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Please show me the steps to calculate current going through R1, R2 and D1 Led (Green color LED), when I tested this circuit through a multimeter it showed as 2.5ma across R1 and R2 but 0ma across D1, and also the voltage drop across D1 was 0.5 Volts.

Any way I can calculate these manually?.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Just a minor correction: The LTL-307EE is a Red LED (2V typical forward voltage drop). Do you mean LTL-307GE Green LED (2.1V typical forward voltage drop)? \$\endgroup\$ – Tut Nov 12 '14 at 14:01
  • \$\begingroup\$ Yes what I used is a Green Led in my Circuit \$\endgroup\$ – Dlect Nov 12 '14 at 14:31
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    \$\begingroup\$ I'm concerned when I hear current described as "across" a component. When using an ammeter, you must put the meter in series with the current path you are measuring. Just a note to the casual reader so as to not mislead. \$\endgroup\$ – vicatcu Nov 12 '14 at 14:51
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Just another method. The key point is to judge if the diode is on.

Assume the LED is on, then the voltage drop on the LED and R2 should be 2V or so, then

$$ V_{D1} = V_{R2} = 2V $$

Then the current on \$R_{2}\$ should be

$$ I_{R2}= \frac{2V}{200\Omega}=10mA $$

Then the voltage drop on \$R_{1}\$ should at least

$$ V_{R1} = (I_{R2} + I_{D1}) \times R_{1} >10mA \times 1k\Omega=10V $$

Apparently, this voltage is greater than the source voltage, so the diode can NOT be on. So we think it's open circuit. Then it's a voltage source in series with two resistor now. And the math will be easy.

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  • \$\begingroup\$ Just a minor correction: 0.7V is too low. The forward voltage drop for a conducting LED is more like 2V. \$\endgroup\$ – Tut Nov 12 '14 at 14:04
  • \$\begingroup\$ Just wanted to point out that the LTL-307EE LED datasheet says its typical forward voltage is 2.0V, not the 0.7V that you used. That, however, did not compromise your reasoning. \$\endgroup\$ – Ricardo Nov 12 '14 at 14:04
  • \$\begingroup\$ Sorry, i didn't notice this. \$\endgroup\$ – diverger Nov 12 '14 at 14:06
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To solve circuits with diodes, you usually assume that the diode is conducting (it's on) and then watch for any inconsistencies.

From page 3 of LTL-307EE datasheet, we know that the LED typical forward voltage (when it's on) is 2.0V. Since D1 is wired in parallel with R2, R2 shares its voltage as well. Since R1 is in series with R2, the voltage drop across R1 and R2 is the same as the voltage provided by the V1 source. So we have:

$$V_{D1} = V_{R2} = 2V$$ $$V_{R1} + V_{R2} = 3V$$

From these two equations above we get that: $$V_{R1} = 1V$$

From Ohm's Law, \$V=RI\$ or \$I=V/R\$, we can calculate the currents: $$I_{R1} = V_{R1} / R1 = 1V / 1k\Omega = 1mA$$ $$I_{R2} = V_{R2} / R2 = 2V / 200\Omega = 2.5mA$$

But from the conservation of current, we know that:

$$I_{R1} = I_{R2} + I_{D1}$$

and

$$I_{D1} = I_{R1} - I_{R2} = 1 - 2.5 = -1.5mA$$

That means there should be a current of \$1.5mA\$ going against the diode. But since \$D1\$ is a diode, we know it cannot conduct when reverse biased (assuming an ideal diode, that is).

There's a contradiction that tells us that the LED isn't on.

Now, assuming the LED isn's on, we can replace it with an open circuit and solve the problem:

$$I_{R1} = \frac{V1}{R1 + R2} = \frac{3}{1000 + 200} = \frac{3}{1200} = 2.5mA$$

$$I_{R1} = I_{R2}$$

$$V_{R2} = R2 \times I_{R1} = 200 \times 2.5mA = 0.5V$$

$$V_{D1} = 0.5V$$

Just like you measured.

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    \$\begingroup\$ Great explanation, Vd1 also is 0.5V isn't it?, and about the conservation of current, can you ever get a minus figure in a real circuit? and when ever we get a minus figure in our calculation , can we safely assume that current is not flowing through it? \$\endgroup\$ – Dlect Nov 12 '14 at 15:04
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    \$\begingroup\$ Thanks! Vd1 is also 0.5V because it's in parallel with R2. Components in parallel share the same voltage. And when you get a negative current, it means that the current actually flows in the contrary direction than initially assumed, but with the same absolute value. \$\endgroup\$ – Ricardo Nov 12 '14 at 15:14
  • \$\begingroup\$ Only now I noticed that I got vd1 wrong. Well spotted!!! \$\endgroup\$ – Ricardo Nov 12 '14 at 15:16
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Ignoring the LED, R1 and R2 form a potential divider reducing the supply voltage by: -

\$\dfrac{R2}{R1+R2}\$ = 0.1666

This means the voltage across R2 (with a 3V supply) is 3V x 0.1666 = 0.5 volts and not enough to start turning on a normal LED.

With 0.5 volts across R2 (200 ohms) the current will be 0.5/200 = 2.5mA

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  • \$\begingroup\$ Thanks Andy for your comment, but can you explain it a bit more on what you meant by "Ignoring the LED, R1 and R2 form a potential divider " , and also why its ignoring the LED in this case? \$\endgroup\$ – Dlect Nov 12 '14 at 13:55
  • \$\begingroup\$ @Dlect with the diode in parallel to the resistance, the voltage drop can only decrease because - in simple terms - the diode provides an alternative path to R2 for the current to circulate, thus reducing the overall resistance. \$\endgroup\$ – clabacchio Nov 12 '14 at 13:58
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    \$\begingroup\$ @Dlect it's called trial and error - if I ignore the LED I can determine the maximum voltage on R2. Then, with the LED in place it cannot be greater. As I have shown the voltage to be 0.5 volts, there will clearly be very little current thru the LED (if any at all). \$\endgroup\$ – Andy aka Nov 12 '14 at 14:08
  • \$\begingroup\$ Diode datasheets have a graph showing current vs. forward voltage. In the LTL-307GE datasheet, it's Figure 2. This graph shows that the typical diode current is effectively zero below about 1.8V. Using Andy's method or your multimeter, you can see that you're nowhere near turning on the diode. \$\endgroup\$ – Adam Haun Nov 12 '14 at 20:16

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