Electronic – Finding current through LEDs in series-parallel circuit

You don't need the series resistors to control the current; the TLC5940 is a current source, and doesn't give a damn about the resistance. If you program it to source 120 mA it will source 120 mA, whether the series resistor is 0 Ω or 15 Ω (15 Ω is the equivalent of six 90 Ω resistors in parallel). Your calculation of the 90 Ω is correct: 3.2 V for the LED plus 20 mA \$\times\$ 90 Ω = 5 V, so a larger resistor would need a higher voltage to keep 20 mA flowing. If the 5 V is all you have increasing the resistance will decrease the current. A 100 Ω resistor for instance will allow

\$ \dfrac{5 V - 3.2 V}{100 \Omega} = 18 mA \$

Further increasing the resistance will also further decrease the current.

There are two reasons why you should use series resistors:

• balance the currents between the LEDs. Small differences in LED voltage will be compensated by the voltage drop across the resistors. Since the resistors will all see the same voltage from the TLC5940 a 90 mV higher voltage for LED 1 will cause 90 mV less for its series resistor, and hence a 90 mV / 90 Ω = 1 mA less current, which won't be noticeable
• but more importantly, they reduce the power dissipation in the TLC5940. If it supplies 120 mA per output, and you wouldn't have series resistors, then it would dissipate (5 V - 3.2 V) \$\times\$ 120 mA = 220 mW. Not too much it seems, but the TLC5940 has 16 outputs, then that 220 mW may become 3.5 W!! That's too much for the poor device (see power rating on page 3 of the datasheet, at 70 °C, not at 25 °C). So the resistors literally take some of that power out of the device. At 90 Ω there would not be any dissipation in the IC at all, but the current source wouldn't have any headroom either. So it's better to choose the resistance a bit lower: 68 Ω will leave 5 V - (3.2 V + 20 mA \$\times\$ 68 Ω) = 440 mV for the IC. 16 outputs driving 120 mA each will then cause an acceptable 850 mW dissipation.

It's good to hear about family projects :)

I'm not sure I understand your description of the circuit topology. I'm assuming that for each 9V battery, he will have four strings of LEDs in parallel to each other, and that each string consists of three LEDs wired in series. Is this correct?

• First, you have to add up the Forward Voltages of each string, and make sure the total is lower than your power source (9V). If there are any string that total more than 8.5V (or so), you'll need to arrange them in a different way.

• Then, you need to figure out your desired current for each string of three LEDs. The datasheets will show what the acceptable Forward Current levels are. Since all three LEDs are in series, they each flow the same current. So you want to limit this current to lowest-current LED. Since each LED in a string is carrying the same current, it won't matter which order you put them in.

• Now, for each string, add up the forward voltages, then divide that sum by your desired current. This will give you an initial resistance value. It's wise to use a larger resistor that you calculated, and then decide if you want to lower it later by measuring the actual current with a multimeter.

• You should use one resistor per string of LEDs. Their forward voltages aren't going to match exactly. If you use only one resistor, and you get a large enough mismatch between strings, then you could burn one of them up.

• Regarding the switch, it won't effect the performance of the LEDs. It will cause the battery to drain a bit faster. Basically, the switch is adding one more LED and a resistor in parallel to your circuit.

• And finally, regarding battery life, that really depends on your LEDs. Lets assume you drive them at 20mA. If you have four strings in parallel, this will take 80mA all together. A typical 9V battery has around 500mA-hours of capacity. Divide 500mA-h by 80mA, and you get over six hours :)

I hope this helps!