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Topic: Circuit Design from UNO board to ATMega328 without board (Read 7305 times) previous topic - next topic

CrossRoads

Possible, yes, you'd have to keep the iron on the pins for more than a few seconds to do that.

Have small vise? Use that, with a spacer on the pin side, to squeeze it together well.
Buzz the connections from pins to LED leg, make sure you have good connections.
Also possible your breadboard connections are just on the opposite side of the clamp-on connector.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

cincy

#16
Aug 28, 2012, 08:27 pm Last Edit: Aug 28, 2012, 08:36 pm by cincy Reason: 1
ok, got everything working :)  I've attached the circuit image again and was hoping I could have some help determining powering options.  I need to power this circuit via battery.

I have 7 RGB LED's (http://adafruit.com/products/302)
I have 3 74HC595  (https://www.adafruit.com/products/450)
I have 1 DFRobot switch (http://www.dfrobot.com/index.php?route=product/product&path=36&product_id=114#.UD0Pq9ZlQbi)

I was trying to compute max current, so came up with this:

7 RGB * 120mA (peak) = 840mA
3 74HC595 * 160uA = 480 uA
1 DFRobot Switch = 100mA
Arduino UNO Board = ?


Is this how I would sum it up?  And then once I got the sum, how do I determine the right power for continuous load?

thanks again!

cincy


CrossRoads

Okay you need current limiting resistors between the LEDs and the 74HC595.
http://www.adafruit.com/datasheets/sn74hc595.pdf
The 74HC595 is only good for 70mA total, so if you were to turn on all 8 LEDs at once connected to a HC595, you'd want to limit the current to each to <8.75mA.

24 * 8.75mA = 210mA total for LEDs max.
Arduino, 20-30mA
IR switch,100mA if you say so link is not opening for me.

So a 1A wallwart would fit your needs easily.

http://www.dipmicro.com/store/DCA-0510

or a 2A, looks like its less:
http://www.dipmicro.com/store/DCA-0520
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

cincy

ok, that makes sense.

However - I need to calculate the max load and determine how long a battery pack (and what battery pack I need) would last :)


CrossRoads

Ok, so 350mA.
How long do you want it to run for?
3 AAs, 4.5V, will run 6-7 hrs.
Get 3 C cells in a battery holder, will run for a long time.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

cincy

ok, I'll try that.  I was thinking to try Lithium, but wasn't sure of the calculation - so didn't know if it would help :)

I will be out from Thursday - Sunday running this thing without AC outlets, so batteries the whole time.  It won't be constant peak, but wanted to base the calculation on peak to be 100% sure I have enough batteries with me to power all (extended) weekend.

cincy

ok, based upon the 350 mA peak (adding resistors), Thursday - Sunday would be 4 days and using 33,600 mA.  this would be 5 C batteries (4.2) or 3 D batteries (2.8).  Sound correct?  I would assume that there will be a lot less than peak unless I am running the RGB's full-time. 

Since it will be turned off during the day - I would realistically expect:

2 Days maximum requirements:  2 C Batteries should work powering the RGB's full-time.

CrossRoads

Wikipedia says
"Alkaline C batteries can hold up to 8,000 mAh,"
So 3 C-cells, 4.5V and 8000mAh, could last ~ 22 hours.
(8000/350)

Alkaline D batteries 12000-18000 mAh
So 3 D-cells, 4.5V and 18000mAh, could last ~51 hours.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

cincy

With 3 C-Cells, wouldn't you have 8000mAh * 3 = 24000mAh?  That's what I was basing my calculations from.

johnwasser


With 3 C-Cells, wouldn't you have 8000mAh * 3 = 24000mAh?  That's what I was basing my calculations from.


Only if you put them in parallel.  You get three times as many watt-hours no matter how you connect them but when they are in series the voltage is three times as high.  This gives you the higher voltage but the same mAh as a single cell.

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