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Alternative Energy Charge Bank

Electric-T

Jun 4, 2017
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Here is the design. There are some redundancies in here. Probably a little over engineered. Let me know what you think
 

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kellys_eye

Jun 25, 2010
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What is the reason behind the inclusion of R1 and R2? They will do nothing other than discharge the battery.

C1 and the LED should be across the output of the regulator IC and to 'ground'. The shorting link between them should be removed.
 

Electric-T

Jun 4, 2017
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What is the reason behind the inclusion of R1 and R2? They will do nothing other than discharge the battery.
Its a voltage divider. Ive never wired one but this is how its supposed to go I guess. I used (Vout=Vin*r2/r1+r2)
 

kellys_eye

Jun 25, 2010
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You don't need a voltage divider. The regulator IC will accept a 12V input quite happily. The resistors will only offer a constant drain on the batteries and operate as a (poor) heater - nothing else.
 

Electric-T

Jun 4, 2017
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You don't need a voltage divider. The regulator IC will accept a 12V input quite happily. The resistors will only offer a constant drain on the batteries and operate as a (poor) heater - nothing else.
It wont cause too much stress on the vr? seems like a big drop
 

kellys_eye

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Get rid of R2 and replace R1 with a wire link.

Add a series resistor to the LED.

The regulator (7805?) will probably require a heatsink - depends on how much current the USB ports are designed to deliver. If it's more than 1A total then you'll need a different regulator than a 7805 (assuming that's what you're using?).

Ideally you would use a small SMPS (buck) converter to deliver the 5V output as it is more efficient.
 

Electric-T

Jun 4, 2017
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Get rid of R2 and replace R1 with a wire link.

Add a series resistor to the LED.

The regulator (7805?) will probably require a heatsink - depends on how much current the USB ports are designed to deliver. If it's more than 1A total then you'll need a different regulator than a 7805 (assuming that's what you're using?).

Ideally you would use a small SMPS (buck) converter to deliver the 5V output as it is more efficient.
7805 5V Regulator is what I'm using. Datasheet says its max current output is 1.5A. So that's as much current as the usb ports will get. (Two ports wired in parelell, so at max draw each port should get 750 mA) I'll use a salvaged heatsink. Other than eliminating the voltage divider, everything looks good?
 

kellys_eye

Jun 25, 2010
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The 7805 requires a capacitor across the input to help prevent oscillation - see the datasheet.

At full load (1.5A) the regulator will dissipate (or try to) ((12-5)*1.5) watts = 10.5 watts. That's a considerable, and unnecessary, loss of energy.

Many buck converters operate at 90% efficiency producing far less heat (waste) and won't require a considerable sized heatsink either.

As already suggested, changing the 7805 to a buck module would extend battery power significantly.
 

Electric-T

Jun 4, 2017
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The 7805 requires a capacitor across the input to help prevent oscillation - see the datasheet.

At full load (1.5A) the regulator will dissipate (or try to) ((12-5)*1.5) watts = 10.5 watts. That's a considerable, and unnecessary, loss of energy.

Many buck converters operate at 90% efficiency producing far less heat (waste) and won't require a considerable sized heatsink either.

As already suggested, changing the 7805 to a buck module would extend battery power significantly.
Id like to keep it small. I will look into a buck converter though. That was the point of the voltage divider though, to not waste so much energy to heat. Though I guess the resistors would do the same thing
 

kellys_eye

Jun 25, 2010
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You're on the right track now! Change the breaker to a 2A (3A) version and make sure the wiring can handle at least the current the breaker is rated for.
 

kellys_eye

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Breakers protect the wiring, not the equipment. Overloading the output may instigate shutdown of the regulator (check to see if it has protection built in, many do) but if your regulator is rated at 3A and has no protection then fusing at 4A won't help.....
 

Electric-T

Jun 4, 2017
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Breakers protect the wiring, not the equipment. Overloading the output may instigate shutdown of the regulator (check to see if it has protection built in, many do) but if your regulator is rated at 3A and has no protection then fusing at 4A won't help.....
I see. Well the usb ports i looked at can supply up to 750mA. The converter 3A. So 1.5 A looks like the max draw since there are two ports. So is a 2 A breaker too high? In theory, current draw will never exceed 1.5A
 
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