Electronics > Beginners

Calculate Current

**etnel**:

Hello, how would I calculate the current at the different points in this circuit?

Vcc = 4v

LED: Vf = 2v, 20mA

R1 = 100 ohms

**Capernicus**:

You dont have to think in exact numbers when you do this... I think that people do too much maths and not enough real thinking when they do electronics.

This circuit is so basic, why bother calculating any number at all.

So you can see we have 1 led on the left with a resistor, and 3 leds on the right, 2 in parallel 2 in series, with no resistance, so the right is definitely alot brighter than the left, and if you used a small enough battery then you wouldnt be able to even see anything on the led with a resistor, regardless of the resistance.

Thats worth more any amount of pages of confusing mathematics, and gets it over and done quicker, and you dont have to worry about pesky details that dont matter.

There is more to conclude here and know than just that, like the led voltage drop, you need to be able to handle that as well... but that probably can be "read" as some approximate laws and the exact numbers of things isnt so important, the general concepts are way more important.

**not1xor1**:

--- Quote from: etnel on October 10, 2021, 02:05:06 pm ---Hello, how would I calculate the current at the different points in this circuit?

Vcc = 4v

LED: Vf = 2v, 20mA

R1 = 100 ohms

--- End quote ---

Supposing ideal (identical) LEDs you wold get 2V/100Ω = 20mA through the left branch and something more than 20mA through the right one (since there are 2 LEDs in parallel and that means less current and less voltage drop).

With real components you might get 15-25mA on the left side and larger variations or even burnt LEDs on the right side >:D

LEDs should always be connected in series to a constant current source/sink or resistor.

**bdunham7**:

--- Quote from: Capernicus on October 10, 2021, 02:13:18 pm ---You dont have to think in exact numbers when you do this... I think that people too much maths and not enough real thinking when they do electronics.

--- End quote ---

Actually, in physics and EE the rule is that if you can't do the math, then you don't really understand it. But...

--- Quote ---This circuit is so basic, why bother calculating any number at all.

--- End quote ---

Probably because it is a homework problem or something like that. But it is a very bad example of a problem because although perhaps you can 'calculate' currents, your calculations won't have any meaning in the real world the circuit design fails to accommodate the non-ideality of the components. And they give Vf of the LEDS only for 20mA, so it can be argued that there is not enough information to solve the problem. So, you're right in this instance--the circuit is so flawed that there's no point in calculating anything.

**mariush**:

You can split that circuit in the middle.

On the left side you have the led and you have the resistor.

Voltage = current x resistance according to ohm's law

So input voltage - (number of leds x forward voltage) = current x resistance

4v (vcc) - (1 x 2v ) = current x 100 = > current = (4-2)/100 = 2/100 = 0.02A

on the right branch, you have two leds in parallel, then one in series with the two in parallel.

So your current would be input voltage - (1 group of leds in parallel x forward voltage) - (1 led in series x forward voltage) = current x resistance

If you had a resistor to limit current on this branch, the section of circuit with 2 leds in parallel will have same current as the single led experiences, so the two leds in parallel will have half the current.

As it is ... you have equivalent of 2 leds in series, so around 4v voltage drop and you don't have any resistance, unless you start to account for the wire resistance as current limiter.

So if we assume ideal circuit where the leds are fully open at 2v and the wire has no resistance, then those leds would simply blow up because there's nothing to limit current going through them.

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