Discrete voltage regulator (review circuit pls!)

Discussion in 'The Projects Forum' started by stefan.54, Mar 28, 2016.

  1. stefan.54

    Thread Starter New Member

    Dec 26, 2015
    Hey guys,
    Long story short: I have to build a voltage regulator (stabilizer it is named in my language) with the following demands:

    *discrete parts only (no op-amps nor IC regulators)
    *Precision: 5%

    I've been looking through a lot of books and on the net, I found this nice circuit (look below).

    What I find odd is the lack of capacitors.
    Also, I have problems calculating the resistances.

    Obs: Qs is the load transistor, Qa is the amplifier tr. and Qprot is the short-circuit protection.

    Any ideas, suggestion or piece of advice is welcomed.

  2. ronv

    AAC Fanatic!

    Nov 12, 2008
    You have a good start -- you have a pretty good spec.
    So let's do it a little bit at a time. I think what I would do first is pick a transistor as the Qs. Since you know your load is 3 watts you can calculate the current thru the transistor. I would make it higher than needed just so you don't have to worry so much about tolerances and "stuff". So go on line and find a good transistor that can handle .5 amps or so and has a Vce of 25 volts or more. You can use one capable of more current and voltage. Since it will dissipate quite a bit of power (needs calculating) it should be in a package that is easy to attache to a heat sink. Here is a site with a pretty good search engine:
    Let's do this first and go from there.
  3. Lestraveled

    Well-Known Member

    May 19, 2014
    Stehan, one of the problems with a linear regulator (stabilizer) is that the power that is not used by the load is dissipated by Qs as heat. So, the power into your circuit is 15V x .375A = 5.6 watts. Your load consumes 3Watts, so, 5.6 - 3 = 2.625 watts. When you are looking for Qs you will need one that can at least this much heat. Also 2.6 watts is a lot of heat for a single part so, Qs would require a heat sink.

    Good luck
  4. ronv

    AAC Fanatic!

    Nov 12, 2008
    There is not much room between 10 volts and 8 volts, so the gain of the transistor is also going to be important. Take a look at the D44h11.
    Last edited: Mar 28, 2016
  5. AnalogKid

    AAC Fanatic!

    Aug 1, 2013
    That is a classic regulator circuit, and your schematic has all of the right pieces in place. It is not super accurate because the current through the zener diode changes when the load current changes. However, it will be much better than your 5% spec.

  6. stefan.54

    Thread Starter New Member

    Dec 26, 2015
    I found around the Uni lab a couple of BD135 and some BC107A which may seem to be good. Any opinions on them?
  7. ISB123

    Well-Known Member

    May 21, 2014
    BD135 should handle the power requirements with heatsink.
  8. ronv

    AAC Fanatic!

    Nov 12, 2008
    If it is the high gain one this circuit is okay.
    To figure the resistor for the big transistor the voltage is 2 volts minus the Vbe ~ .65 volts so 1.35 volts. If the gain is 100 minimum and you want 375 ma out you need 3.75 ma of base current. So the resistor is 1.35 / .00375 or 360 ohms. Use 330 to be safe.
    For the current limit it is similar. If the drop across the resistor is ~ .65 volts it will start to turn on which in turn will turn off the big transistor. A value of 1.5 ohms would start to turn it off at 433 ma.
    The resistor for the zener needs to supply the zener current at the output voltage of 8 volts. If you use a 1/4 watt zener it might be 10 ma or so.
    It would work a little better without the LED to indicate over current, but it will probably survive a short for a short time.
    absf likes this.