# Wide Input Voltage Range Constant Current Source

I would like to design a Constant Current Source of 25mA, With Input varying from 2V to 50V. Please provide just an idea how to get about this.

• What kind of accuracy and output impedance do you need? – John D Dec 17 '15 at 16:25
• Can you use a current sink or does it have to be a source? Make sure you know what each of those means. – KyranF Dec 17 '15 at 16:52
• @Kyran It has to be a source – user1589759 Dec 17 '15 at 17:40
• Specify what load resistance range you need for this to work when the supply (input) is at 2V and at 50V whilst respecting ohms law! – Andy aka Dec 17 '15 at 23:24

LM317HV, 50 ohm resistor, couple of caps, done.

simulate this circuit – Schematic created using CircuitLab

The LM317 (or LMx17) in current source mode works as a current source since the LM317 makes the voltage at its Adj terminal 1.25 V less than the voltage at its Out terminal. So 1.25V = 50 ohms / 0.025 Amps. You need the HV part due to your 50V input range. The capacitors tend to keep the regulator happier - values can vary.

Per the TI datasheet suggested applications for current source mode, C1 could be 0.1uf and C2 could be omitted.

Per the moved goalposts, replace the 55v source with a 55V source consisting of a (probably multi-stage) boost converter to make 55V from 2-50V. Details of implementing that left to whoever wants to see where the goalposts move next. Presumably some sort of initial sense of the applied voltage will be required to determine where to connect the input to the stages of the booster, based on looking at input ranges for a few parts that might nearly do the job, but which have issues with the full range.

• I feel that this answer is not of sufficient effort, nor explanation. – KyranF Dec 17 '15 at 16:57
• Very good, thank you. It may be useful to link the datasheet of an example of this IC, like this – KyranF Dec 17 '15 at 17:10
• This circuit is not able to meet the requirements. The LM317HV has a dropout voltage of roughly 1.5 V at 20 mA, so the minimum input voltage for this circuit is 2.75 V (and then you can drive only a short). – Arsenal Dec 17 '15 at 17:23
• This is true @Arsenal perhaps the OP's requires are not quite so strict that this solution may still be valid – KyranF Dec 17 '15 at 17:30
• @user1589759 A 2K load means you will need a supply voltage exceeding 50V to get 25mA! – Spehro Pefhany Dec 17 '15 at 18:16