In place of the LM7805, using a Low Drop Out (LDO) regulator such as MIC29300-5.0WT would allow operation at as little as 5.37 volts, providing 3 amperes at that voltage, enough for your Beagleboards but not the motors.
The more expensive MIC29500-5.0WT at $5.46 apiece on Digikey would provide the 5 amps you might want instead.
- Tapping between batteries is best avoided with unbalanced loads.
- And yes, the batteries loaded higher would deplete earlier, with unpredictable outcomes.
- In stepping down 6 to 7.5 volts to 5 volts at 5 amps using an LDO, up to 12.5 watts of heat would necessarily be dissipated at the regulator:
Power dissipated as heat = voltage difference * average current
= 2.5 * 5
However, realistically the batteries won't be at 7.5 volts under load, and not for long anyway. Also, the continuous current drawn would not be 5 amperes through the regulator. Your design would ideally provide the motors with unregulated power, so your "step down about 4-5 amps" deserves a revisit. Besides, voltage is stepped down, not current, at least in this context.
Switching (buck) regulators would waste less heat, roughly around 15% that of the LDO, but they usually need more voltage headroom to operate. YMMV, searching DigiKey might yield what you need.
Why is your hub needing to be powered from the same power as your car motors? Your hub should be at your base station, the car would just carry a WiFi dongle of some sort, ideally?
Suggestions:
- Edit: [15-Oct-2012] Provide regulated power only to the logic parts, drive the motors directly off the battery, and add a hefty decoupling capacitor before the regulator to work around motor inductive spikes. If your batteries can handle 4 motors of 2 amp each, they can handle the initial surge the capacitor will need.
- Provide a diagram of some sort, of what is being attempted, and some clarity on how the powered hubs got aboard the moving car. This will enable more helpful answers.
it appears to me that the current generated by 35 V and 2vx will
collide each other
It may be that you are assuming that a voltage source, whether independent or controlled, must source current, i.e., supply power to the circuit.
But, at least in ideal circuit theory, there's nothing "wrong" with a voltage source sinking current, i.e., receiving power from the circuit.
For a real world example, consider that, when a battery is being charged, the current is in the opposite direction than when the battery is being discharged.
I would like to know how the current flows across 5 Ω resistor.
If you're planning to be an EE, don't write or say things like "current across"; current is through, voltage is across.
Now, this circuit is very easy to solve. There are two unknowns so you need two independent equations.
For the 1st, write a KVL equation clockwise 'round the loop:
$$35V = v_x + 2v_x - v_o \rightarrow 3v_x = 35V + v_o$$
Now, you need one more independent equation. Can you find one?
Best Answer
By the time a "car battery" gets down to 12 volts, it is considered dead (mostly dead, Princess Bride). Although I think I would agree with you that my first instinct was to think of 0 volts. When I thought about it scientifically, and looked it up :
I should give credit to : http://www.w8ji.com/battery_and_charging_system.htm
View it this way: The internal resistance of the battery (when discharged) becomes high. You could model the "dead" battery as 12 volts with a high resistance in series with the battery.