First of all can a device that is constant current also be at the same time constant voltage? In my head I just could not grasp how this device can work.

Individually I understand how it works, constant current supplies adjust the voltage to sustain the target current, constant voltage supplies work by having some feedback loop circuit that tries to maintain the voltage most power supplies are these.

Now devices such as these XL4016 are said to be constant current constant voltage. They even have 2 potentiometers to adjust constant current and voltage. For constant current to happen, the voltage must be varied, but we cannot have that as we also need to have a constant voltage.

Assuming that it works, and I set it to 10V and 1A. what will happen if I place a 1k resistor on the output? Will 10V @ 1A will be forced into that resistor? No matter the resistance I placed 10W will always be on that load, which doesn't sound right.

How is this achieved, or am I being misled here?

  • 13
    \$\begingroup\$ Your suspicions are well founded. They have a CC mode and a CV mode and the actual load determines which mode they are in for the potentiometer settings. \$\endgroup\$
    – user16324
    Commented May 1, 2022 at 18:13
  • \$\begingroup\$ @user_1818839 what do you mean by the load determines which mode they are going to be? like what are the parameters for it to be a constant voltage, and what do i need to do with my load for it to be constant current? \$\endgroup\$
    – Jake quin
    Commented May 1, 2022 at 18:21
  • 9
    \$\begingroup\$ TLDR: That power supply lets you set both a voltage limit and a current limit. If you connect a high impedance load, then the voltage will hit the limit, and the current will be controlled by the load. (i.e., it acts like a voltage source.) If you connect a low impedance load, then the current will hit the limit, and the load will determine the voltage (i.e., it acts like a current source.) \$\endgroup\$ Commented May 1, 2022 at 20:22
  • 4
    \$\begingroup\$ Better think of it as voltage source with current limit. \$\endgroup\$
    – Michael
    Commented May 2, 2022 at 9:13
  • 2
    \$\begingroup\$ The most widespread use of CC/CV devices is as battery chargers. This is EXACTLY because it is the load who determines if you are in CC or CV mode, because the battery as a load (to the charger) is expected to change its behavior during the charge process. Most battery designs (Li-whatever, Pb-Acid, vented Ni-whatever) expect CC/CV charger. Sealed Ni-Cd, Ni-MH and some other types are charged in different ways. \$\endgroup\$
    – fraxinus
    Commented May 3, 2022 at 14:21

3 Answers 3


The devices work by not being constant current and constant voltage at the same time.

You set a voltage (V) for the output. You set a current (A)for the output.

It will supply constant V for any current up to A. When the current your circuit draws goes above A, then it will reduce the voltage to keep A at the value set.


  • V set to 10V
  • A set to 1A

If you connect a load of 1000 ohms, then you will measure 10V across it. It will deliver 0.01 amperes of current to the resistor - that's 0.1 watts.

If you connect a load of 1 ohm, then you will measure 1 ampere of current through the resistor. There will be 1 volt across the resistor. That's 1 watt of power.

It would be more accurate to call such power supplies "current limited constant voltage."

  • 1
    \$\begingroup\$ I like how you put it, by NOT being constant current and constant voltage at the SAME TIME. So by default it is a constant voltage , and when maximum current is reach it will go into constant current mode. For analog supplies that uses pots (just like the one i linked) how do i know how much current i am setting it at? would i always need a test load? \$\endgroup\$
    – Jake quin
    Commented May 1, 2022 at 18:29
  • 2
    \$\begingroup\$ Yep. You set the voltage as needed, then turn the current way down. Connect an ammeter straight across the power supply (make sure your meter can handle more current than the power supply can deliver,) then turn the current up to where you want it. \$\endgroup\$
    – JRE
    Commented May 1, 2022 at 18:39
  • 10
    \$\begingroup\$ You could also call them "voltage-limited constant current" to confuse the undergraduates. \$\endgroup\$ Commented May 2, 2022 at 8:47
  • \$\begingroup\$ @Jakequin Some of them even have LEDs on the front showing which mode they are in. \$\endgroup\$ Commented May 2, 2022 at 14:13
  • \$\begingroup\$ @Jakequin Or you could say that by default they are constant current, and when maximum voltage is reached they will go into constant voltage mode. \$\endgroup\$
    – John Doty
    Commented May 3, 2022 at 14:55

It might be better to think of these supplies as "maximum voltage, maximum current" supplies. The potentiometers adjust the maximum current and voltage that the supply will deliver.

With no load, the supply will produce the maximum voltage that you set. As you increase the current, the voltage will remain steady until the load reaches the maximum current setting, then the voltage will reduce to keep the current constant.

  • \$\begingroup\$ What are the behavior of these power supplies on short circuits that are not entirely at 0hms. let say i set the current limit to 3A and the short circuit should produce 6A. So the power supply reduce the voltage to keep the short circuit at 3A burning/heating up the component or that causing the short? \$\endgroup\$
    – Jake quin
    Commented May 1, 2022 at 18:35
  • 3
    \$\begingroup\$ @Jakequin Yes. Though it doesn't know whether the short circuit is burning up. It just knows the limit is 3A. \$\endgroup\$ Commented May 2, 2022 at 8:48
  • \$\begingroup\$ @Jakequin In this video youtu.be/4Iey_qXxAmk?t=2258 they’ve actually shorted the output. It does bring down the voltage a lot. \$\endgroup\$
    – the Hutt
    Commented May 7, 2023 at 7:18

They provide both constant-current and constant-voltage modes of operation, with both setpoints being limits - essentially, the behavior is to deliver "up to 10 V and up to 1 A".

In the case of the 1k resistor, you should expect 10 V and 10 mA to flow. In the case of a 1-ohm resistor you'll expect 1 A to flow with a voltage drop of 1 V.

It's worth noting that this behavior is often desired for things such as battery-charging; many lithium battery charge curves refer to "CC-CV" charging, where you charge at a constant current until the voltage reaches a setpoint, and then charge at constant voltage while the current drops due to the cell becoming more and more charged. This datasheet has an example which could be achieved by setting a power supply similar to yours to 4.2 V, 1.65 A.


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