# How to power a big high voltage 7-segment array with minimum parts, without frying my wimpy power supply?

I am trying to design a clock with a big 8-digit 7-segment display formed by about 250 LEDs, but with a power draw limit of 1A from a 5V USB phone charger. Each stroke is four LEDs in series and requires 9V 50mA to drive, and every digit have a maximum of 8 segments (7 segments for the number and one segment for the dots making up the colons.) There is only one boost converter on the board, and I must minimize the amount of constant current sinks I use.

The rest of the system runs on 5V straight off the phone charger, with ATmega328P being the main microcontroller and DS3231 keeping time.

Questions:

1. Will lighting up all 8 segments in a digit fry my MC34063-based boost converter? If so, how do I prevent it from being fried?
2. How to control the thing with minimum amount of power transistors, N-channel only?
• @DoxyLover I already have to multiplex the digits due to my MCU not having that many pins to spare, so there is zero chance all digits come on at the same time. This is the basis of my this thing with only 5 watts. – Maxthon Chan Oct 24 '15 at 18:58
• Just because the LEDs are rated at 50mA, it doesn't mean that you have to run them at that current. LEDs will work fine at lower currents, they just won't be as bright. – Simon B Oct 24 '15 at 19:40

Since you are already multiplexing your digits, you should have no problem with current drain. 7 segments times 50 mA is only 350 mA. And the number of switches required is simple: 8 segments plus 8 digits equals 16 switches. However, the digit switches must be of the opposite polarity from the segment switches unless you are willing to learn how to make high-side drivers.

For instance, let's say your display is common-anode. Then you would want 8 PNP (for BJTs) or 8 p-type (for MOSFETs) for the digit anodes, and 8 NPN or n-type for the segments.

• I think I have made it clear that all power MOSFETs have to be N-channel devices. – Maxthon Chan Oct 24 '15 at 19:23

If you're going to build a 4 digit clock, then 4 LEDs per stroke times 7 strokes per digit is 28 LEDs per digit, and with 4 digits for the clock that's 112 LEDs. Two LEDs for the colon brings that to 114 LEDS, only, so how big a clock are you planning on building?

1. In any case, 50 mA per stroke times 9V per stroke is 450 milliwatts per stroke. 7 strokes per digit times 450 milliwatts per stroke is 3.15 watts per digit to display an "8", which pretty much blows you out of the water before you even get started. :(

2. If you can surmount the power supply problem, I'd suggest static segment drive using something like a TI http://www.ti.com/lit/ds/symlink/tlc59291.pdf

• There are 8 digits (alternate between yyyy.mm.dd and hh:mm:ss and tt.t°C states) and all strokes including the dots all consists of 4 LEDs, so 64x4=256 LEDs. Also one digit fully on means 9Vx50mAx8/efficiency=4W – Maxthon Chan Oct 24 '15 at 19:47
• @DwayneReid: Yeah, you're right; thanks. I was too eager to get something out there, and didn't dot my "i"s and cross my "t"s the way I should have, even though I had that nagging feeling that something was wrong. Thanks for the reality check; I fixed my error and, hopefully, it won't happen again, :) – EM Fields Oct 24 '15 at 20:55
• @MaxthonChan: With reference to your: 9Vx50mAx8/efficiency=4W, what value of efficiency balances the equation? – EM Fields Oct 24 '15 at 21:33
• 90% efficiency. However my LEDs don't actually use that much power and the control circuit uses at most a milliwatt so I think I am still playing within my 5W energy budget. – Maxthon Chan Oct 26 '15 at 11:56