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There are similar questions asked, but none address this question so specifically.

It is, quite simply the title:

What are the drawbacks/disadvantages of using a voltage regulator as a constant current source?

Cost is the only one I can personally think of.

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    \$\begingroup\$ Voltage regulators are usually cheaper and more available than current regulators (i.e. the 78L05 vs. the LM334Z; the latter costs 10x as much), so if you can use a voltage regulator instead, it's a huge cost advantage. \$\endgroup\$ Commented Apr 22, 2023 at 14:03
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    \$\begingroup\$ Besides the pedantic response that most of the time, a constant current source IS a voltage regulator, this sounds like a homework problem. What do you think about the electrical characteristics? \$\endgroup\$
    – AnalogKid
    Commented Apr 22, 2023 at 14:22
  • \$\begingroup\$ You show me a decent current regulator that doesn't also regulate a voltage. I'm probably going to be embarrassed by the number of options now that I've said it. \$\endgroup\$
    – Andy aka
    Commented Apr 22, 2023 at 14:32
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    \$\begingroup\$ @AnalogKid Not a homework question. I am a hobbyist who likes elegant and ideal solutions. \$\endgroup\$ Commented Apr 22, 2023 at 14:43
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    \$\begingroup\$ Some voltage regulators (wired as a current source) require a minimum output current, below which regulation fails...so these might fail as current sources if your requirement includes low current approaching zero. It might help to suggest a particular part in your question. \$\endgroup\$
    – glen_geek
    Commented Apr 22, 2023 at 17:53

4 Answers 4

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If you are asking about the advantages/disadvantages of an integrated circuit voltage regulator, such as an LM317, compared to a discrete transistor design, I would say that the 317 disadvantages are minimum output current and a relatively large minimum differential voltage across the device (input voltage minus output voltage), and the advantage is superior stability over temperature.

I recently used an LM317 as a constant current regulator, and was impressed with its temperature stability. It might not be so wonderful at higher output currents, but in the 20 mA - 50 mA range it was very good.

For a discrete design, consider these: http://confocal-manawatu.pbworks.com/w/page/82281019/Constant-Current%20Circuits

enter image description here

This is sometimes shown another way:

enter image description here

The minimal solution has only one transistor, which can saturate if necessary. This lowers the minimum input-output differential voltage to less than 1 V, at the cost of poorer initial accuracy and temperature drift:

Cheap, low-quality constant-current sink over wide voltage range enter image description here

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  • \$\begingroup\$ I do something similar, I use a standard LED, they have a very sharp knee and with the appropriate transistor are very stable. It also doubles as a power on indicator. Yes I can use several transistors for seperate sinks or current sources depending on configuration with 1 LED. I started this in about 1980 without any known problems. I keep the LED current in the 2mA range, depending on the knee. \$\endgroup\$
    – Gil
    Commented Apr 23, 2023 at 0:07
  • \$\begingroup\$ This is exactly what I'm getting at. I love jellybean components and truly modular solutions that can be used in a wide variety of situations. This satisfies both of those. Plus the slew rate would be exceptional for particularly demanding loads. I've also found the lm317 is great for low power. It is nice to know that the disadvantages are not related to current stability. \$\endgroup\$ Commented Apr 23, 2023 at 8:08
  • \$\begingroup\$ In all four schematics, the "current stability" comes down to one PN junction and one resistor. and those two components have different temperature coefficients. If you are referring to an LM317 circuit, then yes, the current stability is excellent, basically down to the temperature coefficient of the shunt resistor. \$\endgroup\$
    – AnalogKid
    Commented Apr 23, 2023 at 13:53
  • \$\begingroup\$ I use a 317 in battery testers as a load. It's wired in the standard current-limiting configuration, with a 1.25 ohm (1.3ohm 5W paralleled with 33ohm 2W) resistor, for 1 A current. Then where the load should go, I just tie that to ground. The result is that between the resistors and the (well heatsinked) 317 the circuit draws 1A at any reasonable input voltage (2V - 30V or so). 2.5ohms draws half an amp. \$\endgroup\$ Commented Apr 23, 2023 at 22:22
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One disadvantage I can think of is that maximum compliance voltage is limited by the drop-out voltage of the regulator, and also by its internal reference voltage. For instance, when using the LM317 as a current source:

schematic

simulate this circuit – Schematic created using CircuitLab

One the left you see that it regulates current nicely, since the voltage required at the top of RLa, to pass 100mA, is well below the maximum that the LM317 can put there.

On the right, though, we see the regulator is unable to maintain 100mA, because the required voltage would be +14V, which it cannot achieve. The maximum voltage it can apply at the top of RLb is 11.5V.

This difference between the supply of 15V and the maximum output of 11.5V is the sum of the regulator's drop-out voltage and its reference voltage across R1.

A DIY current source like the one below would not suffer this constraint. Even though both setups are fundamentally voltage regulators, their topology is different:

schematic

simulate this circuit

As you can see on the right, this design is quite able to produce a compliance voltage much closer to the positive supply.

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    \$\begingroup\$ Not all op-amps can work with common mode voltage at the high side supply voltage. \$\endgroup\$
    – PStechPaul
    Commented Apr 22, 2023 at 22:21
  • \$\begingroup\$ I understand where you're going with this one. I do think that there is a wide enough range of linear voltage regulators to mitigate the issues you presented. Sometimes op-amps are just too quirky for a quick solution. Better for designs meant for production than for a truly modular design that can be used in a wide variety of DIY projects. But I do appreciate the quality of the solution you offer. \$\endgroup\$ Commented Apr 23, 2023 at 8:00
  • \$\begingroup\$ @AaronButkovich sure, there are plenty of regulators with \$V_{DROPOUT} \le 0.5V\$ \$V_{REF} \le 0.6V\$ which will perform as well as my DIY "hack job"! I do have more flexibility to fine-tune a DIY closed loop, though. I get that you have to take care with CM input range, and limited output range and so on, though, if these are the "quirks" you refer to. \$\endgroup\$ Commented Apr 23, 2023 at 8:34
  • \$\begingroup\$ @AaronButkovich you have to provide a nice stable voltage reference too, another downside to the DIY approach. \$\endgroup\$ Commented Apr 23, 2023 at 8:38
  • \$\begingroup\$ @SimonFitch It's certainly no hack job, and probably a circuit I need to play around with since I have a strip of 200 LM317s. Current sources for higher than 30V are another can of worms all together so the CM input range probably doesn't matter for most applications. \$\endgroup\$ Commented Apr 24, 2023 at 12:34
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What are the drawbacks/disadvantages of using a voltage regulator as a constant current source.

It is possible to create a constant current source from an adjustable linear voltage regulator by taking feedback across a sense resistor and passing that feedback to the ADJ pin of the voltage regulator, like so

schematic

simulate this circuit – Schematic created using CircuitLab

For some use-cases this is perfectly fine. However, it has some significant drawbacks for other use-cases.

First, in addition to the voltage drop across the linear regulator itself, there must be a voltage drop across the sense resistor equal to the voltage required for correct operation of the ADJ pin. This is typically 1.25 volts. Now a constant current device is meant to have some voltage drop in order to regulate the current. However, in some cases the voltage drop for this circuit is too high for a particular application.

Second, if the current to be regulated is "high", then the power dissipated by the regulator, and by the sense resistor, will also be high. This will necessitate some form of heat removal. It may be advantageous to have only one pass element need heat removal measures rather than two. Sensing a large current through a resistor that has a 1.25 volt drop on it may require a beefy resistor, and possibly a heatsink. It may (or may not) be advantageous to sense the current with a smaller resistor and amplify the voltage to feed back to the linear regulator. Obviously if the current being regulated is small enough, this will not be an issue.

Third, many voltage regulators have minimum current requirements, and therefore are unsuitable if the current specification for the constant current source is less than the minimum for the voltage regulator.

I have taken your description of "using a voltage regulator to create a constant current source" to mean the simple circuit shown in my schematic. But both of the issues I described for that circuit can be mitigated by making modifications to that circuit. So, unless we compare two different concrete circuits, any rule we lay out is subject to exceptions.

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When looking for LED regulators, I have found that current regulators using an external shunt resistor usually have a lower sense voltage (for example 0.2V instead of 0.6V or 1.25V) than voltage regulator.

As all the current goes through the shunt, a lower voltage makes the circuit more efficient as less power is lost in the shunt.

You don't mention if you are looking for linear or switching regulators : Constant current switching regulators can have bad effects when the output is unconnected, real current regulators are protected against open circuit.

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  • \$\begingroup\$ Oh yeah, definitely linear. Switching current sources have their place for higher voltage and power but like you said, require a lot of careful thinking and quick circuits for protection. Shunt regulators are great, but resistor prices are just getting so high. I rather dump the heat through cheap silicon and a heatsink if possible. \$\endgroup\$ Commented Apr 24, 2023 at 12:38

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