vbe's of a darlington will degrade the reulation of a zener: could you please explain that?

    I will consider putting a separate mains module; space is also a consideration though.

    Thks,
    Erik
     
  19. spec

    spec Well-Known Member Most Helpful Member

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    Sorry, post #135

    Afraid not - I have been a bit busy the last few days and today is missus birthday.

    The Zener is a nice stable voltage- that is what it is designed for but if you connect a Darlington's base to the Zener and take the output from the Darlington emitter, the two emitter/base junctions of the two transistors in the Darlington will introduce a varying voltage which will degrade the voltage stability. I hope I have read tour circuit correctly!

    By the way, it is advisable to be cautious when using integrated Darlingtons because of their very low frequency response and and a tendency to generate distortion.

    That's good. If you look on eBay, you will find some very small (and cheap) power supplies.

    spec
     
    Last edited: Dec 15, 2016
  20. earckens

    earckens Member

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    I see, spec; but (apart from the fact that I use a straight zener series regulator) there I still have a 470R R12 between pin 11 (Vc) and emitter of BC546: should that not be eliminated?
    And change R15 (base of pnp TIP42C to emitter BC546) from 2k2 to 1k?

    Grts
    Erik
     
  21. spec

    spec Well-Known Member Most Helpful Member

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    Quite correct earckens.

    It's configuration control that gets you in the end.:)

    spec
     

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