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robot power source circuit [power supply, battery charger, power distribution, etc]

Thread starter #1
Hello everybody,

I am building a robot with 3 wheels, where two of these are powered independently. I am currently building the power source circuits. I made a diagram, which represents the hole idea, and it follows attached to this message. I am thinking to make a single printed circuit board for this circuit.

I would like to hear any comment from anyone that think that can help me somehow, in order to evaluate this idea.

The main idea behind this circuit is that when I need to recharge the batteries, I must only plug a wire on the walls mains source. When it is charged, I would unplug it, and let the robot run. The thing is, to avoid the batteries being discharged thru the charger circuit, I am thinking about using some diodes before the battery. What do you think? But then I think I must increase the voltage due to the tension losses inside the diode. is that right?

Well, I'd appreciate any help on the subject.
Rafael Vasconcelos


you're worried about the voltage sense loop draining your battery? The .5 ohm resistor/transistor pair limit the charge current to 1.2A (or not), so a 3A diode would suffice. However, batteries charge by bringing their input voltage up to a certain level, so a series diode will change the LM317 sense level vs. the actual battery voltage level. You will probably have to adjust the resistors accordingly.
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Thread starter #5
Hi Mike, thanks for the reply.

Yes, I am worried about draining my battery when disconnecting the power source.

I found this circuit [attached], which seems to solve this problem I was facing. What do you think? It looks easier to me, as I am not so good at electronics.


Mike odom

Active Member
nope... for all intents, that is the same circuit, although it will drain your battery faster. The 1N4001 is to drain the 100uF cap in case the input side goes lower than the output of the regulator, and in your case, this will be the battery. When you remove input power, the diode will conduct until the output voltage is lower than the input voltage. This is to protect the LM350. A lot of chips don't like having power on the output pin when the input power is removed. You see this a lot on microcontroller circuits. There will be a 'clamp' diode to 5V to remove reset voltage from the reset cap when 5V power goes away. You have an adjustment on your voltage level anyway (1K pot), so adding a diode inline won't matter. It goes between the output cap and battery, cathode facing the battery (opposite what's shown in your 'new' schematic).
Thread starter #7
yes, you're right.

I have already realized that, as I read again about the circuit. So, If I put a diode right before the battery it will do, if I adjust the potentiometer?

I'm thinking about the 1n5821 to do this job.

thanks a lot.

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