I'm trying to create a schematic for powering 196 LEDs (yellow, 2.1V, 20mA)
by using 2 or 1 of these LED drivers STP16CPP05 which is 16 channel (40mA per channel)

LED series/parallel calculator says:

Solution 0: Source voltage: 14.7V = 7 x 28 array uses 196 LEDs exactly R = 1ohms the array draws current of 560 mA from the source.

Solution 1: Source voltage: 30V = 14 x 14 array uses 196 LEDs exactly R = 33ohms the array draws current of 280 mA from the source.

my input voltage will be 12V battery, so my questions are:

  1. what is more power efficient to boost voltage to 14.7V or to 30V?
  2. do I really need the resistors?
  3. am I better of by designing the LED array for 12V input?

I'm also having some trouble of picking the right switched regulator and calculating the necessary resistor and other values to boost the voltage to 14.7V/30V

Can anyone recommend a good average priced regulator (SMD that can be hand soldered) and help me out with the math? Any help is greatly appreciated.

  • \$\begingroup\$ Seeker, provide us with more information about the specific "12V battery" you intend to use. Do you intend to actually drive the LEDs at 20 mA, or is this just their rating? \$\endgroup\$ – FiddyOhm Feb 7 '17 at 13:05
  • \$\begingroup\$ just their rating, they will never be fully lit, the battery is standard lead acid 12V battery \$\endgroup\$ – Seeker Feb 7 '17 at 19:02
  • \$\begingroup\$ Seeker, do you know what your drive current will be in the actual application? Very important to know this from the get go. \$\endgroup\$ – FiddyOhm Feb 7 '17 at 19:33
  • \$\begingroup\$ if I use 15V it should be about 560mA, I was thinking about using 1A or more IC \$\endgroup\$ – Seeker Feb 7 '17 at 19:47
  • \$\begingroup\$ Seeker, sorry, I was not clear. How much drive current for each individual LED, no matter how you ultimately configure the strings of series LEDs? The LED drive current will determine the brightness of each LED. It's a good idea to know this value, or range of values, before you jump into the actual circuit design. Do this by powering up one, or several LEDs in series, from a power supply with a series resistor and an appropriate ammeter (e.g. a DMM that has a low milliamp range:0-20, or 0-200). Adjust the voltage,and/or resistor to get the brightness you want, note the ammeter current reading \$\endgroup\$ – FiddyOhm Feb 7 '17 at 22:02

Bad news: your design isn't quite realistic exactly as it is. But let's try to make it workable.

First, a basic information: you don't need the resistors. STP16CPP05 is a constant current driver, so it will monitor and limit the current itself. So forget the resistors, they'll just uselessly dissipate power and make your supply voltage requirements higher.

Now, why it doesn't work:

  • Using a 14.7V supply to power a 7 * 2.1V LED string is risky. Unless you bin the LEDS very precisely and maintain the temperature constant, the Vf drop of the LEDS will probably vary between 1.8V - 2.2V. Moreover, the driver itself has a voltage drop (~350mV, see table 11 of datasheet). So you need more margin to ensure you can drive the leds at 20mA. An absolute minimum of 2.2V * 7 + 0.35V is required, but I would advise at least 17V to account for the power supply tolerance.
  • A supply of 30V gets you above the STP16CPP05 maximum output voltage, specified at 20V.

So, now, suppose we use a supply of 17V and a 7 x 28 array. Let's see what will be the worst case power dissipation of the driver chip. If we take the min Vf of diodes (e.g. 1.8V), and max voltage of supply (e.g. 17.85V if we say it has 5% tolerance), it means the voltage drop of the driver must be 17.85 - 1.8*7 = 5.25V. Time 20mA times 16 strings per chip, it means each STP16CPP05 chip dissipates ~1.7W worst case. Wow. But this is workable with the TSSOP24 with exposed pad chip package. You better solder the pad, though.

Note that it would be even worse with the 14 x 14 array (if we could use higher voltages with this driver). Because what makes the sizing difficult is the Vf tolerance, so the more LEDs you have on your string, the higher the worst case voltage drop you'll have at the chip (and consequently the higher the dissipated power at each chip).

So what I would actually advise, to make it safe, is to actually keep your supply voltage at 12V. This removes the need for a boost, and spread the dissipated power on more drivers because you have less LEDs on each string. It means you'd have 5 LEDs per string (2.2V * 5 + 0.35V = 11.35V min supply). I know, this fits less well with the 196 LEDs requirements, because it's not even an integral number of strings (39.2 strings), and you now need 3 driver chips. But this actually makes it simpler (no boost, less power on drivers). To better spread the power, put only 14 strings per chip. Max power for a driver chip is now ~1W (with 12.6V supply max). Workable with SO-24 package. With a 14.4V supply max (lead-acid battery), it would make 1.5W. Much closer to the limit, but still acceptable with this package, if the airflow is sufficient in the case. If in doubt, pour in an additional driver chip and put only 10 strings per chip.

Note: on the third driver chip, you'll have 11 complete strings plus one that has only one LED. This single LED string will higher the dissipation on the chip. This is still fine in this case because you only have 11 complete strings, but you could put a resistor on that string to move some of this dissipation away from the chip to the resistor. Max resistor value is (min supply voltage - 2.2V - 0.35V) / 20mA) = 440Ω. Make it a 1/4W.

  • \$\begingroup\$ Very clearly answered. \$\endgroup\$ – soosai steven Feb 7 '17 at 14:42
  • \$\begingroup\$ That worst case is almost FUD. No way the diodes would display 1.8 Vf at the Vs provided. \$\endgroup\$ – Passerby Feb 7 '17 at 15:08
  • \$\begingroup\$ @Passerby I agree it is totally arbitrary, as I don't have the specs of the diode. This was just to show the logic. But what do you mean, at the Vs provided? The supply voltage? Vf depends on I, not the supply voltage. \$\endgroup\$ – dim Feb 7 '17 at 15:30
  • \$\begingroup\$ Vf typ. of the LED is 2.0 also I will never drive the LEDs fully lit, so even If the battery voltage drops they should still light up, although less bright, right? \$\endgroup\$ – Seeker Feb 7 '17 at 19:08
  • \$\begingroup\$ also 14.7 - 2*7 = 0.7V time 20mA times 14 strings per chip, (14 of 28 strings) is 0.196W \$\endgroup\$ – Seeker Feb 7 '17 at 19:14

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