Andrew StarrToday I did some voltage, current and speed tests on the newly-completed counter section:
| Supply Voltage (V) | Maximum Clock Frequency (MHz) | Average Current Consumption at Maximum Frequency (A) |
| 5.0 | 1 | 1.0 |
| 3.3 | 1.5 | 0.8 |
| 2.7 | 2 | 0.6 |
It's interesting to note the effect of slew rate when comparing supply rail vs maximum clock speed. So lower voltage is better in this regard. However at 2.7V the LEDs are starting to dim, and since 1 MHz is my target clock speed, 3.3V seems a good compromise.
Now I have to make some 2 to 4 decoder modules to generate the row/column signals from the 6-bit address...
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I keep forgetting to hit 'reply'....!
Are you sure? yes | no
Most definitely a big chunk of power is used by the LEDs, but they're an important part of the project, so that's fine. I could probably boost the brightness on 2.7V by reducing the collector resistors that feed them, but I don't need the extra speed and 2.7V is an awkward voltage - a high current 3.3V supply will be easy to source.
Are you sure? yes | no
2.7V is only 3.3V minus a PN diode drop ;-)
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And how much current will be going through that diode? Hint: a lot :)
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The "right" way to power something like that is to distribute power and have step-downs all over the place. Some cheap Cray (mid-90s ?) were like that, with a 48V main rail and DC/DC all over the place.
I could send you several high-power step-down modules from TI that their samples program sent me a decade ago ;-) They are quite bulky (newer generations are even more efficient and compact) but can sustain more than 10A each with easily programmable output voltages.
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Awesome :-) Thanks for the numbers !
it's rare to be able to cut power consumption by 3.
Maybe a lof of power is also drawn by the LEDs ? :-D
Could you change the LED system so you can use 2.7V ?
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