# How to turn on LED with only a cathode pin

I'm looking at the datasheet for some TIL311 hexadecimal displays which I ordered (but have not yet arrived. (Actually they appear to be TIL311 clones, but should be compatible.))

According to the datasheet, the left and right decimal point LEDs have their cathodes exposed via pins 4 and 10. The anodes are not connected to a pin; I assume they are hardwired to Vcc within the IC.

I want to turn the decimal point ON when I have +5v on a certain wire.

Am I correct in thinking that the right way to do this would be to use a discrete transistor to allow current to flow from the cathode to ground when it (the transistor) turns on? If so, what's a good kind of transistor to use?

Or is there a simpler way that I haven't thought of?

• Have you read the datasheet that you link to? It tells you right there. On page 2. – Majenko Dec 13 '16 at 20:17
• Have I missed something? On page 2 I see the note about current-limiting resistors for the decimal point LEDs but no suggestions about how to drive them. – friedo Dec 13 '16 at 20:24
• The decimal point anodes are connected to the LED supply; - did you miss that bit? – Majenko Dec 13 '16 at 20:24
• No, I got that part - that's what I mentioned in my post. I need to control the decimals independently of the other LEDs in the chip. I can't turn the LED power supply on and off. Sorry of that wasn't clear. – friedo Dec 13 '16 at 20:25
• You said: The anodes are not connected to a pin; I assume they are hardwired to Vcc within the IC. - that says you didn't read the bit I pointed to. – Majenko Dec 13 '16 at 20:26

You could use a low-side NPN transistor (e.g. 2N4401) or N-channel MOSFET (e.g. 2N7000). Connect the dot pin to a resistor (~1kOhm), and connect the other end of the resistor into the switch's collector/drain. Connect the switch's emitter/source to ground and feed an on/off signal to the base/gate through a resistor (say 10~100 Ohms).

Unless you want it always on -- then you could just connect a resistor to the dot pin to ground (again, ~1k Ohm).

• Thanks - this is very helpful. I think I have some spare 2N4401s lying around and will try them. – friedo Dec 13 '16 at 20:30
• Great! One quick thing: the 10~100 Ohm resistor was for the MOSFET -- see Spehro's suggestion of ~15k for the base resistor if you're using a 5V signal. – calcium3000 Dec 13 '16 at 20:34

The datasheet schematic and description tell you all you need to know. The anodes of the decimal point LEDs are connected to the supply and you need to provide current limiting. The other LEDs are controlled at 5mA more-or-less. I do not see a spec for decimal point forward voltage, but it should be about 1.8V for a red LED, so a resistor of about (5V-1.8V)/0.005A = 640 ohms (use 680 as the next highest standard value) should be fine. With 5mA collector current, your base should get something like 1/20 of that, so that's 250uA. Vbe is about 0.7V so a base resistor of (5V-0.7V)/250uA = 17.2K (you can use 10K or 15K) will work well.

simulate this circuit – Schematic created using CircuitLab

When the input is at 5V, the 280uA or so base current turns Q1 on, 5mA passes through the decimal point LED and it illuminates.

• Thanks for the answer - I'm aware of the need for current limiting, just not sure how to actually turn the decimal LEDs on and off. – friedo Dec 13 '16 at 20:29
• Actually, go for the next lower value (560 ohm) since the transistor will drop some voltage as well. – JvO Dec 14 '16 at 1:18

Read carefully the datasheet. The info you are looking for is there.

Therefore the two decimal point LEDs are connected between pin 1 (LED supply, common for both) and the two cathode pins (4 and 10).

Then it's up to you how to power them up. If you want to control them independently, you just wire pin 1 to a suitable positive power rail and put any switching device between each cathode and ground. It may be a BJT, a MOSFET or even a relay. If you want to control them via a MCU GPIO pin, you can also connect it to that pin(s), as long as the MCU can sink enough current (of course you must add a limiting resistor in series).

Since the datasheet reports a 5mA recommended operating current for DPs:

you may be confident that many modern MCUs can sink that much current (for example a GPIO pin of the ubiquitous Arduino boards, from its ATmega328P MCU, can handle it without problem).

For example:

drive the GPIO pin LOW to light up the decimal point whose cathode is connected to pin 4 of the display.

• Thanks - I suspected a BJT would probably be a good solution but wanted to make sure. – friedo Dec 13 '16 at 20:32