# Need help

#### Tont

Joined Mar 29, 2006
8
Hi,

I need help with these,

A L.E.D is an indicator used in electronic control systems. These indicators can be damaged if excess current passes through them.Once such device requires a current of 20mA to light brightly without being damaged ; a voltage of 2v is developed across the LED when it is illuminated.

Calculate the size of the resistor R1 that should be connected in series with the LED if the LED is to be connected to a 24v DC supply.

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I've tried dividing the voltage by the current but im not sure if this is the correct way?

Thanks for any help

#### paultwang

Joined Mar 8, 2006
80
Originally posted by Tont@Mar 29 2006, 12:56 PM
I've tried dividing the voltage by the current but im not sure if this is the correct way?
[post=15580]Quoted post[/post]​

Almost. In DC world, if something drops voltage, you can model it as an equivalent resistor. So it drops 2 V and it needs 20 mA.

#### Tont

Joined Mar 29, 2006
8
Thanks what formula would i apply to calculate the resistance of the resistor?

#### paultwang

Joined Mar 8, 2006
80
Originally posted by Tont@Mar 29 2006, 02:12 PM
Thanks what formula would i apply to calculate the resistance of the resistor?
[post=15583]Quoted post[/post]​
Ohm's Law

#### Tont

Joined Mar 29, 2006
8
R = V/I

R = 2/0.02

R = 100 Ohms

Is that correct?

#### paultwang

Joined Mar 8, 2006
80
Originally posted by Tont@Mar 29 2006, 02:22 PM
R = V/I

R = 2/0.02

R = 100 Ohms

Is that correct?
[post=15586]Quoted post[/post]​
You found the equivalent resistance of the diode.

Hint: Figure out how to drop (24 - 2 = 22) volts across the resistor.

#### Tont

Joined Mar 29, 2006
8

#### paultwang

Joined Mar 8, 2006
80
Originally posted by Tont@Mar 29 2006, 02:36 PM
im lost
[post=15590]Quoted post[/post]​

Perhaps you need to know that Ohm's law relates voltage DROP across and current FLOW through two points to equivalent resistance.

Usually you apply Ohm's law across the power supply and the ground. But it can also be applied across any two points.

Components do not see "24 volts". They see the voltage difference between their input pin and their output pin.

#### Tont

Joined Mar 29, 2006
8
There is a volt drop of 22 Volts across the circuit, 24 v from the supply and 2 v to illuminate the LED.

I am tryin to calculate the value of the resistor.

I know the resistance of the diode = 100 Ohms.

I know that i must use the formula R = V/I to calculate the value of the resistor but am unsure about which values to use.

#### paultwang

Joined Mar 8, 2006
80
Originally posted by Tont@Mar 29 2006, 02:53 PM
There is a volt drop of 22 Volts across the circuit, 24 v from the supply and 2 v to illuminate the LED.
[post=15592]Quoted post[/post]​
No. There is a 24 V drop across the entire circuit. 22 V is dropped by the resistor. 2 V is dropped by the diode. Forget about the diode for now. What resistance does it take to drop 22 V while allowing 20 mA through?

#### Tont

Joined Mar 29, 2006
8
Sorry if i seem stupid, still tryin to get to grips with this, i know its basic.

R = V/I

R = 22/0.02

R = 1100 Ohms

#### paultwang

Joined Mar 8, 2006
80
Originally posted by Tont@Mar 29 2006, 03:04 PM
Sorry if i seem stupid, still tryin to get to grips with this, i know its basic.

R = V/I

R = 22/0.02

R = 1100 Ohms
[post=15594]Quoted post[/post]​
Looks reasonable.