I am a newbie trying to finalize my first project (a MIDI foot-switch controller) using a shift register 74HC595.
The issue I am encountering is that the LED I am using require 9V but the output of the 74HC595 is 5V...The LED have a embedded resistor (it is actually a set of LED rings)
Do I need a new shift register able to handle a higher voltage or add some sort of voltage step up after the shift register or is there another way to deal with this type of configuration.
I have a limited knowledge of electronic so any help would be very useful.
Thanks a lot
You can use small MOSFETs to switch the voltage. The + side of the LED ring goes to +9V and the - side goes to the Drian of a MOSFET. The Source of the MOSFET goes to Ground (common to Arduino, 74HC595, and negative side of 9V supply). Gate of the MOSFET goes through a 200+ Ohm resistor to the output of the 74HC595.
Given that you actually do need the 9 V (because LEDs are in series), do not add FETs unless you need more than 250 mA for each LED ring. Of course the 74HC595 is limited to 25 mA anyway.
You use a TPIC6B595 for up to 150 mA per output - there are other versions than the "B" for more current - the LEDs connect with negative to the TPIC output and the positive to the respective supply. Just as you would do if you used individual FETs.
Thanks a lot for the tip. So if I understand correctly the TPIC will act like a sink not a source and I will have to power the LED directly from the 9V source but the ground will be “controlled” by the IC . Correct ?
I think using a transistor is probably what I should do. I don’t think using the TPIC6B595 will be feasible because the ring has two color but only one ground. So if I understand correctly how the TPIC work , I am not sure how I will separate the two color.
Would a BJT work ? Any recommendations on what model?
So transistors or FETs as such will have exactly the same problem.
There is going to be a bit more work involved here. These are designed for 9 V because as I understand it, that is the voltage used by the power packs that are used for effect pedals.
Start by getting out your multimeter and measure the current that they draw (for each colour) when powered by a regulated 9 V supply (that is, one that actually is 9 V), and what happens if you try to power it with 5 V.
That is called a "Common Cathode" configuration (Cathode being the negative side of the LED). That won't work on low-side switches like an NPN transistor or N-Channel MOSFET. To control the two anodes you need high-side switches. You can use a PNP transistor or P-Channel MOSFET but in order to turn them off, you have to raise the Base/Gate voltage to match the Drain voltage. In your case that is 9V so you can't connect it directly to an Arduino pin. One way is to use a pull-up resistor to 9V on the Base/Gate, and then use an NPN/N-Channel to pull the Base/Gate down to turn on the PNP/P-Channel down.
I think you lost me with the pull-up part
Does the schematic below make sense ? (sorry if it doesn't. I am very new with this. - I am a guitarist trying to improve his rig but I don't really understand all the electronic details.
