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I want to make a voltage regulator for a symmetric power rail +30/-30V. The idea is to use an LM317 and an LM337.

I want to be able to regulate with a Vin that can go up to +60V (and -60 and negative rail).) The LM317 can only have a voltage drop between IN and OUT of 40V. In case of a short that value would be exceeded. After some digging I found that I can do a pre-regulator as in the picture below.

enter image description here

Once the voltage VIN is above about 36V, the Zener diode would kick in and the BJT will do it's thing.

My issue is tgat I don't know how to size that pre-regulator.

More specifically, how do I size R9?

I think it's R9= (Vzener-Bbe)/(Izener+Ibase), but Izener depends on temprature and voltage accross the Zener diode. I'm not sure about the Ibase.

Important note (maybe) my load is around 25mA. D1 is just there for reverse polarity protection.

EDIT 19/04. As Vangelo pointed it out, with the current disign there is a issu in case of a short on Vout. So I added this simple circutry and it seems to be working. I found this here https://jeangaillat.wordpress.com/2018/03/29/je-suis-fou-des-alimentations-lectroniques/amp/ the 1200 resistor is to simulate my charge at 25mA. enter image description here

Normal operation: enter image description here

Short: enter image description here

And for the negative rail: enter image description here

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  • \$\begingroup\$ Think about what happens when R9's value is too large: there will not be enough current through the zener (see zener's datasheet what (knee) current is needed). Also, if R9 is too large, Q1 might not have enough base current. You know the supply current so you can derive Ib from that. What if R9 has a value that is too low? Then too much current will flow through the zener, it will get hot. \$\endgroup\$ Apr 16, 2021 at 9:38
  • \$\begingroup\$ Unless you're planning to draw only single, maybe two digits of milliamperes: almost certainly the approach of using a linear regulator here is a wrong way to start, unless you have plenty of heatsink to spare, anyway. \$\endgroup\$
    – mmmm
    Apr 16, 2021 at 9:40
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    \$\begingroup\$ Important note (maybe) my load is around 25mA. That is not a "maybe" the maximum load current can make or break this design. At 25 mA you're probably not going to dissipate that much heat in the components but do a calculation of that anyway! 36 V at 25 mA is 1 Watt so enough to fry a small component. Of course, some power goes into the load but consider where the rest will go. \$\endgroup\$ Apr 16, 2021 at 9:42

2 Answers 2

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By selecting the correct current for the resistor and the Zener for the nominal operation (as you indicated in your text), everything works well, and the power dissipated by these two components can stay comfortably below 1/4 W. Depending on the Zener and resistor power rating, there is a lot of room between a resistor with a value sufficient to provide Zener regulation and value which reaches the power limit in any of these two components. Base current will not be too relevant since the load current is very low (for a low beta, e.g. 50, Ic = 30 mA -> Ib = 600 uA).

enter image description here

But, as you can see @500 ms, when the output is shorted, the current through the resistor is increased (if the input voltage source is capable of providing the short circuit current). The power dissipated by the resistor reaches almost 1 W (beyond the plot scale).

enter image description here

The worst happens to the input BJT (again, if the input source is capable of providing the short circuit current). As the LM317 protects itself, the BJT becomes a soldering iron:

enter image description here

Edit: good to know your circuit is fulfilling your requirements. If the load regulation is affected by the shunt resistor, you may also consider a limitation before the BJT:

enter image description here

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    \$\begingroup\$ Tanks for your great input. I added I few components to do some current regulation. I foud it here : jeangaillat.wordpress.com/2018/03/29/… It seems to be working for me. \$\endgroup\$
    – Neeko
    Apr 19, 2021 at 10:16
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You can fix Izener about 5 mA, and check zener power. Next step is to consider Q1 current. If you select Q1 Ie current of 50 mA, when a short happens, the regulator will drop more current, so most all the power will be dissipated by Q1. In case you select Q1 Ie above LM317A current limit, power will be splitted proportional to voltage drop. Once you have selected Q1 and Ie current, this current divided by Q1 current gain + Izener, will help to calculate R9. When you know R9, check again Izener in normal condition (in case you considered an Ie higher than load current).

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