9th November 2010 07:37 PM
questions about powering leds in large matrix like mikes macmux design
using mike's design shown here I was trying to up the current driving. and was wondering if the 74hc595's with uln2803s in chain with them would work.
Also if the Si2312BDS 5 amp N-fets would need to be better?
The leds will be 20ma 3.1v blue 5mm, in the same amount as his design.
10th November 2010 07:54 AM
He is doing a row scan not a column scan.Positive supply to the LED's will comes from 74HC595 & flow current to GND via row side.So you don't need additional ULN drivers.
I haven't use N-fets as he mentioned because those are very hard to find.
Use darlington on row side.
10th November 2010 08:09 AM
With that "cathode row" design, you could replace the 74HC595's with Micrel MIC5891 serial-to-parallel "sourcing" driver ICs but they're a bit expensive. Use two sets of Si2312BDS N-FET "sinking" row drivers (inputs connected in parallel) with outputs from each set driving the rows for a single 8x64 block.
With an "anode row" design, you could use MIC5821 or TPIC6C595 "sinking" column driver ICs and P-FET "sourcing" row drivers. Or you could use the 74HC595's driving ULN2803A's you mentioned.
Last edited by Mike, K8LH; 10th November 2010 at 11:05 AM.
10th November 2010 08:19 AM
Good Day Mike
Where were you all this time! I was waiting.
I had a similar issue on the current design 16 X 64 multiplex.
For a 16 rows design you are divided into two rows & do the scan in this case brightness will be 1/8.When its comes to real life adding two sets of column shift registers to the top row 8s (8, 595ís) & bottom row 8s (8, 595ís) becomes very hard.
I planed to scan 1/16 using 8, 595ís by adding some 64 PNP transistors to the each 595 outputs.
Whatís your comment sir!!
10th November 2010 08:40 AM
My current LED arrangement is "Cathode row design" that's why when the no of rows gets bigger the hardware part goes hell of a mess.
Thinking to do my LED arrangement like "Anode row design" so all I have to place ULN with each 595 on column side & only need 16 current source transistors on row side.
I need your comments Mike.
10th November 2010 10:49 AM
I prefer to use a lower duty cycle to reduce "peak" current requirements. To get the brightness of 10-ma "average" current per LED you would need to provide about 160-ma "peak" current to each lighted LED on your 1/16th duty cycle design. This is beyond the typical 100-ma "peak" spec' for many LEDs.
The "anode row" design provides a lot more choices for your sinking column drivers. A 74HC595 driving a ULN2803 should work fine. You can also replace that combination with a single but somewhat more expensive IC like the TPIC6C595 or MIC5821 (many others too).
10th November 2010 06:20 PM
Thanks Gayan and Mike,
So is there any reason to use anode row driving vs cathode row driving aside from hardware differences?
Also, would adding shift registers or demux's to the rows to reduce IO lines at the MCU cause a lower refresh rate?
11th November 2010 01:33 AM
mike, another question, my LEDs are 20ma, someone on allaboutcircuits forum informs me that your design, as it says, only sources 4ma to the leds, how can I up this to 20ma?
11th November 2010 01:36 AM
You'll want to drive them at more than 20mA as they're only on 1/8 of the time. 60mA is common and may not damage them as quickly as 160mA. Also HE red are nice and bright with less current that standard red LEDs.
One way is PNP driving transistors.
11th November 2010 01:53 AM
so a pnp on the output of each shift register? or arrays if I can find them?
would I also need to up the 5amp cathode row drivers?
Electronic Circuits |
Page Time: 0.09283 seconds Memory: 7,659 KB Queries: 16 Templates: 0