Help me to finish this project - Page 1
Forum section: electronics / projects

author: LinkZ
author_rank: Contributor
timestamp: June 08, 2015, 01:21:11 pm
content: Hi, I've made this clock:Now let's say that i need to drive a much larger display that needs 12V, so i need a shift register (ULN2003N) as stated here:https://www.eevblog.com/forum/projects/led-clock-out-of-single-3mm-leds/(reply #10)so i will now connect a shift register on the ANODES of the leds, but what to do with CATHODES? They are connected to a transistor acting as a switch (i think) so if i repleace the base resistor of this transistor (4.7K in the schematic) i can make it work with 12V, right? So i simply have to feed 12V in the transistor and it works? Generally, I'm really confused about HOW to drive a larger display starting from an existing circuit like this. I hope someone will me clarify this puzzle

author: Ian.M
author_rank: Super Contributor
timestamp: June 08, 2015, 01:48:24 pm
content: No. The PNP transistors are powered from the 5V rail. If you power them from >5.5V, the MCU driving them wont be able to output a high enough voltage to ever turn them off. To allow the base to reach a high enough voltage to turn off with a 12V supply, you will need to add non-inverting open collector buffers (e.g.74LS07) rated for 12V on their outputs between the base resistors and the MCU, recalculate the resistor values, add pullup resistors on each base and maybe change transistors (depending on the collector current and voltage ratings of the existing ones).

author: LinkZ
author_rank: Contributor
timestamp: June 08, 2015, 02:11:02 pm
content: Each segment of my display is made out of 6 3mm leds (3+3 in parallel), so the voltage is 9V, but i will use a 12V psu. The current is 35nA (yes, nanoamps) that should flow on each series of 3 leds. So, what resistor should I use? is the "shift register" side correct?

author: Ian.M
author_rank: Super Contributor
timestamp: June 08, 2015, 02:21:10 pm
content: I've never heard of a LED display designed to operate with nanoamp LED forward currents. Tens of mA per segment is more usual.ULN2003 is not a shift register, its a seven bit inverting power driver with OC outputs each with a catch diode, optimised for solenoid/relay drive applications.

author: LinkZ
author_rank: Contributor
timestamp: June 08, 2015, 03:34:47 pm
content: So how can I drive this shitty LED display? Can somebody post some calsulations and schematic?

author: DJohn
author_rank: Regular Contributor
timestamp: June 10, 2015, 04:06:23 pm
content: For the digit drivers, you could use an NPN transistor in an emitter-follower configuration. Collector to +12V, emitter to the LEDs, base through a resistor to +12V and also through a ULN2003 to ground.The resistor values will depend on what the LEDs are. 35nA isn't like any LED I've ever heard of either. I think you've misread something. Without knowing what they are, we can't guess. And you'll probably need some trial and error anyway, since it's unlikely you'll want them running at full brightness. I miss the good old days when LEDs didn't require welding goggles.You do know not to connect your two strings of LEDs directly in parallel, don't you? You want LED-LED-LED-resistor, and then parallel those. For this multiplexed application, you can save some of the resistors, since only one digit will be on at a time. Connect the first digit like this (so 14 lots of LED-LED-LED-resistor, with each pair sharing a ULN2003 output), and then the other digits can can connect the bottom of their LEDs to the top of the corresponding resistor from the first digit. That way you only need 14 resistors, not 56.If you use a ULN2003 for the segment driver, then the software will have to be modified to invert the segment outputs. With the digit driver I described above, I think the digit outputs will have to be inverted too. If you're using ready-made software and are unable/unwilling to change it, you could add some inverters. But changing the software is really the way to go. It isn't that hard.