# Voltage Drop Diode Confusion

#### huwilerp

Joined Jan 10, 2013
10
I've attached an image below. I have 1 diode in series, and then the circuit branches out into parallel configuration, each leg has a resistor and an LED. Assume the LED is 1.5V .20 mA.

I am having a hard time trying to figure out what the voltage drop is, and how to accurately calculate circuit, and the V drop for each component.

I am hoping someone could steer me into the right direction.

Sorry for the crappy drawing.

R/S

Patrick

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#### inwo

Joined Nov 7, 2013
2,419
Subtract the 1.5 volts + .7 for the diode from 9 volts. As they are more or less fixed. The rest is across each resistor. 6.8 volts.

You can't assume the 20 ma. It is unknown until calculated by voltage and resistance.

#### GopherT

Joined Nov 23, 2012
8,009
For each leg, you should divide the 6.8 or 6.9V by the resistance of each corresponding resistor to get the current through each LED.

Add up the current of the three LEDs to get the total current drawn from the battery. By the way, most red LEDs are 2V.

4.7k connected LED will see about 3.5 mA
1k connected LED will see about 7 mA
470 ohm connected LED will see 35 mA.

All values from my internal estimator, your results will be more accurate when you measure the actual resistances and use a calculator.

Last edited:

#### THE_RB

Joined Feb 11, 2008
5,438
...
... By the way, most red LEDs are 2V.
...
Most red LEDs are 1.1v (1mA) to 1.4v (20mA) range.

#### GopherT

Joined Nov 23, 2012
8,009
Most red LEDs are 1.1v (1mA) to 1.4v (20mA) range.
RB,

Most electronics books (and websites) describe a red LED forward voltage as a nominal 2 volts at 20 mA.

The random LED I just measured was 2.0 volts at 20 mA and 1.8 at 1 mA

I have never seen a visible red LED with forward voltage that low-even at low current. I just checked Newark.com and the Datasheets of red LEDs are the range of 1.6 to 2.2. But, their selection may be limited.

#### MikeML

Joined Oct 2, 2009
5,444
Here is a simulation of your circuit with some extra real-world complications thrown in. The table shows the computed voltages at every node in the circuit, as well as current through each component.

Note that to figure the voltage across D1, you subtract the voltage at V(a) from V(b), or 7.67 - 6.89 = 0.78V.

Now notice some quirks.

Why is the battery voltage V(b) not 9V?

Why is the forward drop across D1 not 0.6V like the text books say?

Why are the voltages at V(c), V(d), and V(e) not the same?

Why are the currents through the LEDs I(D2), I(D3),and I(D4) not the same?

There are some reassuring things to see:

I(D1) = -I(V1)

I(D1) = I(D2) + I(D3) + I(D4)

I(R1) = I(D2), etc

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#### GopherT

Joined Nov 23, 2012
8,009
Here is a simulation of your circuit with some extra real-world complications thrown in. The table shows the computed voltages at every node in the circuit, as well as current through each component.

Note that to figure the voltage across D1, you subtract the voltage at V(a) from V(b), or 7.67 - 6.89 = 0.78V.

Now notice some quirks.

Why is the battery voltage V(b) not 9V?

Why is the forward drop across D1 not 0.6V like the text books say?

Why are the voltages at V(c), V(d), and V(e) not the same?

Why are the currents through the LEDs I(D2), I(D3),and I(D4) not the same?

There are some reassuring things to see:

I(D1) = -I(V1)

I(D1) = I(D2) + I(D3) + I(D4)

I(R1) = I(D2), etc
ML

Your resistance values are different that the image in the original post. Should be 470, 1k and 4.7k.

#### BobTPH

Joined Jun 5, 2013
8,641
30 Ohms is ridiculously high for the internal resistance of the battery, unless you using a nearly dead one.

Bob

#### MikeML

Joined Oct 2, 2009
5,444
Ok, here it is again with better values....

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