| Electronics > Beginners |
| Potentiometer burning |
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| AngraMelo:
Hello guys, Im making this project, it is an unregulated power supply to use to test amplifiers. The issue is that my 100k pot keeps burning. I have no idea why. I used the 2k2 resistor to limit the current to the pot but after using it for a couple of minutes de pot burns and sometimes the MJE340 goes short as well. I have no idea what is going on. could you help me? |
| GerryR:
I believe you need a different base drive circuit. When the pot gets anywhere above about 8 volts (worst case Vbe for all the transistors set up in the Darlington configuration), you have no base current limit and you load your 47 volt source down. When you get near the top of the pot (actually anywhere above about 8V, the 47 V is bypassed because it is shunted by the base-emitter junctions. It's like putting an 8 V zener in parallel with your 47 volt zener. Just a quick glance :). |
| AngraMelo:
--- Quote from: GerryR on August 27, 2019, 11:32:22 pm ---I believe you need a different base drive circuit. When the pot gets anywhere above about 8 volts (worst case Vbe for all the transistors set up in the Darlington configuration), you have no base current limit and you load your 47 volt source down. When you get near the top of the pot (actually anywhere above about 8V, the 47 V is bypassed because it is shunted by the base-emitter junctions. It's like putting an 8 V zener in parallel with your 47 volt zener. Just a quick glance :). --- End quote --- Thank you Gerry! Could you please explain why is it 8V? Also, I tried putting a 10k resitor between the wiper and the base of the mje340 and now the pot is not burning anymore but the 340 or even a Tip142 are going short. Should I limit the current on the collectors of the 340 and tip41 with resistors? I dont understand why even using a 10k resistor on the base of the 340 can make the all the following transistor going short. They are all mounted on big heatsinks and even with a load around 500mA they are still going bad. |
| GerryR:
You have the emitter of one transistor tied to the base of the following transistor in a darlington configuration. If you look at the worst case Vbe drop for each transistor, they add up to about 8 V. I don't know what you are trying to accomplish(??), a variable voltage output supply (??), and at what current?? If that is what you are trying to do, you need to change your approach. |
| Andy Watson:
You need to allow for the possibility of leakage current between collector and base/emitters of the transistors. This is usually achieved by placing a resistor across the base-emitter junctions of the driving transistors - aim to sink a few milliamps or more such that it takes care of any leakage. The leakage will inccrease with increase in temperature. I believe that this is the mechanism by which the output voltage can rise above the "set" voltage. Whatever the mechanism is, you need to mitigate the effects of the output voltage being greater than the set voltage - the MJE340 can only stand 5V reverse voltage between emitter and base. Attach a diode between the output and the base of the MJE340 to prevent its base becoming negative with respect to the output. And put a series resistor between the potentimeter and the base - to prevent excessive output voltage feeding back, through the diode and into the potentimeter. |
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