Current consumption using a lead acid battery

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Hi,

I have an STK500 based AVR development board that has a 7805 voltage regulator, 4 LEDs for debugging, and an L293D motor driver chip in addition to the ATmega16 chip.

I have developed an application to run 2 DC motors, 1 stepper motor and 1 servo motor. To power all these i have bought a 12 V, 4.5 AH lead acid battery.

My question is whether connecting this battery will fry my AVR development board, since the overall current consumption of just the board should be less than 500 mA. In general my question is if I have a high amperage power source such as the 12 V, 4.5 AH battery, do i need to connect a current limiting resistor in series to the circuit or will the circuit draw just enough current as it requires.

Thanks,
Sumair

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No current limiting resistor is necessary, the circuit will draw as much as it needs if that much is available.

And lead-acid batteries have plenty of current available, as their internal resistance is very small. I think tens if not hundreds of amps.

You seem to be confusing the 4.5AH rating with something else, as it just means the battery can supply 1A for 4.5 hours or 4.5A for 1 hour or 9 hours if 0.5A is drawn.

But what worries more is the regulator, it has to drop 7V at 500mA, so it dissipates 3.5W of power and will heat very much. So in that sense you could add series resistance so that shares the heat. Or use a switching regulator instead of linear regulator. You must keep regulator input at 8V minimum, so at 500mA current and 4V voltage drop you need about 8 ohm resistor, and power rating of about 4V*0.5A=2W.

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I thought STK500 was designed to run off 12V DC...

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Hi,

Thanks for the replies. I just connected the battery and was able to get the 2 DC motors running just fine without anything heating up. And yes i did not connect any current limiting resistor.

Tomorrow I shall try connecting the stepper motor as well and see how it works.

Thanks,
Sumair

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Watch out for the voltage regulator. That is the one that likely to over heat with large loads.

Jim

 

Until Black Lives Matter, we do not have "All Lives Matter"!

 

 

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Does putting a heat sink with the regulator help in mitigating the heating issue?

Sumair

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You aren't saying that the motors themselves are powered via the AVR are you? The usual setup is to have a micro that switches MOSFETs that in turn connect power to the motor. This means that the micro (and the 7805 used to drive it at 5V) are only drawing 10's of milliamps but the main voltage/current to the motors is completely separate and simply controlled by the AVR - that part of the circuit doesn't have to be (and usually is not) 5 volts. The servo, however is a bit different. Again it's current is not supplied via the micro (just the control signal) but as they have a 4.8V..6V operating range it is usual to power them "upstream" of the 5V regulator. If lots of servos are involved then because of the possible current draw if they are all stalled it is usual to use some kind of switch mode supply rather than a linear regulator.

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No the AVR is just setting and clearing the input ports of an L293E motor driver chip. The motor driver is connected to an external power supply which in this case happens to be the 12 V, 4.5 AH battery.
The servo is also connected to this battery source and the control signal is coming from the AVR.

Sumair

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Quote:

The servo is also connected to this battery source

Not directly I hope! As I say servos have a 4.8V..6V operating range - if you connect one direct to a 12V battery you will probably release the smoke.

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Sorry i meant the servo is connected to the output of the 7805 voltage regulator. So the servo voltage is at 5 V :)

Sumair

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Hi Jepael,

You were right, the 7805 does get hot after 5-10 mins of continuous robot operation. After googling I found out the reasoning as you mentioned above.

The difference between the input voltage and the output voltage is quite large and this multiplied by the output current does dissipate 3.5 W of power as heat. I'll first try with a heat sink if this solves the problem, because I'm using a development board to directly drive my robot. Introducing a resistor as you mentioned means i would have to tweak with the existing PCB.

Thanks,
Sumair