# Power suply 300V regulated with LM317

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?

• 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. – Rohat Kılıç Jun 8 '20 at 6:25
• 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. – vtolentino Jun 8 '20 at 6:57