How to calculate the number of LEDs under the power supply and Vice versa?

It is necessary to do illumination small figurines either from batteries or from power supply 5V 1A, now have a ready solution in the form of illumination of the diodes 5 is connected with Arduino. I would like to transplant them to the power supply, how to calculate what resistors to put, if I know that 5 LEDs, har-Ki 3V, 20mA, and connect them to the necessary power supply 5V 1A.
August 23rd 19 at 10:36
August 23rd 19 at 10:38
Solution
There are special calculators for calculating your LEDs.
Please answer another question, as you know, I assume he did not know how, on the links that you took off - the power supply can only specify the voltage and number of Amps no, what does it affect? Can it happen that I have all the diodes will burn up? - buford_Jones commented on August 23rd 19 at 10:41
The number of amps listed on the power supply is the maximum current which it is capable of. For example, in your case, the power supply can power 50 of these LEDs - cxem.net/calc/ledcalc.php?tU=5&tS=3&tI=20&tR=50&ty... . - Mallie76 commented on August 23rd 19 at 10:44
The current consumption of the whole scheme will be 1000 mA. - Mallie76 commented on August 23rd 19 at 10:47
: Now more or less clear, thank you very much for the help ) - buford_Jones commented on August 23rd 19 at 10:50
: 1 Amps on the PSU means that above or much above this current it will not. This is a limitation. So, with nominal resistors(on the website) the led will be at 20mA for a total(sum, because the LEDs are connected in parallel) will give the current in 0.1 A. More simply will not work with this voltage and the resistance, even if you put BP at 5 volts and 10 000 000 Amps. So you have enough power supply 10 times weaker than yours. And the function of resistors to heat up, until they drops the excess voltage, not forgetting to limit the current. If the PSU 5 volt, and the led at 3 Volts, so excessively where it should fall to 2 volts. These 2 volts and will drop on the resistor. - harrison commented on August 23rd 19 at 10:53
: The excellent answer, some of questions have disappeared:) Thanks ) - buford_Jones commented on August 23rd 19 at 10:56
August 23rd 19 at 10:40
5 volts, the extra rastavatsya ballast resistor so (1-0,045)*0,9/0,02= 43 led
of course it is if you consider which will burn all at the same time, otherwise more. Also it is unnecessary to forget that the maximum load current of 40mA output, but is worth more than 25mA. Ie if you turn on multiple LEDs in parallel need the transistor
August 23rd 19 at 10:42
As already mentioned above, there are special calculators. In theory-how to count can read here: ledjournal.info/spravochnik/raschet-rezistora-dlya...
August 23rd 19 at 10:44
The voltage at which the led is × number of LEDs = voltage which will burn all
The output current of the battery is the result
Example:
Led on: 0.5 v
Qty: 10 PCs
Battery: 10 v
0.5 × 10 = 5 (if I'm not mistaken)
10 - 5 = 5
The result of battery

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