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Phibir

Hi,

I've signed up to these forums hoping to get a little help with a basic understanding of electronics.

I have two daughters age 9 who were bored last weekend, so I went to my local electrical shop and purchased a random basic beginners Electrical kit as the shop keeper thought suitable. I have no understanding of electronics and have been searching the net to hopefully get a better understanding.

We have a 9V battery an LED and some Resistors. I have read about Ohm's Law and understand that if I do the equation 9÷0.03=300Ω (9 = 9 Volt Battery, 0.03 = 30mA LED). This presumably means that I need a 300Ω or there about rated resistor?

YouTube videos show that the formula could be changed depending on what's needed, for example..

R=V/I
V=R/I
I=R/V

So if I was to find an already built circuit but with no battery attached. I should be able to look at the parts I have in front of me to calculate the battery that I need, for example..

0.03÷300 = 0.0001 (0.03 = 30mA LED, 300 = 300Ω Resistor). I have no idea what relevance 0.0001 is in Volts.

If anyone has time and can be a*sed to help me out, then I would be very grateful.

Thank you!

Phbir
 
Leds are different from normal ohms law as there are other factors....being a diode they have a forward voltage and limited current...so calculation here is R=(Vs-Vled)/Ied.......Resistance = voltage - voltage drop / forward current.
so normal led should be 1.2v drop 30mA forward current.......R=(9-1.2)/0.03=260 ohms
 
Hi,

I've signed up to these forums hoping to get a little help with a basic understanding of electronics.

I have two daughters age 9 who were bored last weekend, so I went to my local electrical shop and purchased a random basic beginners Electrical kit as the shop keeper thought suitable. I have no understanding of electronics and have been searching the net to hopefully get a better understanding.

We have a 9V battery an LED and some Resistors. I have read about Ohm's Law and understand that if I do the equation 9÷0.03=300Ω (9 = 9 Volt Battery, 0.03 = 30mA LED). This presumably means that I need a 300Ω or there about rated resistor?

YouTube videos show that the formula could be changed depending on what's needed, for example..

R=V/I
V=R/I
I=R/V

So if I was to find an already built circuit but with no battery attached. I should be able to look at the parts I have in front of me to calculate the battery that I need, for example..

0.03÷300 = 0.0001 (0.03 = 30mA LED, 300 = 300Ω Resistor). I have no idea what relevance 0.0001 is in Volts.

If anyone has time and can be a*sed to help me out, then I would be very grateful.

Thank you!

Phbir

Hi. Good to hear that you're getting your daughters interested in some science/technology.

Just to stop you going off track, the formulae are:
V = I x R (or V = IR)
I = V / R
R = V / I

So, using your example, V = 0.03 x 300 = 9V (ignoring the LED volt drop).

As tazz says, LEDs are odd. They are, in effect, a constant voltage device. The voltage drop will be between about 1.8V and 3.5V (reds and yellows low, green and blue higher). 30mA will typically be the maximum continuous current. Often better to reduce this to, say 20mA, if you're unsure. Significant overcurrent will pop an LED pretty quickly.

A pack of short wires with a crocodile clip on each end is useful for connecting components together:
10 X Croc Clip Test Leads Crocodile Clips 50cm | eBay

For more advanced playing, a plug-in breadboard keeps things tidy:
https://learn.sparkfun.com/tutorials/how-to-use-a-breadboard
 
Tazz is spot on with regard to the voltage drop of the particular LED. And as said above it's great to see you getting them interested in electronics. Especially with more and more schools and colleges dropping electronics as a subject.

Be worth also trying substituting different values of resistor in a circuit, and a) measuring volt drop and current and then b) doing the ohms law calculations to see if you get the correct figures.

Also try a potential divider circuit, both in theory and on the workbench. All good stuff.

Daz
 

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