Parallel LEDs driven by LED driver

[FrankT]

I'm using a LP5569 LED driver to illuminate a model car.  It has 9 individually controlled channels where I can set the current from 0.1mA to 25mA. 

Unfortunately, 9 channels is not enough; I need to parallel some of the LEDs (I can't use them in series because the voltage isn't high enough).

If I current limit 2 parallel LEDs to 10mA, do I need use current limiting resistors?  Or is there a danger of one take more current?  If so, what is the minimum current limiting resistors I can use?

[S. Petrukhin]

If you install the LEDs in parallel, they will divide the current among themselves, and this current is limited by the chip, has a finite value. The LEDs will glow about 2 times less or go out altogether if 1/2 of the current is insufficient to glow, if the voltage falls below the threshold. You can turn on LEDs of the same color in parallel, they are about the same, but LEDs of different colors will not work. Putting resistors, hoping for current equalization makes no sense - they will further aggravate the situation with current losses. The CHIP will increase the voltage, but the current will remain constant.

[S. Petrukhin]

I think the easiest and most reliable option would be to increase the power supply voltage step-up and turn on the LEDs in series.

[FrankT]

Thanks.

I should have been clearer - they will always be LEDs of the same colour (from the same tape).

[Benta]

Yes, you'll need to balance the currents if you parallel the LEDs. Absolutely doable, and will work even better if you pair the LEDs for forward voltage.

Another option that's a bit more complicated is multiplexing using two PNP transistors to switch between two positive supplies for two sets of LEDs.
This will give the same result as paralleling, but better current matching and no need for pairing.

[thm_w]

It should be fine, assuming the LEDs are rated to typical 20mA value and of the same type.
If they are really poorly matched LEDs then visually one will be brighter than the other. You can take some, connect many to an adjustable power supply in parallel and test for yourself, to see if the difference is acceptable. If not, add resistors.

If you install the LEDs in parallel, they will divide the current among themselves, and this current is limited by the chip, has a finite value. The LEDs will glow about 2 times less or go out altogether if 1/2 of the current is insufficient to glow, if the voltage falls below the threshold. You can turn on LEDs of the same color in parallel, they are about the same, but LEDs of different colors will not work. Putting resistors, hoping for current equalization makes no sense - they will further aggravate the situation with current losses. The CHIP will increase the voltage, but the current will remain constant.

Its common to use resistors to match the currents: https://www.allaboutcircuits.com/technical-articles/led-arrays-one-resistor-or-many/
Even if OP were using different color LEDs, they could tweak the resistance to compensate for the different voltage drop.

[S. Petrukhin]

If you install the LEDs in parallel, they will divide the current among themselves, and this current is limited by the chip, has a finite value. The LEDs will glow about 2 times less or go out altogether if 1/2 of the current is insufficient to glow, if the voltage falls below the threshold. You can turn on LEDs of the same color in parallel, they are about the same, but LEDs of different colors will not work. Putting resistors, hoping for current equalization makes no sense - they will further aggravate the situation with current losses. The CHIP will increase the voltage, but the current will remain constant.

Its common to use resistors to match the currents: https://www.allaboutcircuits.com/technical-articles/led-arrays-one-resistor-or-many/
Even if OP were using different color LEDs, they could tweak the resistance to compensate for the different voltage drop.

This is true when you power the LEDs from a voltage source, but there is a current source installed inside the chip. That's a big difference. 

[Benta]

Putting resistors, hoping for current equalization makes no sense - they will further aggravate the situation with current losses. The CHIP will increase the voltage, but the current will remain constant.

This is rubbish. Let's make an example with two LEDs in parallel.

LED forward voltage: LED1: 1.6 V, LED2 1.8 V
IC output LED current: 20 mA

No resistors: terrible result.

For each LED a 220 ohm resistor:
Calculating for each branch, the 1.8 V LED will carry a current of 9.5 mA and the 1.6 V LED 10.5 mA.

Note that I haven't even mentioned the supply voltage, just that it's a 20 mA current sink. The only thing I've taken into consideration is, that it's within the voltage compliance of the current source.

[Benta]

This is true when you power the LEDs from a voltage source, but there is a current source installed inside the chip. That's a big difference. 

Kirchhoff's laws apply. Also with current sources.

[S. Petrukhin]

Putting resistors, hoping for current equalization makes no sense - they will further aggravate the situation with current losses. The CHIP will increase the voltage, but the current will remain constant.

This is rubbish. Let's make an example with two LEDs in parallel.

LED forward voltage: LED1: 1.6 V, LED2 1.8 V
IC output LED current: 20 mA

No resistors: terrible result.

For each LED a 220 ohm resistor:
Calculating for each branch, the 1.8 V LED will carry a current of 9.5 mA and the 1.6 V LED 10.5 mA.

Note that I haven't even mentioned the supply voltage, just that it's a 20 mA current sink. The only thing I've taken into consideration is, that it's within the voltage compliance of the current source.

By adding resistance, you will force the current source to raise the voltage, which then equalizes the current due to resistors - this is true. But the author said that it is not possible to connect the LEDs in series, there is not enough voltage - you did not take this into account.

[Benta]

Do the maths yourself.

I've taken all relevant parameters into account.

And Kirchhoff still rules.

[S. Petrukhin]

Do the maths yourself.

I've taken all relevant parameters into account.

And Kirchhoff still rules.

No one argues with Kirgof.
But what did you think?
If the current source does not have enough voltage to turn on two LEDs in series, then its maximum voltage is less than 2*Vf.
Vcc-Vf < Vf < max resistor voltage < 1.6V / 220ohm < 5mA. Ohm's Law rules just as well... 

[ledtester]

Assume you have the following circuit:



V1 and V2 are the voltage drops of the diodes. I is the current set by your driver, I1 and I2 are the currents that will pass through each diode. R is the (common) resistance of your equalizing resistors.

The equations are:

I = I1 + I2
V1 + R I1 = V2 + R I2

Solving them yields: \$2 I_1 = I - \Delta V / R\$, \$2 I_2 = I + \Delta V / R\$ where \$ \Delta V = V_1 - V_2 \$.

So \$ \Delta V/R\$ will be the difference in current between the two branches. This means you want to make it "small" with respect to \$I\$. Let's say small means 1/4. Then you want \$R > 4 \Delta V / I\$.

An example... let's say I = 10mA, and you want to handle a \$\Delta V\$ as large as 0.2 V.  Then R should be at least 4*0.2/0.01 = 80 ohms. The difference in current between the two diodes will be 0.2V/80R = 2.5mA. So, in the worst case the 10mA of current would be split 6.25mA to 3.75mA.

Note that R needs to be at least \$\Delta V / I\$ -- otherwise one branch will be starved of current.

[FrankT]

I will be using a single Lipo, so my voltage will range from about 3.7v to 4.2v.  So only low Vf LEDs will be able to used in series.

I don't expect to run the LEDs at a high current - I expect under 5mA, maybe less for lights that are only indicators.

What kind of Vf difference can I expect between like LEDs?  Or should I just measure them and calculate a small resistor from that?

[ledtester]

If you have more than two branches, say \$n\$ diodes, the formulas is:

$$\Delta I = \frac{ (n-1) \Delta V }{R} $$

where \$ \Delta I \$ is the deviation from the expected current in each branch, \$ I/n \$, and \$ \Delta V \$ is the deviation of the forward voltages from the average forward voltage.

(N.B. that this definition of \$\Delta V\$ here is not exactly the same as the one I used in my previous post.)

As for a good value for \$\Delta V\$, if they are the same product from the same manufacturer I probably would use \$ \Delta V \$ = 0.1V, but measuring their forward voltage with a multimeter to see what variation you get isn't a bad idea. I would be hesitant to use a small value like 0.01V for \$\Delta V\$ even for identically manufactured LEDs.

Example calculation: You have \$n = 4\$ LEDs and want a current of 5 mA through each with a max deviation of +/- 1 mA. Assume \$ \Delta V \$ = 0.1V. Thus, I = 20mA and the condition is \$ \Delta I \$ < 1 mA. Solving for R yields:

  R > 3 * 0.1 V / 1 mA = 300 ohms

[Brumby]

There is a lot of noise above that comes from classic LED "good practice" - but, in truth, you may be able to get away with simple parallel connection - especially....

they will always be LEDs of the same colour (from the same tape).

A lot of consumer products do exactly this.

Reason: LEDs - especially from the same batch - are very often so close in characteristics (V_f) that a simple parallel connection will perform quite acceptably.


Just try a few and see how they go.

[Terry Bites]

You will be lucky if both LEDs make a gentlemens agreement to share the current. A resistor in series with each is the way to go. 220R per LED. Why make a bad circuit with only one resistor to save  penny. Consumer goods use a cheap and nasty approach- that doesn't make it good design. Who says your LEDs are form an identical batch?

[Brumby]

Again - "best practice" noise.  Let's NOT get caught up with IDEALS if things work IN PRACTICE

Please.

If directly paralleled LEDs work (easy to try and zero risk) then just do it.



I fully accept the age-old truth about variations in V_f for LEDs and I have followed this truth for decades - but if taking the easy path ISN'T a problem, then why not do it?

[Zero999]

It's true, LEDs from the same batch share current quite well. It's best practise not to run them at the full rated current anyway, so if one of them passes slighly more current, than the others, it's no big deal. There are numerous commercial products, with LEDs in parallel, without individual current limiting/sharing resistors and they work perfectly.

[mikerj]

Do the maths yourself.

I've taken all relevant parameters into account.

And Kirchhoff still rules.

No one argues with Kirgof.
But what did you think?
If the current source does not have enough voltage to turn on two LEDs in series, then its maximum voltage is less than 2*Vf.
Vcc-Vf < Vf < max resistor voltage < 1.6V / 220ohm < 5mA. Ohm's Law rules just as well... 

Why continue to argue from a losing position?  Any series resistance will help with current sharing, you don't have to drop a full Vf on the resistors to give an improvement.

[mikeselectricstuff]

Two LEDs of the same type will share current quite happily. Remember that Vf is dependent on current, so there is effectively already some internal resistance which will help balance any small differences.
Also, for individual LEDs directly viewed ( as opposed to through a diffuser, diffuse reflected etc.) you won't usually notice a small difference in current as the brightness of small points tends to saturate the eye, making even quite significant differences (10-20%) indistinguishable.
 

[John B]

Lighting applications will typically have LEDs in parallel banks, however they are usually many series LEDs (sometimes >20) in each parallel branch. These tend to share current quite well, as I would guess that the manufacturing tolerances average out when many are used in series.

Also consider the conduction curve. They share current more safely down at the lower currents. 10mA is already quite high on the conduction curve of a small LED.

For single LEDs in parallel I would use some resistance. Is the application extremely frugal with power consumption?