AC to DC power supply without an IC

Discussion in 'The Projects Forum' started by MHDS, Jul 24, 2015.

  1. MHDS

    Thread Starter New Member

    May 25, 2015
    20
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    Hello everyone! I want to make a 12 V regulated dc power supply without IC

    The first thing I thought out is to use a zener diode after installing a CLC filter. And then use a pot to vary it from 5V to 12 V.
    My first Q is: Can there be a much better approach without an IC?
    And the 2nd one is: How can I make some arrangements in this ckt so that when I adjust the pot from till exactly 6V, an Led lights up? Similarly exactly at 5V, another one lights up???

    Plzz help... Am a newbie in the world of electronics
     
  2. Veracohr

    Well-Known Member

    Jan 3, 2011
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    You can make your own regulator with discrete components but it probably won't work as well as an IC regulator.
     
  3. crutschow

    Expert

    Mar 14, 2008
    13,013
    3,233
    Using a pot to directly regulate the output voltage does not work well due to the impedance of the pot, which cause poor regulation of the output voltage versus load current.

    The common, easy way to make such a supply is to use an LM317 adjustable regulator.
    It can deliver over an amp of current and is internally protected against over current and over temperature.

    To cause the LEDs to light up at a particular voltage you could use a comparator, such as the LM339. It comes 4 comparators to a package.
     
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  4. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    Don't forget the transformer...
    As required by the terms of this site.. "transformerless" power supplies are a banned topic (unless that stopped too when the automotive restrictions were lifted)

    What you are asking about is VERY dangerous for newbies and should only be attempted by professionals only..
     
  5. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    IF you have a transformer with a secondary voltage of 9 VAC or more, THEN this can be done for each output with one zener and one transistor for poor regulation, or one zener and two transistors for better regulation (plus some R's and C's).

    No transformer, no help.

    ak
     
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  6. MHDS

    Thread Starter New Member

    May 25, 2015
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    No, I wasn't thinking of a transformerless power supply, but an IC less. And yeah, I was thinking of zener diode but the idea of transistors wsn't known to me, as supported by " Analog Kid". Can you please help me out in this regard in terms of its biasing & the schematic of the circuit??
     
  7. ScottWang

    Moderator

    Aug 23, 2012
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    How many current you need?
     
  8. absf

    Senior Member

    Dec 29, 2010
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  9. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    A couple of questions:

    1. No IC's but - IC like? That is, a voltage reference, differential amplifier, current amplifier, output voltage feedback divider, all of the major circuit functions of an LM78xx series regulator. Or do you want to use a feedback circuit that is familiar but not nearly as common, and definitely not a clone of IC innards?

    2. feedback or no feedback? Both circuits in #1 use negative feedback to stabilize the output voltage. Another approach is much more simple, but has less precise regulation. This approach has a zener diode with a pot across it driving a Darlington emitter follower/current booster.

    ak
     
  10. ScottWang

    Moderator

    Aug 23, 2012
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    I tested that kind of circuit before, it's easy, but since the TS don't want to use the IC and this circuit is more complicated than IC, so I think the TS will give up this.
     
  11. absf

    Senior Member

    Dec 29, 2010
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    MHDS likes this.
  12. ScottWang

    Moderator

    Aug 23, 2012
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    2SC1061 spec:
    Pc = 25W
    Ic = 3A
    If you want to use this bjt then the heatsink must be bigger or use the Ic of bjt = 5A or some more, even counts the Pc to 25W and when the output = 30V and then the current will not enough 1A, so the Pc should be counts it less than 8W, it means that when you want to get a 1A output and then the Pc at least must be has 90W or some more, otherwise the bjt will be getting hot or the heatsink must be use bigger one, probably using the fan to reducing the heat.
     
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  13. MHDS

    Thread Starter New Member

    May 25, 2015
    20
    1
    Approx 500mA
     
  14. MHDS

    Thread Starter New Member

    May 25, 2015
    20
    1
    Okay.. Thanks. I got your approach.
     
  15. MHDS

    Thread Starter New Member

    May 25, 2015
    20
    1
    Okay. I got better circuits than mine from all of you. One last doubt remains.. I want to use it in circuits so I would require less than 1A. If I use resistors to limit the output current, will I need additional heat sinks so as to stop any physical damage by excess heat dissipation? Or shall I require none??
     
  16. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    It's good that you are thinking ahead to issues like this. Resistors are difficult to add heatsinks to, plus a series resistor makes for lousy voltage regulation with changing currents. You already have a power transistor or two handling the voltage regulation. With some additional parts you can have them handle the current limiting also.

    If this power supply is to have an adjustable output voltage, you can add a current limit control circuit that is independent of the output voltage setting, so for example you could have a 1/2 A max output current at 3 V, 5 V, 9 V, etc. This can take as little as one transistor and one resistor, depending on which voltage regulator circuit you decide to use.

    The worst case power dissipation in the power supply happens when the supply output is shorted to GND. For a maximum output current of 1 A, the current limit circuit would dissipate about 0.7 W, so a 2 W resistor will be fine without any additional heatsink.

    P = E x I (Watt's Law)

    For example, if the power supply is adjusted to 12 V out and 1 A max. current, and has 18 Vdc across the bulk filter capacitor, that is 6 V across the regulator for 6 W of heat to be dissipated in the pass transistors. When you short the output to GND, now there is 18 V across the regulator at 1 A, for 18 W to be dissipated (minus the power dissipated in the current limit circuit as above, less than 1 W). These are completely reasonable numbers for the circuits we are discussing.

    ak
     
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  17. ScottWang

    Moderator

    Aug 23, 2012
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    You may want to know if you just using the pot to do the voltage adjust, how the situation is?
    Wirewound Variable Resistor.

    If you using 10Ω/25W to cross on the 12V and then the resistor will flows through current as:
    I = V/R
    = 12V/10Ω
    = 1.2A
    Although you can get the current from the pin 2 of pot to get the current, but you still have to waste the current flows through from 12V to Ground, if you choose the other pot as 100Ω and then the current will not enough.

    Another method is that you can in series the pot with the load, but this method will related with the voltage divider, when the pot values is too small and then you can't adjust it drop down too much, it means that maybe you can't get a lower voltage, if you using a 100Ω pot and then the current will not enough to use.

    And if you want to input 12V through the voltage regular to get 12V output, that is hard to use IC to do or even using the bjt, probably using one or two stages of bjts and one stage of 5A P mosfet can be considered.
     
  18. MHDS

    Thread Starter New Member

    May 25, 2015
    20
    1
    Ok.. Thanks. I got your views. I will design this ckt & post it here.
    But will you clear one more doubt, that why the first capacitor from the input side should have more values than the output one? And what will be the optimum values of the capacitor and the inductor?

    My ckt simulator Multisim is giving an ideal output in case of high values & low values too.
     
  19. ScottWang

    Moderator

    Aug 23, 2012
    4,854
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    Where is the 12V came from? (AC110V/220V to 12Vdc?)
     
  20. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,538
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    The bulk filter capacitor (after the rectifier diodes, before the regulator) usually is the largest capacitor in a power supply. You determine its size by working backwards from the maximum desired output voltage, plus the minimum headroon across the pass transistors in the regulator. This gives you the minimum capacitor voltage; if the negative peaks of the AC ripple on the bulk capacitor dip below this number, that ripple shows up on the output of the regulator. Next is the capacitor's peak voltage, the upper peaks of the ripple. this comes from the transformer's secondary voltage (RMS x 1.414) minus the voltage drop across the rectifier diodes. The positive and negative peaks of the ripple plus some arithmetic gives you the minimum value of the filter capacitor. Most power supplies do not have series inductances because of size, weight, and cost. But a series inductance "vens out" the ripple waveform peaks and can reduce the size of the filter capacitor.

    A large filter capacitor on the output can cause stability problems for the regulator circuit in some cases. Also, it is an unregulated source of huge surge currents. For example, a power supply with a 1 A current limit and a 10,000 uF capacitor on its output easily can supply 10 A or more into a wiring problem for a few milliseconds, more than enough time to blow some expensive parts.

    ak
     
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