Powering the BBB

I am building a robot that uses an 11.7V battery. I tried attaching a 5V regulator to the T5(+)/T8(-) (pins with the T6(SENSE) bridged to T5 and a 10K resistor between T8(TS) and ground. The board only stays on for about thirty seconds then shuts down. Is it because of the TS? What type of thermistor does it use? Or should I try to use the barrel jack since it’s not directly connected to a battery?

Can you share a schematic with me? I can’t visualize what you are doing.



That particular regulator can have different pinouts. I have been bitten before. So double check that based on the supplier.

What voltage are you seeing on the output? I believe the best you can get out of that regulator without a big heatsink is maybe 500mA which is borderline for the BBB. You are dropping 6.7V across the regulator.


Addicore 5V 1.5A Positive Voltage Regulator L7805CV

Are you using a heatsink? Voltage?


No heatsink 5V

OK. Then what you are describing as your issue, i would expect to happen.


Because the IC can’t provide the required current without a heat sink? Or because the BBB is going into thermal protection? A little insight into why you would expect that would be appreciated."

Because the regulator is going into thermal limit due to excessive heat dissipation. Make the 11.7V supply say 8V and it will run longer. The hole in the tab of the regulator is for the heat sink.

Now, if the regulator is putting out 7V, then it could be the board.


11.7V supply shouldn’t be a problem because it can take up to 25V, but I’ll try a heat sink and see if that helps.

This is a linear regulator. Very old technology. A watt waster. It is dissipating the difference between the input voltage and the output voltage. That creates a lot of heat.


11.7V supply shouldn’t be a problem because it can take up to 25V, but I’ll try a heat sink and see if that helps.

Gerald is exactly right about this. You must consider all the parameter of the regulator, not just one. The data sheet specifies the heat that can be dissipated with and without a heat sink. Look at the operating envelop. BTW, you should be using a switching regulator as it is much more efficient and won’t generate that much heat.


in hopes of explaining what is being talked around is this. You want 5V from 11V so there’s 6V difference. Those linear regulators will give you the 5V at some current level(let’s say 1A for simplicity). So you get 5V at 1A and that’s 5Watts(5V*1A) but that 1A of current is also involved in that 6V drop from 11V to get you 5V. The energy/power wasted in that 6V drop is calculated by the current of your load( 1A ) times the drop( 6V ) which is 6Watts in the example.

This is why Gerald is correct in his statement that the linear regulator is wasteful since in the example you only need 5W and have 6W of waste so have a total energy cost of 11W. A better solution would be a DC->DC converter and you can get those on amazon with free shipping from china(if you can wait) for less than $10. The take the input power turn it into a AC signal which can then be chopped and reassembled into a lower voltage with little energy loss.

for example:


The most important detail here is the 6W of waste heat. A linear regulator regulates the Voltage by generating heat. The heat generated in a linear regulator is measured in Watts and can be calculated as the Voltage dropped across the regulator times the current through it. From this discussion I understand that to be 6W (6V * 1A). The regulator will have a specification for the temperature rise from the die to case (or ambient) per Watt. A common LM340 in a TO-220 case, for example, has a 4C/W junction/case and 54C/W junction/ambient rating. In this example that means with no heatsink the junction will be 324C over ambient (54C/W * 6W). The 24V limitation is not the factor here, it’s dissipating the heat. You will need a heatsink capable of disipating at least 17.6C/W to ambient just to keep the LM340 TO-220 from reaching its rated 150C limitation with an ambient temperature of 20C (20C + (4 + 17.6C/W) * 6W = 150C). At elevated operating temperatures you will need a better heatsink than that. I don’t know what regulator you are using as you have not stated that publicly yet, nor do I know the maximum temperature for your application so I can’t guess what size heatsink you will actually need. Short answer: Use a big heat sink or a switcher. Artie