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| Potentiometer burning |
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| Ian.M:
N.B. A potentiometer's power rating assumes a uniform current through the whole track. If you are only using 10% of the track length it only has 10% of the power rating. Divide the wattage by the total resistance and take the square root to get that max rated current. If you exceed that current through any terminal (including the wiper) you'll burn out the pot sooner rather than later. Assuming a 1W rated potentiometer, with a 100K pot, its good for max. 3.16mA. With 47V across it your pot only has 0.47mA goint through its track. That isn't much to supply base current for your triple Darlington emitter follower, especially as one generally doesn't want to draw more than 10% of the track current at the wiper. If you switch to a 10K pot, its going to have 4.7mA through it and is good for an abs. max current of 10mA. The down side is the greater load on the Zener shunt regulator circuit, as the 2K2 resistor only provides 8.6mA. Looking at the OnSemi 1N4756A datasheet, its tested at a nominal 5.5mA Iz, and 3.9mA remaining for the Zener will be good enough. When you rapidly turn the output voltage down, the previous voltage the output capacitor was charged to causes Vout to exceed Vwiper. If that's by more than 15V or so, all the B-E junctions will Zener, and pop goes your pot and probably the weakest transistor as well. The easiest fix is to put a diode in series with the pot wiper to block reverse current. As Andy points out above, you'll also need a B-E resistor for each transistor to handle its worst case C-B leakage current, and with the wiper diode in place, the B-E resistor also prevents the B-E junction zenering. You've still got the problem of protecting the transistors against the surge current during startup, or if the load draws too much current. However adding effective current limiting to the design becomes quite tricky, as it would need to be fold-back limiting to protect the transistors effectively, and it rapidly turns into a full-blown high current high voltage bench PSU design exercise which is far from being beginner friendly. As an alternative, if all you need is variable 'raw' DC for amplifier testing, consider scrapping everything to the right of the fuse and simply putting a variac on the AC input! |
| GerryR:
--- Quote from: Jwillis on August 28, 2019, 05:24:21 am ---The MJE340 has a broad range of gain (30- 240) with a 10k resistor at the base gives around 4.7mA at the base when 100k pot is at zero ohms....... --- End quote --- You just lost me here. When the 100k pot is at zero ohms, isn't the base, even through a 10k resistor at ground, therefore no base current?? Just asking. |
| GerryR:
You can get a hi-voltage, hi-current op amp, OPA548, or similar, for ~$10 to $15, set up as a unity gain amp. Tie the input to the wiper of your pot and use the output to drive your transistor. You will have to make some changes to your circuit, but this will isolate the pot from the driver section. You should be able to get within a couple of volts of the rails on each end. Just a suggestion. |
| Ian.M:
An OPAMP isn't an easy option with a 66V supply rail. If you are going to the trouble of either getting a HV OPAMP, or providing a floating supply for a regular one, you might as well design a proper regulator using the OPAMP with feedback from the output. |
| jmelson:
--- Quote from: Ian.M on August 28, 2019, 06:05:23 pm ---An OPAMP isn't an easy option with a 66V supply rail. If you are going to the trouble of either getting a HV OPAMP, or providing a floating supply for a regular one, you might as well design a proper regulator using the OPAMP with feedback from the output. --- End quote --- There are such items, like the LTC2057HV that can go about that high. Very nice op amp for high voltage supplies. Jon |
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