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I need to generate a fair bit of current (up to 2A peak, much less average) in the 6V-15V range for driving MOSFET gates, and also generate 5V (less than 100 mA) for an MCU, from an input that may be between 6.5V and 60V. I'd like the output of the 6-19V regulator to be ~6V when input is 6.5V, and ~15V when input is >= 16V, and then not go over 15V as voltage goes up.

I've found a number of promising candidates on Digi-Key, but none that has the combination of ultra-low drop-out and high current capacity that I want. For example, the MAX5024 can generate my 100 mA for 5V from a high input voltage, and only has 0.9V drop-out; it can also be used to generate 11V output (which will be less with less input voltage) but only up to 100 mA, and using two of this becomes expensive. 11V is also in the lower end of the gate drive voltage I want to use when the input voltage is high -- 15V would be better.

There are also some cheaper parts (like some Intersil part) that generate 5V and 15V, but require 12V and 20V input respectively, and only generates 30 mA for the 15V output.

Perhaps I can use a reservoir capacitor after the 11V output to supply the transient current of switching the MOSFET gates on/off, and get around the 100 mA limit? As long as the time switching is less than a handful of percent of the total duty cycle?

An alternative would be to use a lower-voltage ULDO (with 16V or 35V max input) and somehow extend its working range by using some offsetting Zeners or resistors in some clever combination -- but I can't quite work out what that would be that would let me extend like that while still having the very wide (> 50V) input range span.

Any advice would be appreciated!

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  • \$\begingroup\$ 2A through an integrated linear regulator is not advised, especially with such a large output range. I'd recommend building it from discrete components instead. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Apr 22, 2014 at 2:42
  • \$\begingroup\$ @JonWatte In the first sentence, you wrote "... for driving MOSFET gains...". Did you mean ... for driving MOSFET gates..." ? \$\endgroup\$
    – Nick Alexeev
    Commented Apr 22, 2014 at 3:06
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    \$\begingroup\$ If you have 60V input generating 15V @ 2A through a linear regulator, said regulator will be dropping 90W. You'll need quite a heat sink \$\endgroup\$
    – DoxyLover
    Commented Apr 22, 2014 at 3:12
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    \$\begingroup\$ @DoxyLover I think Jon needs 2A peak only for gate driving. The average current would be smaller, and the heat dissipation requirement could be more manageable. \$\endgroup\$
    – Nick Alexeev
    Commented Apr 22, 2014 at 3:17
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    \$\begingroup\$ Why don't you simplify this question to the bare basics. Say what you want and don't clutter it with how you think it can be done. \$\endgroup\$
    – Andy aka
    Commented Apr 22, 2014 at 7:25

2 Answers 2

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To increase input voltage range of your supply with linear regulator you should use some step-down converter.

Thats the only reasonable method to "loose" more than 10-20V at 2A peak current.

Correctly designed simple step-down converter will decrease voltage with little power losses and heat generation.

MAX5024 has low current because you get 6W of heat at 100mA and 60V difference between in and out pin. Thats a lot. For 2A it would be 120W :) So forget about loosing 30-60V on linear regulator with 2A current requirements. Thats insane.

There are many integrated switching regulators (like easy to use LM2574 family), but I can't find any capable to deliver 2A output with 60V input. Maybe someone will suggest something.

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    \$\begingroup\$ As already stated in the question, the 2A draw is peak while switching gates, not sustained. \$\endgroup\$
    – Jon Watte
    Commented Apr 23, 2014 at 15:26
  • \$\begingroup\$ LM2574 has internal, pretty fast current limiter that limits current to 0.6A or something like that. Maybe it would work with huge capacitor after regulator, but its better to find something bigger. \$\endgroup\$
    – Kamil
    Commented Apr 23, 2014 at 15:29
  • \$\begingroup\$ @JonWatte Maybe create another question about "60-80V input, 15V@2A output easy to implement switching regulator" \$\endgroup\$
    – Kamil
    Commented Apr 23, 2014 at 15:32
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    \$\begingroup\$ @JonWatte If 2A is peak - you should estimate your RMS power on device and base your calculation on this. \$\endgroup\$
    – Kamil
    Commented May 23, 2014 at 22:07
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Use a switch-mode buck regulator to get 6 volts from your 6.5 to 60 volt input. From that, a simple linear regulator gives you 5 volts for your MCU. Use a switch-mode boost regulator to get your 15 volt gate drive and you'll have it whatever the input voltage. The boost regulator will need large output capacitors anyway, so right there you have the transient current reservoir.

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  • \$\begingroup\$ I am familiar with switch-mode regulators, and quite like them for many cases (there are ready-made modules for a variety of cases.) In this case, board space, robustness, and cost are all different from the cases where I've normally used buck regulators in the past. \$\endgroup\$
    – Jon Watte
    Commented Apr 23, 2014 at 15:27
  • \$\begingroup\$ Fair enough, but don't discount the cost and board space needed to get rid of the heat from a linear regulator. \$\endgroup\$
    – Graham Davies
    Commented Apr 23, 2014 at 16:52

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