Control a 12V LED circuit with a Raspberry Pi

Question

I have 22 lights that I am trying to control. Each light consists of a resistor and LED, and the manufacurer recommends running the lights at 12V. I want to be able to switch these lights on and off indivdually from a Raspberry Pi. I am aware that there are LED driver ICs, but I am trying to make a circuit out of components that I have access to.

Below is the circuit I have designed and following that is a description.

I have a 12V power supply that I will convert to 3.3V using a ld1117v33 regulator. The shift registers will run on 3.3V as the Raspberry Pi GPIOs opperate on 3.3V, and this voltage is within the tollereance of the 74hc595 shift register. The Raspberry Pi's ground pin will be tied to the ground of the 12V power supply, and the Pi will be powered by its own power supply. To control the light, there is a transistor operating as a switch. \$A_{R1}=\frac{3.3V}{1000}=3.3mA\$, which is within the tollerance per pin of the shift register.

I intend daisy chaining 3 shift registers together, and will have a transistor and resistor pair per light.

Will this circuit work as I intend? I am worried about damaging my expensive Raspberry Pi.

Answer 1

What you have should work, but could be improved:

• R.pi GPIO is 3.3V, so the HC595 also has to be powered 3.3V to have compatible levels (so, ok.)
• Don't draw power from the Pi. Instead, make local 3.3V for the HC595's using an LDO from 12V (so, also ok.)
• You have the option of running the HCT595s on 5V, which is compatible with 3.3V logic so you could use USB power and save the regulator. That's kind of messy, wiring-wise assuming you want to tap from the same USB as the Pi, so I don't recommend it.
• Add series damping resistors to the shift register controls. Make sure there's a good ground return for them.
• Add bypass caps to the HC595s
• Use n-FETs instead of NPNs to drive the LEDs. This will reduce Idd to the HC595's.

The latter point is important. Worst-case, you have 3.3mA * 22 NPN bases = 72mA. This base drive comes from the 3.3V linear regulator, which will dissipate up to 631mW when all 22 lamps are on. That's kind of wasteful.

With FETs instead of NPNs there will be only the HC595 static draw, so the regulator dissipation is far less: just 80uA (696uW) per shift register shed in the regulator.

What kind of FETs? Any common n-FET like 2N2002A or BSS138 can work.

Answer 2

The schematic as shown will work fine.

You could replace the transistors with ULN2003. It is easier to solder 3x SOIC chips than 22 transistors and resistors.

You could also replace the whole thing with TPIC6C595, also exists as NPIC6C595 and 596. The 596 version is preferable as it shifts data out on the negative clock edge. It's basically a 74HC595 with integrated open drain power MOSFETs that can drive your 12V LEDs directly. It runs on 5V so you can power it from the Pi's 5V supply or replace the 3V3 regulator with a 5V one and add a 74HCT chip powered from 5V to do voltage translation between 3V3 and 5V.

If you're feeling extra paranoid and want to protect the Pi's outputs from anything that can happen on the 12V board:

Use a MOSFET that is fully on with 2.5V Vgs. This will also handle voltage translation between 3V3 and 5V. MOSFET source is on the bottom. If something goes wrong and the output goes to 12V, the MOSFET will limit voltage at the Pi to a bit less than 3V3.