Can someone explain "constant-current" circuits?

  1. When are they necessary?
  2. What happens without one?
  3. What's the simplest version of such a circuit?

I bought some 20W LEDs off ebay intending to make a bike light. "Simple," I thought, "The LED and my laptop battery are both 11Vdc, so I'll just connect them."

I haven't actually tried this, but an EE friend suggested that if I do so, it will draw too much current and burn out -- but he didn't know why; he doesn't do power electronics.

  • 4
    \$\begingroup\$ Can you provide a part number or link for the LED? \$\endgroup\$ – tyblu Dec 24 '10 at 7:03
  • \$\begingroup\$ Note that battery voltage drops as soon as you provide a load, modeled by an equivalent series resistance (ESR) or internal resistance (IR). One must obey the datasheet -- stay below the maximum allowed constant current and voltage ratings. There will be graphs to illuminate these values. \$\endgroup\$ – tyblu Dec 24 '10 at 7:05
  • 2
    \$\begingroup\$ Some engineer, that friend of yours! :-) \$\endgroup\$ – Federico Russo Jun 13 '11 at 15:38

Most LED control circuits just have a series resistor to control the LED current. This is OK if the supply voltage is more or less constant, but it the voltage varies so will the LED current and luminosity. If you want constant luminosity you also want constant current. This means that the series resistance must vary with supply voltage. This is about the simplest circuit for it:

constant current source

This is a constant current source dimensioned for 1mA, which is too little for your application, but the principle remains the same. The base-emitter voltage of the transistor is 0.6V, so given that the zener creates a drop of 5.6V between \$V_+\$ and the transistor's base there remains 5V for the resistor. \$\dfrac{5V}{5k\Omega} = 1mA\$.

If your LED is 20W @ 11V then you need 1.8A of current, so I'll calculate the other components for this value. Replace the zener by 3 x 1N4148, this gives you a 0.6V drop over R1 (three diodes because we'll use a darlington, which has 2 x 0.6V between base and emitter). \$R1 = \dfrac{0.6V}{1.8A} = 0.33\Omega / 1W\$.
The TIP125 darlington transistor has an \$H_{FE}\$ of 1000, R2 should be \$\dfrac{10V}{1.8A / 1000} = 5k\Omega\$.
Since \$V_{CE(SAT)}\$ for the transistor is 2V, and you have an additional voltage drop over R1 of 0.6V, you'll need a 14V power supply to be sure that there's enough left for the LED!
The darlington will dissipate \$2V \times 1.8A = 3.6W\$, so you'll have to mount it on a heat sink.

a note on current sources:
Current sources are the dual of voltage sources, which are more common; we use them all the time in power supplies. This duality means that certain parameters are each other's opposites.
Whereas a voltage source will try to keep the voltage at its output constant regardless of the load, a current source will keep the output current constant regardless of the load. That means for a current source the output voltage will vary, like for a voltage source the current will be variable.
An ideal voltage source will have zero output impedance, an ideal current source will have infinite output impedance. And while an ideal voltage source should never be short-circuited because the current will go to infinite, an ideal current source should never be left open, because the voltage which determines the set current will go to infinite.


P = V * I. In your case for the LED, P = 20 W, V = 11 V and therefor I = 1.82 A.

To drive the LED correctly you will need to limit the current to 1.82 A. (you will also need to provide adequate heat sinking as 20 W is a lot of power to dissipate and most of it will be in the form of heat!!!! .... but that is another matter)

Diodes, using a PN junction have forward voltage drop, in your case that drop is 11 V. When referring to the diode datasheet they will include a V I curve, this will relate how much current flows for a given voltage. When you examine this curve you will notice that the current heads toward infinite for the given voltage drop. If you allow this to happen, you diode will simply burn out.

The simplest method to limit the current is to use a resistor. V = I * R. For a given voltage drop, a fixed resistor will limit the current to a fixed amount.

Assuming you have a 18 V battery and an 11 V LED, the difference, 7 V will need to be dropped across the resistor. This is fixed. Also fixed is the desired current, of 1.82 A. Rearranging the equation you get R = V / I => R = 3.8 ohm. This resistor will need to be rated to a minimum of 12 W. That is a large resistor that will get hot!!!!

The brings up the second issue - heat. At a total of 32 W to dissipate, you will need some serious heatsinking.

You can probably do away with the resistor and replace it with a controlled current circuit based on transistor and or mosfets but I will leave that to someone else to answer. The same issue of heat dissipation may still be present depending on the circuit design!!!

I know you mentioned you have an 11 V battery and an 11 V LED. Unless the LED has built in current limiting circuit, or the battery has, it will be difficult to drive it at the same voltage (depending on the VI curve of the diode) without a boost converter.

  1. When you are using non-linear power device, like a LED. Other notable example are gas-discharge lamps & lasers.

  2. Something burns or explodes. LED or power supply. Or both :-)

  3. For 20W the only reasonable solution is constant-current DC-DC circuit, but it's not simple. Simplest solution is linear regulator with powerful BJT transistor, but this will dissipate at least 10W of heat. Simple resistor will not give you linear current, but acceptable in low-power cases (let's say for 0.1W).


If you want a simple constant current LED driver, on All About Circuits we've been designing one.

However, it's only rated to 1W. You'd probably need to upgrade the fet and driver to get it to run at 20W, the fet would also need heatsinking and you'd probably need a bigger inductor.


ad. 1) Whenever you use power leds and don't want them to burn.

ad. 2) The key is the heat-current interdependence. In short, when diodes heat up, their dynamic resistance decreases, which leads to further increase of current. Sort of a chain reaction. So if you fail to dissipate all the heat, your LED can easily burn.

Another thing, from end-user perspective, is that the amount of light emitted by LED is proportional to current and the current-voltage dependence is highly nonlinear. Thus, it is much harder to precisely control the brightness by controlling the voltage.

ad. 3) The simplest, though inefficient way to do that is to connect a series resistor that will compensate some of the negative heat-current dependence of the LED. That is because for resistors, that dependence is opposite in direction (but not in shape). So you won't really have constant current, but your LED will be safer. For basic tests, you can also simply add a fuse to that to protect your LED.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.