# Confused with LED circuit.

Discussion in 'General Electronics Chat' started by Voltboy, Jul 26, 2010.

1. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
Hello, I'm currently trying to do a little LED lamp. The LEDs are blue, with a measured 3.2V drop each. My source is 5V.
I have 100Ω, 80Ω and 20Ω resistors.

The left schematic has a 20Ω and 3 80Ω resistors, and the right one has 3 100Ω resistors.

What I want to know is if both circuits will behave the same way.
I tried to do some math and got different results, but my logic says me that both circuits are the same. Heres my math, correct me please:
LED "resistance" = 3.2V/20mA = 160Ω

For the left schematic:
1/Rparallel = 3*(1/240Ω) ---> Rparallel = 80Ω
Rt= 20Ω + 80Ω = 100Ω
Current = 5V/100Ω = 50mA

For the right schematic:
1/Rt = 3*(1/260Ω) ---> Rt= 86.7Ω
Current = 5V/86.7Ω = 57mA

This is clearly a different circuit, but for me, logically, both should be the same because the 20Ω resistor in the left "works" for all 3 of the branches, which would give each branch a 100Ω resistance. Or was my math wrong?

Thanks and appreciate any help.

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Since the three LEDs are all pretty much the same, then it is safe to estimate that the current flowing in each of the LEDs is the same. That means that the left hand circuit will behave as though they had approximately 87 ohms. That is 80 ohms plus 20/3 ohms.

hgmjr

May 11, 2010
241
38
The circuits are not the same... the left circuit will be dimmer.

The 20 Ohm resistor on the top is squeezing *all* the current through it, while the imaginary top 20 Ohms on each of the right-side resistors is passing only the current for one LED.

So the voltage drop over the 20 Ohms on the left will be higher, and therefore the current passed to the LEDs will be lower.

Another little point: An LED isn't like a resistor -- it has a nonlinear response to voltage/

4. ### dsp_redux Active Member

Apr 11, 2009
182
5
The right circuit will give you $\frac{5V-3.2V}{100\Omega} = 18mA$ in each LED (or a total of 54mA from the power supply), not 57mA. To do a go analysis of the left one, we need the saturation current (datasheet) of the diode. If you drop 3.2V, it must be really small (like x10^(-53)). Look at the Shockley equation $I=I_S(e^{\frac{V_D}{n V_T}}-1)$... rings a bell?

Last edited: Jul 27, 2010
5. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
That would mean the left circuit is brighter? (87Ω < 100Ω)