The LT3080 is a current source based generator, that is, the reference voltage is set by the fixed 10uA flowing through the adjust pin and its external resistor. If so, I don't see (according to their schematic) why they say the input voltage is limited to 36V. Suppose for example I want to regulate a supply with 5V - 10V ripple at about 700V, and that I connect a 70MOhm resistor (approx.) to the adjust pin; how could the IC know that the input voltage is 700V since its reference is near the top voltage ?

  • \$\begingroup\$ Only the difference of voltages between the in and out pin imports here no ? \$\endgroup\$ – MikeTeX Aug 26 '19 at 18:32
  • \$\begingroup\$ Oh, I see what you mean. This is one of those ICs that has no ground. \$\endgroup\$ – DKNguyen Aug 26 '19 at 18:32
  • \$\begingroup\$ @MikeTeX It's basically a 3-pin device. All three pins must be within a range of at most \$36\:\text{V}\$. How would you ensure that with the recommended schematic? \$\endgroup\$ – jonk Aug 26 '19 at 19:08
  • \$\begingroup\$ It might have something to do with conditions on power-up. This is the only thing I think of. This would limit thing even if the 36Vmax is all technically relative to Vout. You might be able to push the startup envelope a bit, but I would think that the LT3080 could not withstand 700V even for a few microseconds. \$\endgroup\$ – DKNguyen Aug 26 '19 at 19:30

The part actually supports 40V, it says 0 to 36V because that is the voltage range that the specifications of the part will work. The part has been tested for 0 to 36V.

The absolute maximum ratings are ratings for which the part will fail you can't go more than 40V relative to Vout

enter image description here

So you could do what you suggest if the voltage drop across a resistor was always more than 660V, thus keeping the voltage drop across the LT3080 under 40V.

Looks like it could work depending on the load:

enter image description here

  • \$\begingroup\$ See my comments (and DKNguyen's) elsewhere on this question. \$\endgroup\$ – jonk Aug 26 '19 at 19:17
  • \$\begingroup\$ I don't understand your last edit with the picture: you haven't put the output capacitor and your load is only 100 Ohm (at 700V you get 4900W Uh). Also the set resistor is only 1kOhm. How can it work with these settings? \$\endgroup\$ – MikeTeX Aug 26 '19 at 19:39
  • \$\begingroup\$ There is nothing in the post about what current you need to regulate at or what your load is. The output cap adds nothing to the simulation, but would be required in real life, adding it doesn't make sense without the load. I was testing the idea of keeping the voltage within the range from a 700V source, which I didn't think was possible. Thats why I said "I think it could work", but I should edit my answer to I think it could work depending on the load \$\endgroup\$ – Voltage Spike Aug 26 '19 at 20:17
  • \$\begingroup\$ @VoltageSpike What about referencing the capacitor to the high side? Just as a starting thought..... \$\endgroup\$ – jonk Aug 26 '19 at 20:19
  • \$\begingroup\$ @jonk I guess it would depend on if you put the 3080 as a low side or high side current reference. If high side then that might be a good idea, if low side (like in the sch above) I'd keep it to ground. \$\endgroup\$ – Voltage Spike Aug 26 '19 at 20:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.