Project: Double adjustable power supplies

Discussion in 'The Completed Projects Collection' started by cumesoftware, Apr 27, 2007.

  1. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    I have two projects about similar power supplies. They are internally regulated, adjustable between 2 to 13,5V. The output is symmetrical, that is, one is positive and the other is negative. Both can supply up to 1,25mA (give or take). I used one LM317 and one LM337 to regulate the output voltage. Also, to protect them from shorts, I had to add fuses (incredibly, the LM317 and LM337 are not shortable, unlike other regulators (althought it is refered in their respective datasheets that they are) - I tested myself with regulators from National Semicondutor).
    The version 2 has some improvements relative to version 1, such as the LED's are driven by transistors, so their luminance will not vary very much. Also, in this version we have internal adjustable resistors, so the voltages can be locally fine tuned (mostly because of different room temperatures).

    View attachment FADR0213V1A.pdf
    List of components (version 1):
    C1/3 – 6,8mF electrolytic capacitor (35V);
    C2/4 – 1μF tantalum electr. capacitor (25V);
    D1-4 – 1N5400 rectifier diode;
    D5/7/8/10 – 1N4001 rectifier diode;
    D6 – Red LED (GaAsP/GaP);
    D9 – Green LED (GaP);
    F1 – 630mA slow fuse;
    F2/3 – 1,25A fast fuse;
    IC1 – LM317T voltage regulator (with 1,2°C/W heatsink);
    IC2 – LM337T voltage regulator (with 1,2°C/W heatsink);
    Lp – 230V~ resistorized neon bulb;
    R1/12 – 71,5Ω metal film resistor (1% tol., 1/8W);
    R2/13 – 42,2Ω metal film resistor (1% tol., 1/8W);
    R3-7/14-18 – 115Ω metal film resistor (1% tol., 1/8W);
    R8-10/19-21 – 28,7Ω metal film resistor (1% tol., 1/8W);
    R11/22 – 680Ω carbon resistor (5% tol., 1/2W);
    S1 – DPST switch;
    S2/4 – 6 position SP selector;
    S3/5 – 4 position SP selector;
    Tr – Transformer 230V~ 2x15V~ 80VA.

    View attachment FADR0213V1A(2).pdf
    List of components (version 2):
    C1/3 – 10mF electrolytic capacitor (35V);
    C2/4 – 1μF tantalum electr. capacitor (25V);
    D1-4 – 1N5400 rectifier diode;
    D5 – Red LED (GaAsP/GaP);
    D6/7/9/10 – 1N4001 rectifier diode;
    D8 – Green LED (GaP);
    F1 – 630mA slow fuse;
    F2/3 – 1,25A fast fuse;
    IC1 – LM317T voltage regulator (with 1,2°C/W heatsink);
    IC2 – LM337T voltage regulator (with 1,2°C/W heatsink);
    Lp – 230V~ resistorized neon bulb;
    Q1 – 2N3904 NPN transistor;
    Q2 – 2N3906 PNP transistor;
    R1/16 – 680Ω carbon resistor (5% tol., 2W);
    R2/17 – 1,5KΩ carbon resistor (5% tol., 1W);
    R3/18 – 95,3Ω metal film resistor (1% tol., 1/8W);
    R4/6/19/21 – 20Ω preset resistor (5% tol., 1/8W);
    R5/20 – 48,7Ω metal film resistor (1% tol., 1/8W);
    R7-11/22-26 – 169Ω metal film resistor (1% tol., 1/8W);
    R12-14/27-29 – 42,2Ω metal film resistor (1% tol., 1/8W);
    R15/30 – 18KΩ carbon resistor (5% tol., 1/8W);
    S1 – DPST switch;
    S2/4 – 6 position SP selector;
    S3/5 – 4 position SP selector;
    Tr – Transformer 230V~ 2x15V~ 80VA.
     
  2. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    Althought this is a finished project, comments and constructive critics are apreciated so they can be applied in future versions.
     
  3. lightingman

    Senior Member

    Apr 19, 2007
    374
    22
    Hi...The bench PSU is probably the most usefull piece of equipment, yet professional variable bench supplies are expensive (even on E-bay).The project shown here is is very good, and ideal for most applications..I will be designing soon a fully digitaly controled dual PSU with constant current control, controled by a PIC, with LCD readout of volts, amps and watts.It can be used both as single or trackable output mode.I will place all of the details here when it it done, including the schematics, PCB layouts and code...Be patient... Rome wasn't built in a day (that's cos I wasn't the foreman !!).Daniel.
     
  4. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    Thanks for your comment!

    Well, I tried to cover the most used voltages. The problem with these two power supplies it that they are big in size for most applications (and also they tend to spend more energy than the necessary). Probably I will bet on having a switch mode power supply, so I can cut down in the size of the transformer. Also it would be ideal for the transformer having 5 or 7 terminals, so I can reduce also the size of the heatsinks.
    As you can see, I looked little to component costs, so this power supply has maximum safety under the working limits. It was designed to work at 250V AC, and prepared for the maximum load of 1,25A in both terminals, having the positive terminal 2V and the negative terminal -2V. In these conditions, the heatsinks will heat up to 70ºC with a room temperature of 40ºC.
     
  5. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    Nevertheless, if I were using other regulators like the LM7805, I would use fuses not to protect them, but to protect the transformer and the diodes from the bridge. They can past 1A with no problem without burning or, more seriously, without shutting down. They only have a thermal shutdown. Controling this shutdown by using a "correct" heatsink (not to big or to small) it is a problem since the room temperature is unpredictable. So, it is better to use fuses or some kind of external protection, so the shutdown happens with a precise current. Also, it is a bad idea to lead a big heatsink associated with such regulator to extreme temperatures (up to 120ºC, the shutdown temperature). Better to have a small heatsink doing that.
     
  6. mozikluv

    AAC Fanatic!

    Jan 22, 2004
    1,437
    1
    what's the lowest load you expect to connect to your psu?

    moz
     
  7. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    That question is a bit odd, but I would say a load of infinite resistance (the PSU is prepared for all loads up to 1.25A with no problems).

    Nevertheless, the maximum heat dissipation will occur if you have two loads of 1.25A, and if you have both terminals set at the lowest voltage possible (2V for the positive terminal and -2V for the negative terminal). That situation is predicted and the PSU is prepared to hold such a stress and for a long time (days if needed).
     
  8. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    I think Moz was asking about lowest impedence of load.
     
  9. mozikluv

    AAC Fanatic!

    Jan 22, 2004
    1,437
    1
    thingmaker3,

    thanks for the clarification, it seems my fingers can't keep up anymore with my mind.


    cumsoftware,

    since you're making an all purpose bench supply, i believe you should also highly consider the lowest load impedance you can connect. say you have a load of 10ohms and need to have a ripple voltage of say only 200mv would your psu still be sufficent?

    moz
     
  10. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    Well, since this is a regulated power supply (IC1 and IC2 are voltage regulators, LM317 and LM337 respectively), the ripple won't pass to the output, even with the highest output voltage possible, In fact, the fuse will be blown before the output gets unregulated (that is, the lowest voltage between ripples goes below 13.5V + 2.5V (I don't remember the minimum voltage differencial but I think its 2.5V)). So, the lowest load impedance would be 1.6Ohm at 2V. If you raise the voltage the fuse gets blown.
     
  11. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    Here are further instructions, for both version 1 and version 2:

    Positive terminal:
    S2 - Coarse voltage selector (base voltage) - +2V, +4V, +6V, +8V, +10V, +12V
    S3 - Fine voltage selector (add to base voltage) - +0V, +(+0.5V), +(+1V), +(+1.5V)

    Negative terminal:
    S4 - Coarse voltage selector (base voltage) - -2V, -4V, -6V, -8V, -10V, -12V
    S5 - Fine voltage selector (add to base voltage) - +0V, +(-0.5V), +(-1V), +(-1.5V)

    So, if we put S2 in the position 3 and S3 in the position 0, we will have 6V + 0V = 6V in the positive terminal. In the same way, if we set S4 in the position 3 and S5 in position 4, we will have -6V + (- 1.5V) = -7.5V.

    Notice:
    All voltages are relative to the ground terminal.
    In version 1 the potentials may differ a little from the "should be" values and they cannot be corrected.
    In version 2, fine adjustments will be needed throught the presets R4, R6, R19 and R21. When presets are properly adjusted, maximum precision will be achieved (an enhancement relative to version 1), but if the presets are not even adjusted, the output voltages will differ largely from the "should be" values.

    Warning:
    For the sake of the circuitry being fed, never switch any of the selectors with the circuit connected. This will cause spikes in the output voltage because the selectors are breaking the circuit when they are being switched (unless you use expensive make-before-break selectors).
     
  12. mozikluv

    AAC Fanatic!

    Jan 22, 2004
    1,437
    1
    hi cumesoftware,


    "So, the lowest load impedance would be 1.6Ohm at 2V. If you raise the voltage the fuse gets blown."

    that's a valuable piece of info for guys out there who might want to use your circuit:)

    moz
     
  13. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    You have to be more specific. I don't quite get your sarcasm.

    Also, the PSU is rated for amps, not for ohms. It is said the limit is 1.25A. You don't even have to worry about ripple since the PSU is regulated. What more information you might need? Did you looked at the circuit?
     
  14. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    To calibrate the positive end of the power supply, take the following steps:
    - Switch the selector S2 to position the 1 (2V) and S3 to position 3 (+1V);
    - Adjust the preset R6 until you get a tension of 3V between the positive terminal and the ground terminal (if that is not be possible, adjust until you get the nearest value possible);
    - Switch S2 to position 6 (12V) and keep S3 in the same position (+1V);
    - Adjust the preset R4 until you get a tension of 13V between the same terminals;
    - Repeat the previous steps until the tension is 3V with S2 in the position 1, and 13V with S2 in position 6 (you will have to repeat these steps at least once).

    The steps will be similar when calibrating the negative end, having in account that:
    - The tension is now measured between the negative terminal and the ground terminal;
    - The selectors S4 and S5 are homologous, respectively, of the selectors S2 and S3;
    - The presets R19 and R21 are homologous, respectively, of the presets R4 and R6.

    Translated with some strange fish and corrected over. Sorry!!
     
  15. mozikluv

    AAC Fanatic!

    Jan 22, 2004
    1,437
    1
    sorry if you feel that way,:( there is no sarcasm intended with my questions, am just extracting more infos from your project.:)

    moz
     
  16. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    Sorry about the misunderstanding then.
     
  17. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    I'm here to announce that these projects have errors, unnoticed until now. By the time, I considered the junction to case thermal resistance (TO-220) to be 0.3ºC/W instead of 5ºC/W. This means that, even considering an infinite heatsink, the temperature of the junction will rise above 125ºC, and the heatsink temperature will not rise above 70ºC, seeming to work fine (I normally consider that Tcase < 70ºC and Tjunction < 110ºC for an ambient temperature of 40ºC. This also give me margin to work at Tamb = 50ºC, since with an increase of 10ºC, the temperatures will still be acceptable.).

    All other projects requiring heatsinks are not affected by this error, due to the nature the following considerations that I normally use:
    Heatsinks are calculated for Tcase < 70ºC as long as Tjunction doesn't go higher than 110ºC. The formula is:
    Rs-a= (Tc - Ta) / q - Rc-s

    If Tjunction does go higher than 110ºC, then:
    1- I consider Tjunction to be the limiting factor, so it should be < 110ºC (this happens with higher wattages, and in those cases, Tcase might be well below 70ºC). The formula is:
    Rs-a= (Tj - Ta) / q - Rj-c - Rc-s;
    2- I reduce the dissipated power when step 1 is impossible.

    So this means that for lower wattages, any similar error around junction to case resistance does not affect the calculations. That's why all the other projects were unaffected. The solution here is to reduce the dissipated power of the regulators, with the implied redesign.
     
  18. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    cumesoftware,

    Thanks for the clarification.

    Would you like me to copy this into the "Double adjustable power supplies" thread as a record of this information?

    Dave
     
  19. cumesoftware

    Thread Starter Senior Member

    Apr 27, 2007
    1,330
    10
    Yes. This is a really severe error and the thread should be copied.
     
Loading...