What Wattage Rating Is Needed for a 33Ohm Resistor in a 9V, 1A Circuit?

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In a 9V, 1A circuit powering an Arduino and multiple LEDs, the appropriate wattage rating for a 33-ohm resistor is crucial for preventing overheating. The correct formula for calculating power dissipation is P = I² * R, which indicates that a 33-ohm resistor will dissipate approximately 0.132 watts, making a 1/4 watt resistor sufficient. When connecting both LEDs and the Arduino to the same power supply, the total current drawn will be the sum of both components' requirements, not exceeding the supply's 1A limit. It's recommended to group LEDs in series with appropriate resistors to manage current effectively, ensuring each LED receives sufficient voltage without exceeding its maximum rating. Proper calculations and resistor ratings are essential for safe and efficient circuit design.
  • #31
In the link, with an NPN, you keep the Emitter and Arduino close to 0V (GND) .

12+
Current Limiting Resistor
LEDs
Transistor
GND
 
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  • #32
Windadct said:
In the link, with an NPN, you keep the Emitter and Arduino close to 0V (GND) .

12+
Current Limiting Resistor
LEDs
Transistor
GND
So I want my wire that is at the end of the LEDs to go into the Collector part of the Transistor. Base to the Arduino, and the Emitter to Negative?
Virtual Example(Drawings of circuits really help me):

FidCnxE.png


Is this how it is supposed to work? The arduino just completes the circuit, correct?
 
  • #33
HAP...Thank you for taking a few minutes to review ( so many people that come here for answers -- well just want answers!) Yes your sketch is what we need as a concept. -- I have not done the math...
As a good practice the 12V and the arduino should share the same GND or Negative Bus - in this schema these details are not shown...
 
  • #34
OK back - from my sketch pad... 4 x 2.2 V + 0.2V ( transistor) = 9V... 12 V supply... the difference is ? ... 3 V ... series ckt is 30 mA ..

3 V and 30 mA --- what is the necessary resitance to drop that V?...

Then use the I^2 * R or V*I to get W, and that is the resistance and size (wattage) of the resistor you are looking for... now repeat for the other colors...
 
  • #35
Windadct said:
OK back - from my sketch pad... 4 x 2.2 V + 0.2V ( transistor) = 9V... 12 V supply... the difference is ? ... 3 V ... series ckt is 30 mA ..

3 V and 30 mA --- what is the necessary resitance to drop that V?...

Then use the I^2 * R or V*I to get W, and that is the resistance and size (wattage) of the resistor you are looking for... now repeat for the other colors...
Thanks for this. And yes, I am looking to learn how to do this on my own, instead of coming here every time I need an answer to something.

So when you were calculating the transistor, you used: (Vf * ledcount) + 0.2V = 9V
What exactly is this meaning? Why do you add the 0.2V?

So then when there is 3V and 30 mA left over, you add a resistor just to drop the voltage to completely 0? (3*30 = 90) This would mean I would put a 100Ω(Closest to 90 rounding up) to the negative then to ground or negative?
Note: Power source in diagram is meant to be DC, not AC
fwuWAVF.png
 
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  • #36
Only one resistor. It's R1 that soaks up the spare voltage (and in the process sets the circuit current). Don't put a resistor where you show R2.
 
  • #37
NascentOxygen said:
Only one resistor. It's R1 that soaks up the spare voltage (and in the process sets the circuit current). Don't put a resistor where you show R2.
Other than that, it is fine? What about the output of the arduino. Should I limit the milliamps or voltage going into the transistor? And does the voltage going into the base affect the current/voltage going through.
 
  • #38
HeyAwesomePeopl said:
Other than that, it is fine? What about the output of the arduino. Should I limit the milliamps or voltage going into the transistor? And does the voltage going into the base affect the current/voltage going through.
Oh, I see what you were up to. Yes, you'll probably need a resistor in series with the base to limit the current from the Arduino. So you had just shown it in the wrong place. How many volts will the port deliver for output HIGH? Anyway, choose R to limit the base current to about 3 or 4mA.

What is the maximum current the Arduino port can supply?
 
  • #39
Are you definitely going to power the Arduino board from the same voltage source as the LED array? If so, as well as connecting the +12V to the Arduino, you need to connect the 0V of the supply to the 0V of the Arduino board. It might be labelled GND or (-).
 
  • #40
NascentOxygen said:
Oh, I see what you were up to. Yes, you'll probably need a resistor in series with the base to limit the current from the Arduino. So you had just shown it in the wrong place. How many volts will the port deliver for output HIGH? Anyway, choose R to limit the base current to about 3 or 4mA.

What is the maximum current the Arduino port can supply?

The arduino can supply 3.3 or 5 volts and a max of 40mA per pin. Meaning I would need a

NascentOxygen said:
Are you definitely going to power the Arduino board from the same voltage source as the LED array? If so, as well as connecting the +12V to the Arduino, you need to connect the 0V of the supply to the 0V of the Arduino board. It might be labelled GND or (-).
0V of supply, the output of the transistor?
 
  • #41
HeyAwesomePeopl said:
0V of supply, the output of the transistor?
The emitter of the transistor, which is also the 0V of the power supply, connects to the 0V of the Arduino.

The output is 3.3V or 5V ... how does it choose?
 
  • #42
Using the 3.3 volt output of the arduino, a 100Ω would make it 0.033 amps
 
  • #43
HeyAwesomePeopl said:
Using the 3.3 volt output of the arduino, a 100Ω would make it 0.033 amps
The transistor base won't need that much current. Around 4mA will do.

The base uses 0.6V, so that leaves 2.7V across the base resistor. You can calculate the Ohms needed from that.
 
  • #44
NascentOxygen said:
The emitter of the transistor, which is also the 0V of the power supply, connects to the 0V of the Arduino.

The output is 3.3V or 5V ... how does it choose?
There is a 3.3 and 5 volt version boards. In this case its 5 volts
NascentOxygen said:
The transistor base won't need that much current. Around 4mA will do.

The base uses 0.6V, so that leaves 2.7V across the base resistor. You can calculate the Ohms needed from that.
So using a 5 volt board, it would come out 4.4 volts, and to get it around 4 milliamps (3.8 milliamps) it needs a 1200Ω resistor.
 
  • #45
HeyAwesomePeopl said:
So using a 5 volt board, it would come out 4.4 volts, and to get it around 4 milliamps (3.8 milliamps) it needs a 1200Ω resistor.
Sounds good. Are you going to update your circuit diagram now?

With the reds figured out, have you researched how many volts you'll see across your green and yellow LEDs? When you do this, you may decide to go with just 3 LEDs of each colour. :)
 
  • #46
NascentOxygen said:
Sounds good. Are you going to update your circuit diagram now?

With the reds figured out, have you researched how many volts you'll see across your green and yellow LEDs? When you do this, you may decide to go with just 3 LEDs of each colour. :)
Updated diagram with 3 colors of LEDs and correct resistors:
hdKcKel.png

Does everything look right? I'm going to double check the calculations right now to make sure everything is right.

NascentOxygen said:
When you do this, you may decide to go with just 3 LEDs of each colour. :)
Why 3? 5 too many? They are pretty cheap at 5 for a buck.
 
  • #47
I was thinking over 3V drop for one colour, but just checked and that's for whites.

You might end up tweaking the LED resistors, anyway, if it seems one colour is too bright in comparison with the others.

Don't forget to connect the power supply gnd to the circuit's gnd.

Good luck!

Thanks to Windadct for Arduino insight.
 
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  • #48
NascentOxygen said:
I was thinking over 3V drop for one colour, but just checked and that's for whites.

You might end up tweaking the LED resistors, anyway, if it seems one colour is too bright in comparison with the others.

Don't forget to connect the power supply gnd to the circuit's gnd.

Good luck!
Thanks. I used the data sheet to provide the recommended voltage and amperage per LED type. The Luminous Intensity varies for the LED colour, and would have to overamp/volt the LEDs which could significantly reduce life expectancy.

Don't forget the connect the power supply ground to the circuits ground? The power supply doesn't have a ground build in, no third pin(American Style). The ground is digital and is maintained by the Arduino. Does the whole ground to board still apply now?EDIT: Also, this transistor should work, correct? https://www.futurlec.com/Transistors/PN2222pr.shtml
 
  • #49
Instead of making the dimmer LEDs brighter, you could make the over-bright ones dimmer. :oldwink:

I wasn't talking about mains ground. I meant the "negative" terminal of the 12V supply should be connected to the negative or ground on the Arduino board and your LED board.

The 2222 transistor will be okay. It's always a good idea to place a finger on these switching transistors when you first switch the power on, they should not get even warm. If they get noticeably warm, you've made a wiring or design blunder.
 
  • #50
NascentOxygen said:
Instead of making the dimmer LEDs brighter, you could make the over-bright ones dimmer. :oldwink:

I wasn't talking about mains ground. I meant the "negative" terminal of the 12V supply should be connected to the negative or ground on the Arduino board and your LED board.

The 2222 transistor will be okay. It's always a good idea to place a finger on these switching transistors when you first switch the power on, they should not get even warm. If they get noticeably warm, you've made a wiring or design blunder.
The power to the board is a 2.1mm plug. That is why I question you plugging everything into the boards ground. The boards ground is usually only used if you use the boards 5V power pins.

And thanks for the warning!
 
  • #51
You can connect the LED board directly to the +12 and 0V terminals of the supply, and connect the Arduino's plug to the two terminals of the supply also.
 
  • #52
NascentOxygen said:
You can connect the LED board directly to the +12 and 0V terminals of the supply, and connect the Arduino's plug to the two terminals of the supply also.
Confused. Like this?

AbRMWrd.png
 
  • #53
Two wires from the 12VDC supply have to go to the Arduino. You show only one, and it's +12V so should not be going to GND.

Separate question: you show something going to pin 5 of the board. You need the base resistors to each connect to a different pin on the board. Do you have this sorted? You seem to have the base resistors going to a constant +5V on the board---this would keep the LEDs lit all the time, not under software control.
 
  • #54
NascentOxygen said:
Two wires from the 12VDC supply have to go to the Arduino. You show only one, and it's +12V so should not be going to GND.
That one line includes a postive and negative. It's kinda confusing, I cannot seem to split the line.

NascentOxygen said:
Separate question: you show something going to pin 5 of the board. You need the base resistors to each connect to a different pin on the board. Do you have this sorted? You seem to have the base resistors going to a constant +5V on the board---this would keep the LEDs lit all the time, not under software control.
Oh yes, I just forgot to remove that. Each "+5" sign resembles another pin used. In this case, I am using 3, one for each color.
 
  • #55
thumbsup.png
 
  • Like
Likes HeyAwesomePeopl
  • #56
NascentOxygen said:
thumbsup.png
And 56 replies later, I have a solution! Thanks for everything.
 
  • #57
Better not close this thread 'till you have it working ... :nb) :nb)
 
  • #58
NascentOxygen said:
Better not close this thread 'till you have it working ... :nb) :nb)
Haha I won't. It'll be a few weeks until I get the parts in.

Just to confirm, to calculate the wattage rating I should get for the resistors, I should use P = I^2 * R which is Watts = Current^2 * Resistance. In this case, I would just need a half watt for this circuit.
 
  • #59
HeyAwesomePeopl said:
Just to confirm, to calculate the wattage rating I should get for the resistors, I should use P = I^2 * R which is Watts = Current^2 * Resistance. In this case, I would just need a half watt for this circuit.
No.  I suggest that you use resistors rated at at least 3 times the power you will require they dissipate. Some of your higher value resistors will dissipate close to 0.5 watts, so you should use nothing less than 1.5W. I guess this means going for 2W.

If you were to divide your LEDs up into strings of 3 LEDs and 2 LEDs you might need only 1W resistors. Do the math!

(An alternative to a single 2W resistor is to use 3 resistors, either: each 0.5W but 3 times the Ohms that you need all in parallel,
or
each 0.5W but one-third the Ohms you calculate, connected in series. Or something equivalent.)

 
  • #60
Kudos for your diagrams - they are the language of electronics! ( Edit - not that they are perfect or follow the best practices(still good) , but that you took the time to do them with a proper schematic tool...)

Now how does it turn on and off:)
 
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