I am using a Beaglebone Black development board (BBB) by Texas Instruments as part of a control unit for a race car at university.
My task is to develop a circuit to activate the reset function on the dev board when the power is cut suddenly, because if it's not properly powered down (OFF button), the system gets stuck or hangs or something.
This is essentially the schematic I've come up with:
I'm thinking of using a NOR gate, with one input tied to Vcc (5V), and the other to ground. So basically when Vcc goes LOW, the output of the logic goes high and triggers RESET.
The Beaglbone has the ability to charge an external battery to 3.7V, so this pre-charged battery will keep the logic gate powered.
One thing I haven't indicated in the drawing is that I will also use a reservoir capacitor as a buffer between Input A and Vcc.
I just want to check that my thought process here is correct. BBB uses roughly 60mA at 5V during the reset process. That's roughly 0.3W or 0.3J/s. I was informed that the reset operation is instantaneous, but as a worst case scenario I just assumed it would take 5 seconds for instance. If that's the case, then it would require 1.5J from a reservoir capacitor.
Using E = 0.5 * C * V^2, I calculated C= 120mF
To get a resistance model of BBB to calculate the RC time constant:
R = 5V/60mA = 83 Ω
Therefore τ = 83 x 120mF = 9.96 seconds
Using the capacitor discharge equation, V = Vo.e^(-t/RC) and the fact that BBB will operate with at least 4.8V supply
I calculated t = 0.407 seconds for the voltage across the capacitor to drop to 4.8V which I figure should be enough to allow the reset operation to complete, if not I can just bump up the capacitance rating.
Thoughts?
My task is to develop a circuit to activate the reset function on the dev board when the power is cut suddenly, because if it's not properly powered down (OFF button), the system gets stuck or hangs or something.
This is essentially the schematic I've come up with:
I'm thinking of using a NOR gate, with one input tied to Vcc (5V), and the other to ground. So basically when Vcc goes LOW, the output of the logic goes high and triggers RESET.
The Beaglbone has the ability to charge an external battery to 3.7V, so this pre-charged battery will keep the logic gate powered.
One thing I haven't indicated in the drawing is that I will also use a reservoir capacitor as a buffer between Input A and Vcc.
I just want to check that my thought process here is correct. BBB uses roughly 60mA at 5V during the reset process. That's roughly 0.3W or 0.3J/s. I was informed that the reset operation is instantaneous, but as a worst case scenario I just assumed it would take 5 seconds for instance. If that's the case, then it would require 1.5J from a reservoir capacitor.
Using E = 0.5 * C * V^2, I calculated C= 120mF
To get a resistance model of BBB to calculate the RC time constant:
R = 5V/60mA = 83 Ω
Therefore τ = 83 x 120mF = 9.96 seconds
Using the capacitor discharge equation, V = Vo.e^(-t/RC) and the fact that BBB will operate with at least 4.8V supply
I calculated t = 0.407 seconds for the voltage across the capacitor to drop to 4.8V which I figure should be enough to allow the reset operation to complete, if not I can just bump up the capacitance rating.
Thoughts?