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Topic: 12V but what Ampere? (Read 992 times) previous topic - next topic

tsunaliew

As already explained,the buck converter output @5V will draw < 266mA @5V , which is only 100mA @12V.

Either you are not paying attention or you have not
done  the math:
P= I*V
(Ohms Law)
Continued ignorance of this is going to get you
into trouble later.
All this talk about the buckconverter supprting 2A is
pointless because it is only going to draw 100mA @12V. (266mA @5V).

How many times do we have to tell you this ?

So if my arduino will draw around 800mA @5V and the converter around 266mA @5V plus solenoid 1.2A @12V, is it a good idea to get 12V 4A (48W) power supply? Since someone told me that it is a good practice to get 2 times the current I need.

raschemmel

#31
Nov 04, 2019, 02:06 pm Last Edit: Nov 04, 2019, 02:32 pm by raschemmel
P= I * V => I = P/V ( devide both sides by V)
V cancels on right side leaving P/V = I
Swap sides to get I= P/V
OHM's LAW !
266mA @5V drawn by the arduino is ONLY  190mA
@7V coming from the buck converter that is supplying the arduino.
You're the one who said you were going to use the buck yo power the arduino because the arduino
heats up too much with a 12V input.
PS=>BUCK=>Arduino
GOT IT ?
And we've already established that what you plan to use doesn't draw 800mA so why are you using that value. Stick to reality, not fantasy.
And I've already stated that 333mA@5 is only
100mA@12V.
I had hoped you would be able to the math to
realize that's only 237nA@7V (buck output voltage)
(You can't make any statements about the buck@5V because the buck is set for 7V, REMEMBER ?

tsunaliew

#32
Nov 04, 2019, 02:27 pm Last Edit: Nov 04, 2019, 02:30 pm by tsunaliew
P= I * V => I = P/V ( devide both sides by V)
V cancels on right side leaving P/V = I
Swap sides to get I= P/V
OHM's LAW !
266mA @5V drawn by the arduino is ONLY  190mA
@7V coming from the buck converter that is supplying the arduino.
You're the one who said you were going to use the buck yo power the arduino because the arduino
heats up too much with a 12V input.
PS=>BUCK=>Arduino
GOT IT ?
And we've already established that what you plan to use doesn't draw 800mA so why are you using that value. Stick to reality, not fantasy.
I enquired from the shopkeeper who sold me the buck converter (DC-DC Step Down Buck Converter with Display 3.2 - 40V to 1.25 - 35V 10W (3A) LM2596), she say if my supply is 4A for the circuit, the components will only take what it needs from the current, so in a sense if my arduino need 800mA, the arduino will get the supply from the buck converter and will give me 800mA. As long as the output current is not over the max rating, it is fine.

So am I being misled?

raschemmel

#33
Nov 04, 2019, 03:37 pm Last Edit: Nov 04, 2019, 04:25 pm by raschemmel
You absolutely refuse to learn ohm's law ?
You've learned nothing.
You've understood nothing we've told you.
You post on an International forum and we tell
you everything you need to know and after all
that you consult a vendor ?
Do you think the vendor knows more about it
than all the engineers who have answered your
post ?
Why do you need a vendor's advice when you
have you have the Geman Physicist  Georg Ohm?
After all I've said you STILL haven't Googled
"Ohm's Law"
You're hopeless.
I give up.
How old are you ?

tsunaliew

#34
Nov 04, 2019, 04:19 pm Last Edit: Nov 04, 2019, 04:21 pm by tsunaliew
You absolutely refuse to learn ohm's law ?
You've learned nothing.
I give up.
How old are you ?
Calm down, no need to be angry all I am doing is ask, you said 1.5A is not enough so I increase to 4A and is there any datasheet or link that you can kindly provide can show me why LM2596 cannot output the ampere I want. I know there is the efficiency of the converter is calculated by dividing the output power (Pout) by its input power (Pin) and then times the input current to get output current.

So for example 7V divide 12V is around 58.3% x 4A = 2.33A. Even if the solenoid takes 1.2A @12V, the current left is 2.8A, so the current is 1.63A, @ 7V. So if arduino current draw what it needs, it should be sufficient isn't it?

raschemmel

#35
Nov 04, 2019, 04:47 pm Last Edit: Nov 04, 2019, 05:30 pm by raschemmel
Quote
you said 1.5A is not enough
List the Reply # where I said that .
Not only did I NOT say that , I said 1.3A was enough (1.2 for the solenoid,+0.1 for the buck converter.

Quote
so the current is 1.63A, @ 7V
I said 1.65W, NOT 1.65A.
1.65W/12V = 0.1375A@12V, NOT 1.65A.

Do you even know what Watts is ?

Watts is "P" in "P=I*V". (Again, Ohm's Law)

The solenoid draws 1.2A @12V.
Everything you said you planned to use draws
LESS than 333mA @5V which is ONLY 237mA @7V, which is only 0.138A@12V.

(I may have said 0.1A but it I meant to say 0.138A@12V , which, while different , is still a far cry from
1.2A@12V (which you claim I said) for the buck converter)

from the buck converter. Your math is all wrong.
The buck converter efficiency is probably 95% or better..
The power dissipated by the buck converter is the output voltage (7V) times the output current 237mA, (1.65W) , because it based on what you
are powering (the devices you listed) snd NOT
the arduino maximum (800mA@5V), so 1.2A@12V=14W + 237mA@7V (1.65W) = 15.65W
15.65W/12V= 1.3A (as I stated before).
The buck converter current is only 100mA @12V.
You don't need 4A@12V because YOUR
application (what you SAID you plan to use
only draws 100mA (0.10A) @12V.
0.1A+1.2A=1.3A@12V, and as Georg has tod us,
P=I*V
power =1.3A *12V =15.65W
current I = P/V
         =15.65/12V
         =1.3A (@12V)
NOT 4A
After everything you STILL Have NOT
Googled "Ohm's Law".
Instead you choose to consult the vendor.

tsunaliew

The solenoid draws 1.2A @12V.
Everything you said you planned to use draws
LESS than 333mA @5V which is ONLY 237mA @7V
from the buck converter. Your math is all wrong.
The buck converter efficiency is probably 95% or better..
The power dissipated by the buck converter is the output voltage (7V) times the output current 237mA, (1.65W) , because it based on what you
are powering (the devices you listed) snd NOT
the arduino maximum (800mA@5V), so 1.2A@12V=14W + 237mA@7V (1.65W) = 15.65W
15.65W/12V= 1.3A (as I stated before).
The buck converter current is only 100mA @12V.
You don't need 4A@12V because YOUR
application (what you SAID you plan yo use
only draws 100mA (0.10A) @12V.
0.1A+1.2A=1.3A@12V, and as Georg has tod us,
P=I*V
power =1.3A *12V =15.65W
current I = P/V
              =15.65/12V
              =1.3A (@12V)
NOT 4A
After everything you STILL Have NOT
Googled "Ohm's Law".
Instead you choose to consult the vendor.
Oh ok so thats how you calculate, I get your point now, sorry that I misunderstood what you said and my fault lies on taking wrong value. Thanks for correcting me in everything, I will use the calculation you shown me as reference. To be honest, I thought need to use max rating so that is the part I have been misunderstanding all along.

I did studied ohm's law, just the basic and the formulas.

raschemmel

#37
Nov 04, 2019, 06:17 pm Last Edit: Nov 04, 2019, 06:51 pm by raschemmel
Quote
I did studied ohm's law, just the basic and the formulas.
This is Reply #37 .
There is not a single reply from you where you have demonstrated (or even posted calculations) , that
you have learned anything about Ohm's Law.
Not one equation.
Not one formulat
Not one Ohm's Law calculation.
If you studied it , then you learned nothing and you need to review it.
Prove that you know the basics.
Post a reply that contains Ohm's Law calculations and explain what they mean.
Talk is cheap.
If you had replied: "You know what ? You're right.
I still don't get it."
then I would be posting this reply.
But you said you "studied" it.
What exactly is you're definition of "study" if
it does nit include understanding ?
Nothing you have said indicates even the remotest
level of understanding of Ohm's Law.
Is it so important that I need to hammer away at it
again and again ?
 Some would say yes.

tsunaliew

This is Reply #37 .
There is not a single reply from you where you have demonstrated (or even posted calculations) , that
you have learned anything about Ohm's Law.
Not one equation.
Not one formulat
Not one Ohm's Law calculation.
If you studied it , then you learned nothing and you need to review it.
Prove that you know the basics.
Post a reply that contains Ohm's Law calculations and explain what they mean.
Talk is cheap.
If you had replied: "You know what ? You're right.
I still don't get it."
then I would be posting this reply.
But you said you "studied" it.
What exactly is you're definition of "study" if
it does nit include understanding ?
Nothing you have said indicates even the remotest
level of understanding of Ohm's Law.
Is it so important that I need to hammer away at it
again and again ?
 Some would say yes.
You need to calm down alright? Look, mostly of my post look dumb to you but those are my insecurities, I just need someone to correct me just in case. Like I said this is my first project and I am not really confident enough. I do not want my time and hard work spent on this project to be wasted because of miscalculation.

I know my ohm's law basics well and since you provided so much help, I feel bad that you deem me as hammering my way out and being ignorant. So to clear all these misunderstandings, how about you provide me questions and test me instead?

The reason why I did not post equation or formula is because I think I stated my calculation well enough to be understood. I tried my best to state it well but if it is still not enough for you, I do apologise.

raschemmel

#39
Nov 04, 2019, 08:42 pm Last Edit: Nov 04, 2019, 11:05 pm by raschemmel
"
The reason why I did not post equation or formula is because I think I stated my calculation well enough to be understood. "
All your calculatios are wrong whether they are
understood or not.

You still have not indicated you know what
the power is for anything and you continue to use
maximum values instead of values that reflect the
stated application you outlined at the beginning.
"Hi guys, i am not sure if my plan will work so i need people to correct me if i am wrong. My plan is to get a 12V 3 pin adapter plug for the whole circuit, everything here is powered by Arduino. 16x2 LCD with backlight on (120-160mA), 3-24V piezo buzzer (20mA), 5V relay module (7-10mA), 4x4 Matrix Membrane Keypad (30mA), total 220mA."

Remember saying that ?
If so , why do keep quoting 800mA @5V ?

Based on what you said above the 12V current
supplying the buck converter is only 92.5mA@12V.
(+ a little overhead for converter operation)

Show me one ohm's law equation or formula or post where you used the ohm's law formulas.
There is not one in 38 posts.
You want a test ?
List ohm's law calculations for the following loads:

Solenoid
buck converter
Arduino
lcd
relay (if there is one)
List Power
     Current
      Voltage

List total Power for all of the above.
Show your work.

in there respective units of measurements
List ohm's law formula you are using as it is common written with letter designations.


raschemmel

#40
Nov 04, 2019, 11:05 pm Last Edit: Nov 04, 2019, 11:31 pm by raschemmel
Test Elapsed Time:  2 hours, 20 minutes...(and counting)

tsunaliew

#41
Nov 05, 2019, 01:10 am Last Edit: Nov 05, 2019, 01:13 am by tsunaliew
Let this be the final post, since the main point of concern has been answered.

Like I said before, I took the wrong value because I thought it has to be max value.

P=V×I
I=P/V
V=P/V

Solenoid 12V × 1.2A = 14.4W
Arduino & Buck converter 12V × 0.1A = 1.2W (7V × 0.171A)
lcd 5V × 0.07A = 0.35W
relay 5V × 0.077A = 0.385W

Total Wattage = 14.4W + 1.2W = 15.6W

Reason why I add only Solenoid and Arduino is because the rest of the components is supplied by Arduino.

raschemmel


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