# how to calculate current with led

Discussion in 'The Projects Forum' started by Gadersd, Jan 1, 2013.

Dec 8, 2012
98
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I am stuck on this circuit. I cannot seem to find the total current of it. How would I do it?

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2. ### MrChips Moderator

Oct 2, 2009
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I suppose this is a homework question.
In any case, it cannot be done simply because an LED or diode is a non-linear device.
It can be solved if you knew the current-voltage characteristic or you make an assumption about the turn on voltage of the diode.

Dec 8, 2012
98
1
The forward voltage of the diode is 1.7v and the voltage source voltage is 5v. Couldn't it be solved with 5/(((5-1.7)/0.1+5/0.1)+0.1)?

4. ### MrChips Moderator

Oct 2, 2009
12,651
3,460
No.
You have two unknowns hence you need two equations.
See if you can use KVL to find two equations.

5. ### WBahn Moderator

Mar 31, 2012
18,092
4,918
No. Let's take a look at why.

First, let's put units on the quantities.

5V/(((5V-1.7V)/0.1kΩ+5V/0.1kΩ)+0.1kΩ)

Without even considering where the pieces of this equation come from or what it is supposed to be determining, I can tell you that it is incorrect. Why? Because the units don't work out. In the denominator you have three terms and the first two are voltage divided by resistance, which is current, while the third term is resistance. You can't add current and resistance -- it's meaningless. So there is no point going any further because you KNOW that the result is meaningless, too.

Always, always, always track and check your units.

Now let's look at some of the pieces parts and see if they make sense.

(5V-1.7V)/0.1kΩ

This would appear to be an application of Ohm's Law. But Ohm's Law requires the voltage across a resistor and (5V-1.7V) is not the voltage across anything. The same is true for the second term,

5V/0.1kΩ

What 0.1kΩ resistor has 5V across it?

You can't just pull some voltage from somewhere and divide it by some resistance from somewhere else and get a result that means anything. Ohm's Law is a relationship between a resistance and the voltage across THAT resistance and the current through THAT resistance.

Dec 8, 2012
98
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How would I find the voltage across the parallel resistors when the bottom resistor affects the voltage and current within them?

7. ### Dodgydave AAC Fanatic!

Jun 22, 2012
5,166
775
Well why dont you build the circuit and measure it with a DVM.

8. ### THE_RB AAC Fanatic!

Feb 11, 2008
5,435
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With the LED removed there is 2.5v across the top resistor and 25mA.
Attach the LED you get 2.5-1.7v = 0.8v across the LED resistor, an additional current of 0.8 / 100 or 8mA.
adding 8mA raises the current about 32% which increase voltage on the bottom resistor to 3.3v, which would drop the top resistor voltage to 1.7v which would extinguish the LED so the final LED current result MUST be somewhere between 0-8mA.
Allowing for top:bottom balancing I'd punt on the low side of average so I'm guesstimating about 3 to 3.5mA LED current.

9. ### WBahn Moderator

Mar 31, 2012
18,092
4,918
Let's label the three resistors Ra, Rb, and Rc from top to bottom. Let's call the currents in them Ia, Ib, and Ic with the direction coming from the positive side of the battery toward the negative side.. For the voltages, let's call the node at the bottom of the battery GND (0V), the voltage at the top of the battery V0, and the voltage at the remaining node (the top of Rc) V1. Finally, let's call the forward voltage drop across the LED Vf.

This gives you five unknowns (V0, V1, Ia, Ib, and Ic). We can solve for one of them really easily -- V0=5V. Vf is not an unknown since it is given that Vf=1.7V. So you have four unknowns and you need four independent equations.

Q1) In terms of the two node voltages, Vf, and the resistances (use labels, not the actual values), what are the currents in each of the three resistors?

Q2) What is the relationship between the three currents (think KCL)?

You now have your four equations.

Q3) What the total current?

10. ### MrChips Moderator

Oct 2, 2009
12,651
3,460
How did you arrive at this?

11. ### WBahn Moderator

Mar 31, 2012
18,092
4,918
With the LED removed, all that remains is a voltage divider having equal resistances and 5V applied across them.

12. ### MrChips Moderator

Oct 2, 2009
12,651
3,460
He said top resistor.

According to my calculations the current through the LED is 5.333mA. Total current is 27.666mA.

13. ### WBahn Moderator

Mar 31, 2012
18,092
4,918
Good point. I guess I just took it to mean top resistor in the surviving circuit.

I got the same answer you did, but preferred to try to lead the OP toward figuring it out on his own.