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Is it possible to deliver a linear voltage regulator with the following specficiations:

  1. input reference voltage is from 0.8V to 1.4V
  2. output voltage has to follow the input reference voltage, it is the voltage connected to the load
  3. deliver about 1mA to the load regardless of the output voltage, that is, when the output voltage changes from 0.8V to 1.4V, the output current has to be constant at 1mA.

In my understanding, if the loading is unchanged, say fixed at 6 ohm, then increasing the output voltage will cause an increase in output current as well. So how to make the output current constant at 1mA?

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    \$\begingroup\$ You want a linear current regulator. Don’t confuse issues by mentioning linear voltage regulation. \$\endgroup\$ – Andy aka Sep 6 '20 at 22:15
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    \$\begingroup\$ The Ohm's Law police will come for you if you attempt this. \$\endgroup\$ – Peter Bennett Sep 6 '20 at 22:15
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    \$\begingroup\$ Depending on the load, current is a function of voltage, so for a fixed load you can only regulate voltage or current, not both, at one time. \$\endgroup\$ – Spehro Pefhany Sep 6 '20 at 22:29
  • \$\begingroup\$ how will you vary the load so that 1 mA current is maintaned when voltage changes? \$\endgroup\$ – jsotola Sep 6 '20 at 22:31
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Simple: You don't.

What you're asking is physically impossible. In order to drive 1 mA of current through a 6 ohm resistance, you must have a voltage of 6 mV.

A linear current regulator achieves constant current by adjusting the output voltage in proportion to the load. So, for instance, if you replaced the load with a 10 ohm one, the regulator would increase the voltage to 10 mV to maintain the specified 1 mA of current. So, if all you're after is a constant current, that's the way to go. But just be aware that the output voltage isn't going to "follow" the input voltage, it's going to be reduced to whatever will provide the specified current to the load (subject to the device's operating limits, of course).

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  • \$\begingroup\$ Thanks for your reply. Suppose now I want a voltage regulator with adjustable output voltage from 0.8V to 1.4V delivered to the 10-ohm load. In this case, as voltage changes from 0.8V to 1.4V, the output current will change from 80mA to 140mA. Correct? That is, if I want to regulate the output voltage, I will have no control over the output current, right? \$\endgroup\$ – Learner Sep 6 '20 at 23:02
  • \$\begingroup\$ @PhaeLeung Right! Be aware that every voltage-regulated supply has an upper limit of current that it can deliver to a load. Above that limit, it overheats, or something breaks. \$\endgroup\$ – glen_geek Sep 6 '20 at 23:11
  • \$\begingroup\$ Thanks all. I now understand. \$\endgroup\$ – Learner Sep 6 '20 at 23:20
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You describe a current regulator, not a voltage regulator. These exist.

A typical (albeit pretty old) example is using the adjustable LM317 in a constant current configuration.

Other options include using a voltage source and a opamp voltage-controlled current sink or source to get a regulated current supply.

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  • \$\begingroup\$ Thanks for your comments. \$\endgroup\$ – Learner Sep 6 '20 at 23:02
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These things are called "constant current source". They take input voltage and burn as much of it as they need to set such an output voltage that it will produce specific current. They are self-regulated - if the load changes, they change output voltage to output the correct current. All of this, of course, works only if the output voltage has to be lower than input voltage. So I see where this linear regulator analogy comes from, but they're not linear regulators.

I would strongly suggest LM334 constant current source. A 3-pin device both SMD and TH that takes 0-40V input and can output stable and very precise current of 1uA - 10ma. All it needs is 1 external resistor that sets current value (you can have potentiometers there too, of course), everything is in the datasheet. It's extremely stable with temperature and everything as it is, but the datasheet suggests ways to improve even further (although I'll be honest unless you're making some ultraprecision device to detect gravitational waves, there is no need really)

Check out its datasheet: HERE

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