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I try to build a 300V voltage regulator using a LM317, which should be designed for 100 mA, but the circuit does not behave as I thought.

I loaded the circuit with 27K which is a current of 11 mA, and set the voltage regulator to 300 volt over the planned 22K in parallel with 1K. It took me a few tries until I found the right resistor. Some heat development at the transistor (heat sink) is not noticeable.

Then I bridged the output with a 5 kΩ resistor parallel to the 27 kΩ resistor which gives 4.2 kΩ and loads the circuit with 71 mA. But instead of the voltage remaining at 300 V, it breaks down by more than 100 V, and the resistor before the Zener diode planned 3 kΩ used 3.3 kΩ gives smoke signals.

Also with this load no heat development at the heat sink. At the transistor, 0.6 volts drop between base and emitter. Can someone give me an assistance there?

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  • \$\begingroup\$ What capacitor does it have at the output? Diagram doesn't show any, so maybe there's at least one at the remaining section(s). Nevertheless, I'd put at least 22uF/400V right across the output terminals. It'll help the output to be regulated better. Besides, make sure the voltage at the input terminal does not drop as you increase the load current. \$\endgroup\$ – Rohat Kılıç Jun 8 at 6:25
  • \$\begingroup\$ At startup the 3.3kOhm + zener + output resistor 27kOhm (not shown) + BJT form a voltage regulator, until the 317T ramps up the output voltage to 300V. During start up, the peak power across the 3.3kOhm equals approx \$((390V-15V)/(3k\Omega+27k\Omega))^2\cdot 3.3k\Omega = 377mW\$. With the lower output resistance (4.2kOhm), the peak power dissipation in the 3.3kOhm increases to \$6.15W\$. Although it is only the peak power, if the ramp up is slow enough and the resistor is not rated, it could be the problem you are facing. For the total power loss, you would have to integrate it. \$\endgroup\$ – vtolentino Jun 8 at 6:57
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If the 3k (3.3k) resistor1 is getting hot, it means that all of the load current is passing through it.2 If the transistor is working correctly, only a tiny fraction of the load current should be passing through this resistor, so verify that all of the connections to the transistor are correct. It's behaving as if its collector is open-circuit.


1 In the future, please add reference designators to your schematics. It makes it a whole lot easier to talk about specific components.

2 It's dropping about 36 V (0.4 W) with the 27 kΩ load, which (just barely) allows the LM317 to regulate. But it's dropping more like 150 V (6.7 W → smoke!) with the 4.2 kΩ load.

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  • \$\begingroup\$ easider -> easier \$\endgroup\$ – muyustan Jun 8 at 12:04
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    \$\begingroup\$ @muyustan: Thanks, fixed. I was typing that on my tablet, which is frustrating experience on the best of days. \$\endgroup\$ – Dave Tweed Jun 8 at 12:07
  • \$\begingroup\$ np, I wanted to fix myself but "edits must be more than 6 characters" was what I ended up with. \$\endgroup\$ – muyustan Jun 8 at 12:08
  • \$\begingroup\$ Dear Dave, That the transidtor is not properly controlled I thought so too. Is it possible that the Z-diode current is too low for the base current? Besides I use a transformer 230V sec X 1,414 =325V input voltage. That means: 300V output +15 V Zdiode -325V input = 10V for the transistor. Current through 3,3K =3,03 mA. I suppose that is too little. I have the possibility to add 20V on the secondary side of my transformer. That would be 230+20=250*1,414=353,5 V as input. Ps. I am not a professional i On the autodidact Translated with www.DeepL.com/Translator (free version) \$\endgroup\$ – RiSo Jun 8 at 12:55
  • \$\begingroup\$ No, the zener diode should not be conducting at all, except during start-up, when it protects the LM317 from overvoltage while the load charges up, or if the output is otherwise overloaded. During those times, the excess voltage is dropped across the transistor. Measure the voltage at the "in" pin of the LM317 to verify. \$\endgroup\$ – Dave Tweed Jun 8 at 13:13

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