Help a beginner understand if a design is correct

Question

I'm fairly familiar with Ohm's law and I did some calculations I'd like experienced engineers to confirm if it is correct or point where I made a mistake

Here's the idea

• I want to power a self-made strip of 20 LEDs preferably in series (assuming 2.5Vf @ 20ma)
• I got a 4.5W, 5V, 0.9ampere DC power supply
• I calculated the required resistor for the circuit like this:

P = V * I
P = (20 * 2.5) * 0.02
P = 1W
---
V = I * R
50 = 0.02 * R
R = 2500ohms

From that I assumed I could power 20 LEDs in series with that power supply and a 2.5Kohm resistor. I also learned that circuits in series have voltage drops but current is the same, and in parallel circuit, voltage is the same but current drops. Is that correct or did I miss something?

Again, I'm an absolute beginner in electrical engineering. I graduated in CS but never got into the physics of circuits.

EDIT

From the answers I gathered that a resistor is used to drop the voltage that isn't dropped by the LEDs, so in this case, my 5V power supply would be able to power at most 1 LED at full brightness or 2 albeit dimmer.

Could I use a step-up to get the voltage higher and power those 20 LEDs in series?

EDIT 2

All the answers helped me understand a lot more about electronics and how they work as well as to always design circuits for margin and not for the nominal values. I realized my mistake by trying to go for LEDs in series instead of parallel, it didn't come to me I'd need just two wires and not 2 for each LED, I know, dumb mistake.

Thank you all for the help, tips and information.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

You can use the power supply with your LEDs ... if you connect each LED in series with its own resistor, and 20 of these in parallel.

As you have 0.9A and 20 LEDs at 20mA each, the LEDs will take 400mA total so you have plenty of power to spare.

You don't mention the type of LEDs, but blue/white/violet will take about Vf = 3V each, red/green/yellow about Vf = 2V each (for more exact values, read their datasheets).

So the resistor should be (5V - Vf) / 20mA = (5V - 3V)/0.02 = 100 ohms for white, or 150 ohms for red; increase these values for a safety margin (120 or 180 ohms) and it's perfectly safe to increase further if too bright.

There's more wiring involved, but it may still be simpler than a boost converter.

Answer 2

Your resistance calc.is incorrect. You have 20 leds with a nominal (note nominal- ie it maybe smaller or larger) forward voltage of 2.5V. So you need at least 50V for your power supply. Lets say the power supply is 60V to cope with variation of the nominal Vf. The series resistor calculation would then be R = (60-50)/0.02 = 500 Ohms. P = I2R = 0.2W wasted in the resistor. So you’d need a 500Ohm resistor with a rating in excess of 0.2W.

Answer 3

You need the resistor to drop the voltage that isn't dropped by the diodes. Your diodes actually drop all 50V (Assuming that's the voltage of the power supply), so if the diodes are exactly as you describe (hint: they won't be) you don't need or want a resistor. That would be a cruddy design, though. Better to use a supply with some headroom so you can take up the difference with a resistor, or use a smaller supply and split you diodes into two parallel strings.